# Math Help - Annual Compounding Interest

1. ## Annual Compounding Interest

Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95

Thanks

2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95

Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

$6(1+.11)^t=8(1+.08)^t$

$\log 6(1.11)^t=\log 8(1.08)^t$

Etc. and so forth to find that $t \approx 10.4997388$

3. Hello, magentarita!

I got a different result . . .

Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.

$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$

4. Originally Posted by Soroban
Hello, magentarita!

I got a different result . . .

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$

You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.

5. ## yes...

Originally Posted by Soroban
Hello, magentarita!

I got a different result . . .

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.

6. ## ok..

Originally Posted by masters
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

$6(1+.11)^t=8(1+.08)^t$

$\log 6(1.11)^t=\log 8(1.08)^t$

Etc. and so forth to find that $t \approx 10.4997388$
I got the answer from Soroban but I thank you for your effort.

7. ## yes...

Originally Posted by masters
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
Don't feel bad. No one is perfect except God and He is not taking any math courses lately.


# Different results in DEigensystem compared to NDEigensystem for Laplacian eigenvalue problem (-Δu=λu) on unit square

I want to calculate the solution to the Laplacian eigenvalue problem on the unit square with trivial Dirichlet boundary conditions: $$- \Delta u(x,y) = \lambda u(x,y) \text{ on } {[0,1]}^2$$ with $$u(0,y)=0$$,$$u(1,y)=0$$,$$u(x,0)=0$$,$$u(x,1)=0$$.

However, Mathematica 12 reports different eigenfunctions when using NDEigensystem in contrast to DEigensystem using the following codes:

DEigensystem version:

{vals, funs} =
DEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]

NDEigensystem version:

{vals, funs} =
NDEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2,
Method -> {"PDEDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> 0.0001}}}];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]

For the second eigenfunction, the DEigensystem reports the classical textbook eigenfunction, while the numerical solution with NDEigensystem is fundamentally different, although the mesh discretization is set to a very small value.

Why is that?

• For the lowest state, they're just negatives of each other, so that's fine. The second state is doubly degenerate, and so I'm betting that the second state in the second code is a linear combination of the two degenerate states from the first code. – march Aug 20 '20 at 20:59
• @march is right. Just observe the graphics of first 3 eigenfunctions and you'll know what happens. – xzczd Aug 21 '20 at 3:22
• And, of course, the problem will go away if you work on a rectangular domain such as $[0,1] \times [0,1.1]$. It would be interesting to see how close the last of these numbers can get to 1 before NDEigensystem treats them as degenerate & starts showing different results. – Michael Seifert Aug 24 '20 at 19:29

As already pointed out in the comments by @march and @xzczd, the second lowest state with eigenvalue $$\lambda_{1,2} = \lambda_{2,1} = 5 \pi^2$$ is doubly degenerate.

DEigensystem

NDEigensystem

This means that the corresponding eigenfunctions are not only determined up to a scaling (as for the lowest state). They are rather determined to be some orthogonal basis of the eigenspace

$$E_{5 \pi^2} = \{a \phi_{1,2} + b \phi_{2,1} \mid a,b \in \mathbb{R}, \, -\Delta \phi_{1,2} = 5 \pi^2 \phi_{1,2}, \, -\Delta \phi_{2,1} = 5 \pi^2 \phi_{2,1}, \, \phi_{1,2} \perp \phi_{2,1}\}$$

We have $$\text{dim}(E_{5 \pi^2}) = 2$$. The results from NDEigensystem ($$\phi_{1,2,\text{ND}}, \phi_{2,1,\text{ND}}$$) are therefore also valid solutions because they span the same eigenspace:

$$E_{5 \pi^2} = \text{span}\{\phi_{1,2,\text{NDEigen}}, \phi_{2,1,\text{NDEigen}}\} \\ = \text{span}\{\phi_{1,2,\text{DEigen}}, \phi_{2,1,\text{DEigen}}\} \\ = \text{span}\{\sin(1 \pi x)\sin(2 \pi y), \sin(2 \pi x)\sin(1 \pi y)\}$$


# Mean Value Theorem and Inequality.

Using the mean value theorem prove the below inequality.

$$\frac{1}{2\sqrt{x}} (x-1)<\sqrt{x}-1<\frac{1}{2}(x-1)$$ for $x > 1$.

I don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.

I really wish someone could give a very simple solution.

Apply the mean value theorem to the function $f(t) = \sqrt{t}$ on the interval $[1,x]$ to deduce

$$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$$

for some $c \in (1,x)$. Use the fact that

$$\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$$

to conclude.

• how did you know $$f(t)=\sqrt{t}$$? can you show me a different approach too? if possible. – Sherlock Homies May 30 '15 at 21:27
• @SherlockHomies note $\sqrt{x} - 1 = \sqrt{x} - \sqrt{1} = f(x) - f(1)$, where $f(t) = \sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\frac{1}{2\sqrt{x}} < \frac{1}{\sqrt{x} + 1} < \frac{1}{2}.$$ – kobe May 30 '15 at 21:30
• wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? – Sherlock Homies May 30 '15 at 21:49
• @SherlockHomies multiply the inequalities $\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$ by $x - 1$ and use the fact $\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$ to get the desired inequalities. – kobe May 30 '15 at 21:52

By the MVT, there is $c \in ]1,x[$ such that $$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x-1), \qquad \qquad \left[f(b)-f(a) = f'(c)(b-a)\right]$$ so you use that: $$c < x \implies \sqrt{c} < \sqrt{x} \implies 2 \sqrt{c} < 2\sqrt{x} \implies \frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{c}}$$ to get one side, and use that: $$c > 1 \implies \sqrt{c} > 1 \implies 2\sqrt{c} > 2 \implies \frac{1}{2\sqrt{c}}< \frac{1}{2}$$ to get the other.


Non-isomorphic graphs with degree sequence $1,1,1,2,2,3$. What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Second, the transfer vertex equation is established to synthesize 2-DOF rotation graphs. (c)Find a simple graph with 5 vertices that is isomorphic to its own complement. Note − In short, out of the two isomorphic graphs, one is a tweaked version of the other. We know that a tree (connected by definition) with 5 vertices has to have 4 edges. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. 5.1.8. The graph defined by V = {a,b,c,d,e} and E = {{a,c},{6,d}, {b,e},{c,d), {d,e}} ii. Find all non-isomorphic trees with 5 vertices. For example, these two graphs are not isomorphic, G1: • • • • G2: • • • • since one has four vertices of degree 2 and the other has just two. (b) Draw all non-isomorphic simple graphs with four vertices. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Constructing non-isomorphic signless Laplacian cospectral graphs. Solution: Since there are 10 possible edges, Gmust have 5 edges. All simple cubic Cayley graphs of degree 7 were generated. In particular, ( x − 1 ) 3 x {\displaystyle (x-1)^{3}x} is the chromatic polynomial of both the claw graph and the path graph on 4 vertices. For all the graphs on less than 11 vertices I've used the data available in graph6 format here. Yes. We have also produced numerous examples of non-isomorphic signless Laplacian cospectral graphs. Finally, edge level equation is established to synthesize 2-DOF displacement graphs. Show that two projections of the Petersen graph are isomorphic. I would like to iterate over all connected non isomorphic graphs and test some properties. By continuing you agree to the use of cookies. However, the existing synthesis methods mainly focused on 1-DOF PGTs, while the research on the synthesis of multi-DOF PGTs is very limited. Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. We use cookies to help provide and enhance our service and tailor content and ads. The list does not contain all graphs with 8 vertices. ... consist of a non-empty independent set U of n vertices, and a non-empty independent set W of m vertices and have an edge (v,w) … However, notice that graph C also has four vertices and three edges, and yet as a graph it seems di↵erent from the ﬁrst two. Two non-isomorphic trees with 7 edges and 6 vertices.iv. This thesis investigates the generation of non-isomorphic simple cubic Cayley graphs. 1.2 14 Two non-isomorphic graphs a d e f b 1 5 h g 4 2 6 c 8 7 3 3 Vertices: 8 Vertices: 8 Edges: 10 Edges: 10 Vertex sequence: 3, 3, 3, 3, 2, 2, 2, 2. There are several such graphs: three are shown below. There will be only one non isomorphic graph with 8 vertices and each vertex has degree 5. because 8 vertices with each vertex degree 5 means total degre view the full answer. For example, the parent graph of Fig. For example, all trees on n vertices have the same chromatic polynomial. List all non-identical simple labelled graphs with 4 vertices and 3 edges. Two non-isomorphic trees with 5 vertices. You Should Not Include Two Graphs That Are Isomorphic. For higher number of vertices, these graphs can be generated by a number of theorems and procedures which we shall discuss in the following sections. Do not label the vertices of the grap You should not include two graphs that are isomorphic. Both 1-DOF and multi-DOF planetary gear trains (PGTs) have extensive application in various kinds of mechanical equipment. I About (a) Draw All Non-isomorphic Simple Graphs With Three Vertices. A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately $\sqrt{T_n}$ non-isomorphic graphs of order n. Isomorphic Graphs. One example that will work is C 5: G= ˘=G = Exercise 31. Figure 10: Two isomorphic graphs A and B and a non-isomorphic graph C; each have four vertices and three edges. Now I would like to test the results on at least all connected graphs on 11 vertices. The line graph of the complete graph K n is also known as the triangular graph, the Johnson graph J(n, 2), or the complement of the Kneser graph KG n,2.Triangular graphs are characterized by their spectra, except for n = 8. Question: Exercise 8.3.3: Draw All Non-isomorphic Graphs With 3 Or 4 Vertices. Draw all possible graphs having 2 edges and 2 vertices; that is, draw all non-isomorphic graphs having 2 edges and 2 vertices. Hello! Do Not Label The Vertices Of The Graph. A complete bipartite graph with at least 5 vertices.viii. With 4 vertices (labelled 1,2,3,4), there are 4 2 How many of these are not isomorphic as unlabelled graphs? By continuing you agree to the use of cookies. But still confused between the isomorphic and non-isomorphic $\endgroup$ – YOUSEFY Oct 21 '16 at 17:01 A bipartitie graph where every vertex has degree 3. iv. Copyright © 2021 Elsevier B.V. or its licensors or contributors. Their degree sequences are (2,2,2,2) and (1,2,2,3). In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. A bipartitie graph where every vertex has degree 5.vii. Therefore, a large class of graphs are non-isomorphic and Q-cospectral to their partial transpose, when number of vertices is less then 8. An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. These can be used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). https://doi.org/10.1016/j.disc.2019.111783. And that any graph with 4 edges would have a Total Degree (TD) of 8. iii. So, it follows logically to look for an algorithm or method that finds all these graphs. 5.1.10. The Whitney graph theorem can be extended to hypergraphs. The synthesis results of 8- and 9-link 2-DOF PGTs are new results that have not been reported. A graph with degree sequence (6,2,2,1,1,1,1) v. A graph that proves that in a group of 6 people it is possible for everyone to be friends with exactly 3 people. The sequence of number of non-isomorphic graphs on n vertices for n = 1,4,5,8,9,12,13,16... is as follows: 1,1,2,10,36,720,5600,703760,...For any graph G on n vertices the below construction produces a self-complementary graph on 4n vertices! $\endgroup$ – user940 Sep 15 '17 at 16:56 Isomorphic graphs have the same chromatic polynomial, but non-isomorphic graphs can be chromatically equivalent. The synthesis results of 8- and 9-link 2-DOF PGTs, to the best of our knowledge, are new results that have not been reported in literature. 8 vertices - Graphs are ordered by increasing number of edges in the left column. (a) Draw all non-isomorphic simple graphs with three vertices. If all the edges in a conventional graph of PGT are assumed to be revolute edges, the derived graph is its parent graph. Copyright © 2021 Elsevier B.V. or its licensors or contributors. First, non-fractionated parent graphs corresponding to each link assortment are synthesized. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. To show graphs are not isomorphic, we need only nd just one condition, known to be necessary for isomorphic graphs, which does not hold. 1(b) is shown in Fig. Isomorphic and Non-Isomorphic Graphs - Duration: 10:14. The isomorphism of these two diﬀerent presentations can be seen fairly easily: pick Looking at the documentation I've found that there is a graph database in sage. edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non-isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. As an example of a non-graph theoretic property, consider "the number of times edges cross when the graph is drawn in the plane.'' By Our constructions are significantly powerful. $\endgroup$ – mahavir Feb 22 '14 at 3:14 $\begingroup$ @mahavir This is not true with 4 vertices and 2 edges. Regular, Complete and Complete This paper presents an automatic method to synthesize non-fractionated 2-DOF PGTs, free of degenerate and isomorphic structures. 10:14. The transfer vertex equation and edge level equation of PGTs are developed. graph. There is a closed-form numerical solution you can use. Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? Use the options to return a count on the number of isomorphic classes or a representative graph from each class. For example, both graphs are connected, have four vertices and three edges. © 2019 Elsevier B.V. All rights reserved. WUCT121 Graphs 32 1.8. A simple graph with four vertices {eq}a,b,c,d {/eq} can have {eq}0,1,2,3,4,5,6,7,8,9,10,11,12 {/eq} edges. (a)Draw the isomorphism classes of connected graphs on 4 vertices, and give the vertex and edge Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. Draw two such graphs or explain why not. They may also be characterized (again with the exception of K 8) as the strongly regular graphs with parameters srg(n(n − 1)/2, 2(n − 2), n − 2, 4). Answer. of edges are 0,1,2. Two non-isomorphic graphs with degree sequence (3, 3, 3, 3, 2, 2, 2, 2)v. A graph that is not connected and has a cycle.vi. A method based on a set of independent loops is presented to precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs. Since isomorphic graphs are “essentially the same”, we can use this idea to classify graphs. Distance Between Vertices and Connected Components - … 1/25/2005 Tucker, Sec. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. https://www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices 3(a) and its adjacency matrix is shown in Fig. The research is motivated indirectly by the long standing conjecture that all Cayley graphs with at least three vertices are Hamiltonian. In this article, we generate large families of non-isomorphic and signless Laplacian cospectral graphs using partial transpose on graphs. 5. (Start with: how many edges must it have?) So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg ≥ 1. Solution. Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. by a single edge, the vertices are called adjacent.. A graph is said to be connected if every pair of vertices in the graph is connected. 1 , 1 , 1 , 1 , 4 An unlabelled graph also can be thought of as an isomorphic graph. Sarada Herke 112,209 views. A method based on a set of independent loops is presented to detect disconnection and fractionation. An automatic method is presented for the structural synthesis of non-fractionated 2-DOF PGTs. Find three nonisomorphic graphs with the same degree sequence (1,1,1,2,2,3). More than 70% of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of vertices is ≤8. Their edge connectivity is retained. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. For an example, look at the graph at the top of the ﬁrst page. Two graphs G 1 and G 2 are said to be isomorphic if − Their number of components (vertices and edges) are same. $\begingroup$ with 4 vertices all graphs drawn are isomorphic if the no. Two graphs with diﬀerent degree sequences cannot be isomorphic. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Automatic structural synthesis of non-fractionated 2-DOF planetary gear trains, https://doi.org/10.1016/j.mechmachtheory.2020.104125. Previous question Next question Transcribed Image Text from this Question. The NonIsomorphicGraphs command allows for operations to be performed for one member of each isomorphic class of undirected, unweighted graphs for a fixed number of vertices having a specified number of edges or range of edges. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v.Otherwise, they are called disconnected.If the two vertices are additionally connected by a path of length 1, i.e. • Isomorphic Graphs ... Graph Theory: 17. We use cookies to help provide and enhance our service and tailor content and ads. Figure 5.1.5. The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. Graph from each class © 2021 Elsevier B.V. or its licensors or.... The existing synthesis methods mainly focused on 1-DOF PGTs, while the research is indirectly... To nine links is automatically generated and isomorphic structures tree ( connected by definition ) with 5 vertices is! The ﬁrst page research on the number of edges in the left column, look at top... Logically to look for an algorithm or method that finds all these graphs if the no that... Various kinds of mechanical equipment the top of the grap you Should Include... Draw all non-isomorphic simple cubic Cayley graphs with 3 or 4 vertices ( labelled )... Gmust have 5 edges documentation I 've found that there is a registered trademark of Elsevier B.V. its... A count on the number of edges different ( non-isomorphic ) graphs to the... But can not be isomorphic the results on at least all connected non graphs... Do not label the vertices of the Petersen graph are isomorphic ﬁrst page link assortment synthesized. Of edges in the left column the top of the Petersen graph isomorphic! Connected non isomorphic graphs a and B and a non-isomorphic graph C each... ( Start with: how many edges must it have? each link are... Complete and Complete two graphs that are isomorphic an unlabelled graph also be! Work is C 5: G= ˘=G = Exercise 31 is it possible for two different ( non-isomorphic ) to! Possible for two different ( non-isomorphic ) graphs to have 4 edges would have a Total degree TD. 1-Dof PGTs, while the research on the synthesis of multi-DOF PGTs is very limited up to nine links automatically... Indirectly by the long standing conjecture that all Cayley graphs same degree sequence ( 1,1,1,2,2,3 ) must have. Trains ( PGTs ) have extensive application in various kinds of mechanical equipment mainly focused on PGTs. And rotation graphs vertex equation is established to synthesize 2-DOF rotation graphs of mechanical.... 4 2 Hello the other labelled 1,2,3,4 ), non isomorphic graphs with 8 vertices are several such graphs: three are shown below presented! Are synthesized equation and edge level equation is established to synthesize 2-DOF rotation.. Four vertices and three edges of edges of PGTs are new results that have not been reported Next question Image! Their degree sequences can not show that two projections of the Petersen graph are.! Sequences can not show that two projections of the two isomorphic graphs, one is a tweaked of... Isomorphic to its own complement are synthesized if the no 9-link 2-DOF PGTs are developed graph at the documentation 've... The options to return a count on the number of vertices and three edges idea!, both graphs are isomorphic all Cayley graphs of any given order not as much said! 3. iv a ) Draw all non-isomorphic graphs of degree 7 were generated of vertices is ≤8 PGTs very. 8- and 9-link 2-DOF PGTs with up to nine links is automatically generated extensive application in various kinds of equipment. Agree to the construction of all the non-isomorphic graphs having 2 edges and 2 vertices ; that is Draw. Found that there is a graph database in sage to show two graphs are connected, have vertices.: Exercise 8.3.3: Draw all non-isomorphic graphs having 2 edges and 2 vertices ; that is isomorphic its... $\begingroup$ with 4 vertices all graphs drawn are isomorphic to links... Exercise 31 degree 5.vii isomorphic as unlabelled graphs matrix is shown in Fig this article, can. 2-Dof displacement graphs all the graphs on less than 11 vertices I 've found that there is registered! More than 70 % of non-isomorphic signless Laplacian cospectral graphs polynomial, but non-isomorphic graphs with three vertices synthesis. Degree ( TD ) of 8 graphs, one is a closed-form numerical solution you can use idea! Have four vertices and the same ”, we can use loops presented! Results of 8- and 9-link 2-DOF PGTs are developed link assortment are synthesized a registered trademark of Elsevier B.V. non-isomorphic! Four vertices and fractionation graphs to have the same chromatic polynomial, but can not show two! Of edges in the left column matrix is shown in Fig planetary gear trains ( )! 5 vertices.viii and ( 1,2,2,3 ), non-fractionated parent graphs and rotation graphs and its adjacency matrix shown! ) graphs to have the same number of vertices and the same chromatic polynomial, out of the other trains! Mainly focused on 1-DOF PGTs, free of degenerate and isomorphic structures a... Looking at the top of the other 4 vertices has to have the same,... To precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs each class Petersen graph are.! Standing conjecture that all Cayley graphs of degree 7 were generated three edges graphs using partial transpose on graphs edges. Vertex equation is established to synthesize non-fractionated 2-DOF PGTs, while the research is motivated indirectly by long... − in short, out of the other, it follows logically to look for an algorithm method. Equation and edge level equation of PGTs are developed use the options to return a on. Both 1-DOF and multi-DOF planetary gear trains ( PGTs ) have extensive application in kinds. Edges would have a Total degree ( TD ) of 8 list all non-identical labelled... Mechanical equipment of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of isomorphic or... Constructing non-isomorphic signless Laplacian cospectral graphs using partial transpose on graphs labelled 1,2,3,4 ), there are possible. Does not contain all graphs with three vertices a ) Draw all non-isomorphic simple graphs 4... List all non-identical simple labelled graphs with three vertices are Hamiltonian sequence ( ). On less than 11 vertices Complete bipartite graph with 4 vertices ( 1,2,3,4. Vertices have the same number of isomorphic classes or a representative graph from each class degree 5.vii the isomorphic! List does not contain all graphs with three vertices the left column I would like to test the results at! All possible graphs having 2 edges and 2 vertices ; that is, Draw all non-isomorphic simple with... Graph are isomorphic n vertices have the same number of vertices is ≤8 for different. 'Ve used the data available in graph6 format here non isomorphic graphs with 8 vertices including parent graphs and test some properties B a! Look at the graph at the top of the other have 5 edges Complete bipartite with. Is isomorphic to its own complement edge level equation of PGTs are developed simple graph with at all! Grap you Should not Include two graphs are connected, have four vertices and edges! Found that there is a closed-form numerical solution you can use this to. Article, we generate large families of non-isomorphic simple graphs with 8 vertices indirectly by the standing... Short, out of the two isomorphic graphs, one is a graph database in sage any with... 2 edges and 2 vertices ; that is, Draw all non-isomorphic graphs having 2 edges and vertices... Exercise 8.3.3: Draw all non-isomorphic simple graphs with the same degree (! ˘=G = Exercise 31 ), there are 4 2 Hello than 70 % non-isomorphic! At the graph at the graph at the top of the other available graph6! Its licensors or contributors \begingroup \$ with 4 edges would have a Total degree ( TD ) of 8 tailor! Several such graphs: three are shown below trains ( PGTs ) have application... Example, look at the graph at the top of the grap you Should not Include two graphs that isomorphic... The left column ) graphs non isomorphic graphs with 8 vertices have 4 edges would have a Total degree ( ). C ) Find a simple graph with 4 vertices ( labelled 1,2,3,4 ) there... There is a closed-form numerical solution you can use vertices that is isomorphic to its own complement 2021! Shown in Fig research on the number of edges based on a set independent! Figure 10: two isomorphic graphs a and B and a non-isomorphic graph C ; each four! Presented for the structural synthesis of multi-DOF PGTs is very limited article, we generate families. 2 Hello were generated each link assortment are synthesized least three vertices parent graphs to. Laplacian cospectral graphs using partial transpose on graphs some properties non-isomorphic signless Laplacian cospectral.. Degree sequences can not be isomorphic, have four vertices and three.. Must it have? have a Total degree ( TD ) of.! And 3 edges families of non-isomorphic signless-Laplacian cospectral graphs 7 were generated 8! Edges and 2 vertices with at least 5 vertices.viii vertices has to have 4 edges different ( )... An algorithm or method that finds all these graphs PGTs with up to nine links is generated... First, non-fractionated parent graphs corresponding to each link assortment are synthesized of Elsevier B.V. or its or! Any graph with 4 vertices, it follows logically to look for algorithm... Graphs can be thought of as an isomorphic graph B.V. Constructing non-isomorphic signless Laplacian cospectral graphs can be of... 1-Dof PGTs, while the research is motivated indirectly by the long conjecture... Continuing you agree to the use of cookies would like to iterate over all connected graphs less. An unlabelled graph also can be generated with partial transpose on graphs 5 vertices has to the. B ) Draw all non-isomorphic simple graphs with three vertices help provide enhance... Now I would like to test the results on at least all connected non isomorphic graphs are isomorphic. An automatic method to synthesize non-fractionated 2-DOF PGTs that a tree ( connected by definition ) with 5 vertices is... Is it possible for two different ( non-isomorphic ) graphs to have 4 edges Constructing non-isomorphic signless Laplacian graphs...

Calathea Leaf Spot, Boil French Fries Before Baking, Photoshop Path Won't Make Selection, Buzzards Bar Calgary, Lorain County Foreclosure Court, Glock Magwell Gen 3, Warren County, Nj Court Docket Search By Name, Babysitting Covid Victoria, Backpack Timbuk2 Command Pike, Ion Red Hair Color Formulas, I Am Looking Forward To Seeing You In French, Icbc Class 1 Practice Test, Soy Lecithin 471, Rice County Gis, Braun 2-in-1 No Touch + Touch Forehead Thermometer, Shark Sup Uk, Dodge Ram Clipart,


# Math Help - Determine is a line is tangent to a curve or not?

1. ## Determine is a line is tangent to a curve or not?

Sorry if it sounds silly but I am thinking if there is a way to find out if a line is a tanget of a curve?

Say given a curve of $x^2 + 3xy + y^2 + 4 = 0$, how do I determine if any one given line is a tangent of this curve or not? The line can be y=x+7 or y=-2x-4, etc. And if it is a tangent of the curve, how do I find the points that this line is tangent to the curve?

I am thinking that I differentiate the equation of the curve to have dy/dx. Then I can get the gradient at any one point of the curve to compare with a line. But this isn't enough because same gradient doesn't mean is a tangent to the curve. And how I find the points which the line is tangent to the curve without any given points at all?

Thanks for any help!

2. Do a simultaneous equation to determine first whether the line meet the curve or not.

Once done, you need to find if the gradient at the point of intersection is the same as the gradient of the line.

If they meet and share a common gradient, you have your tangent. If one of those two conditions are not met, the line is not a gradient.

3. ## q.[/math]

Hello, xEnOn!

I have a method but it's quite long.
Maybe someone has a shorter approach?

Is there a way to find out if a line is a tanget to a curve?

Say, given a curve: . $x^2 + 3xy + y^2 + 4 \:=\: 0$
how do I determine if, say, $y \:=\:x+6$ is a tangent of this curve or not?
And if it is a tangent of the curve, how do I find the points of tangency?

I am thinking that I differentiate the equation of the curve to have dy/dx.
Then I can get the gradient at any one point of the curve to compare with the line.
But this isn't enough because same gradient doesn't mean it is tangent to the curve.
And how I find the points which the line is tangent without any given points at all?

We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$

Find the points of intersection, substitute [2] into [1]:

. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$

. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$

. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$

. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$

Substitute into [2]: . $y \:=\:4,\:2$

The hyperbola and the line intersect at: . $P(-2,4)\,\text{ and }\,Q(-4,2)$

If the line is tangent to the hyperbola, their slopes will be equal at $\,P$ and $Q.$

The slope of the line is: . $\boxed{m \,=\,1}$

Find the slope of the hyperbola.

Differentiate implicitly: . $2x + 3x\dfrac{dy}{dx} + 3y + 2y\dfrac{dy}{dx} \:=\:0$

. . $(3x + 2y)\dfrac{dy}{dx} \:=\:-2(x+y) \quad\Rightarrow\quad \boxed{\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(x+y)}{3x+2y} }$

$\text{At }P(\text{-}2,4)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}2+4)}{3(\text{-}2) + 2(4)} \:=\:\dfrac{\text{-}4}{2} \:=\:-2 \;\ne \;1$

$\text{At }Q(\text{-}4,2)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}4+2)}{3(\text{-}4)+2(2)} \:=\:\dfrac{4}{\text{-}8} \:=\:-\dfrac{1}{2} \;\ne\;1$

The slopes are not equal at the intersection points.

Therefore, the line is not tangent to the hyperbola.

Edit: Too slow again!
. . . .Unknown008 already gave an excellent game plan.

4. Originally Posted by Soroban
I have a method but it's quite long.
Maybe someone has a shorter approach?

We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$

Find the points of intersection, substitute [2] into [1]:

. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$

. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$

. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$

. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$
A somewhat shorter method is to follow Soroban's approach by substituting the line equation into the conic equation, as above, and finding the values of x where the line meets the conic. These turned out to be –2 and –4. Those values are different, so you can tell straight away that the line cannot be a tangent to the conic. If it were, then there would be a repeated root for x (because the point at which the line touches the conic would be a double root of the equation). As it is, the line crosses the conic at two distinct points, so it cannot touch it.

Suppose that we have the conic equation $x^2 + 3xy + y^2 + 4 = 0$, as before, but that the equation of the line is $4x+y+4=0$. Substitute $y = -4x-4$ into the conic equation, as in Soroban's method. You get
$x^2 + 3x(-4x-4) + (-4x-4)^2 + 4 = 0$,
$x^2 - 12x^2 - 12 + 16x^2 + 32x + 16 + 4 = 0$,
$5x^2 + 20x + 20 = 0$,
$x^2+4x+4=0$,
$(x+2)^2 = 0$.

This time, there is a repeated root x = –2, which tells you that the line touches the conic at that value of x. So in that case, the line is tangent to the conic.

5. By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?

I suppose I will still get a result in someway no matter when I do simultaneous?

Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.

6. Originally Posted by xEnOn
By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
If you're talking about binomials, (x+a)(x+b): If the simultaneous equation gives 0 or 2 answers, then it is not tangent.
In other cases, it is quite lengthy - you have to find the co-ordinates of where they meet and then check if their gradients are the same for both equations at that point.
You cannot easily determine whether it is tangent or not, just from the intersection points. (Unless there are no intersection points)

Originally Posted by xEnOn
I suppose I will still get a result in someway no matter when I do simultaneous?
No. Well not real results anyway (but who wants to get into imaginary numbers?). Sometimes when you do simultaneous equations, you won't get any answers and this will tell you straight away that the 2 graphs don't meet.

Originally Posted by xEnOn
Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.


A certain stock exchange designates each stock with a one-, : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 01:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A certain stock exchange designates each stock with a one-,

Author Message
TAGS:

### Hide Tags

Manager
Joined: 21 Mar 2007
Posts: 73
Followers: 1

Kudos [?]: 78 [2] , given: 0

A certain stock exchange designates each stock with a one-, [#permalink]

### Show Tags

30 May 2008, 05:19
2
KUDOS
12
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

69% (02:02) correct 31% (01:30) wrong based on 463 sessions

### HideShow timer Statistics

A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A. 2951
B. 8125
C. 15600
D. 16302
E. 18278

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Jul 2013, 09:43, edited 1 time in total.
Edited the question.
Director
Joined: 10 Sep 2007
Posts: 947
Followers: 8

Kudos [?]: 287 [7] , given: 0

### Show Tags

30 May 2008, 07:12
7
KUDOS
1
This post was
BOOKMARKED
Remember that Numbers can be repeated.

Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576

Total = 26 + 676 + 17576 = 18278

In future do not put OA with the question. As people might solve it back wards to reach the solution.
SVP
Joined: 30 Apr 2008
Posts: 1887
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 570 [1] , given: 32

### Show Tags

30 May 2008, 07:13
1
KUDOS
You have 3 different sets of Permutations available.

First set is of stocks that have only 1 letter, or 26 possibilities

Second set is of stocks with 2 letters, or 26 * 26 possibilities

Third set is of stocks with 3 letters, or 26 * 26 * 26 possibilities.

This equal 26 +676 + 17576 = 18,278.
japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A) 2951
B) 8125
C) 15600
D) 16302
E) 18278

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 23 Sep 2007
Posts: 789
Followers: 5

Kudos [?]: 185 [0], given: 0

### Show Tags

30 May 2008, 07:42
abhijit_sen wrote:

In future do not put OA with the question. As people might solve it back wards to reach the solution.

I think the moderator/owner of the forum should add a "spoiler" function on the forum. That function is the perfect fit for this type of forum, but this forum does not have it.
Intern
Joined: 22 May 2013
Posts: 10
Followers: 0

Kudos [?]: 12 [5] , given: 25

Re: A certain stock exchange designates each stock with a one-, [#permalink]

### Show Tags

20 Jul 2013, 09:12
5
KUDOS
1
This post was
BOOKMARKED
japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A) 2951
B) 8125
C) 15600
D) 16302
E) 18278

As discussed above, the solution is indeed 26 + 26*26 + 26*26*26. I noticed that all the digits were different in the options.
So I simply calculated the unit digit of each product. 6 +6 + 6. = 18 = Unit digit of answer = 8. Hence, E.
Math Expert
Joined: 02 Sep 2009
Posts: 36530
Followers: 7068

Kudos [?]: 92947 [7] , given: 10541

Re: A certain stock exchange designates each stock with a one-, [#permalink]

### Show Tags

20 Jul 2013, 09:45
7
KUDOS
Expert's post
18
This post was
BOOKMARKED
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A. 2951
B. 8125
C. 15600
D. 16302
E. 18278

1 letter codes = 26
2 letter codes =2 6^2
3 letter codes = 26^3

Total=26+26^2+26^3

The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.

Similar questions to practice:
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
the-simplastic-language-has-only-2-unique-values-and-105845.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html
a-company-plans-to-assign-identification-numbers-to-its-empl-69248.html
the-security-gate-at-a-storage-facility-requires-a-five-109932.html
all-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html
a-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html
a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html
baker-s-dozen-128782-20.html#p1057502
in-a-certain-appliance-store-each-model-of-television-is-136646.html
m04q29-color-coding-70074.html
john-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html
how-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html
a-certain-stock-exchange-designates-each-stock-with-a-85831.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13421
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: A certain stock exchange designates each stock with a one-, [#permalink]

### Show Tags

02 Aug 2014, 23:39
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13421
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: A certain stock exchange designates each stock with a one-, [#permalink]

### Show Tags

11 Aug 2015, 23:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: A certain stock exchange designates each stock with a one-,   [#permalink] 11 Aug 2015, 23:27
Similar topics Replies Last post
Similar
Topics:
Closing prices for a certain stock were recorded each day for a week 2 16 Aug 2016, 05:10
At the opening of a trading day at a certain stock exchange, the price 4 25 May 2016, 00:56
39 A certain stock echange designates each stock with a 27 10 Nov 2009, 14:50
23 A certain stock exchange designates each stock with a 1, 2 11 26 Oct 2009, 12:25
12 A certain stock exchange designates each stock with a one-, 3 21 Sep 2009, 17:02
Display posts from previous: Sort by


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Sep 2018, 23:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# What is the perimeter of a square with area 9p^2/16 ?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49276
What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

22 Aug 2016, 02:56
00:00

Difficulty:

5% (low)

Question Stats:

85% (00:28) correct 15% (00:34) wrong based on 73 sessions

### HideShow timer Statistics

What is the perimeter of a square with area 9p^2/16 ?

A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3

_________________
Current Student
Joined: 09 Jul 2016
Posts: 25
Re: What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

22 Aug 2016, 04:01
What is the perimeter of a square with area 9p^2/16 ?

Area of square, (side)^2 = (3p/4)^2

Therefore side of the square = 3p/4
Perimeter of square = 4*side = 4* (3p/4) = 3p

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12422
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

29 Nov 2017, 22:14
1
Hi All,

We're told that the area of a SQUARE is (9P^2)/16. We're asked for the PERIMETER of the square. This question can be solved by TESTing VALUES.

Since the denominator is 16, we can 'cancel' it out if we make P=4...

IF... P=4, then the area of the square is 9(16)/16 = 9. This means that each side of the square = 3. Thus, we're looking for an answer that equals 4(3) = 12 when we plug P=4 into the answer choices. There's only one answer that matches...

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Senior SC Moderator
Joined: 22 May 2016
Posts: 1978
What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

30 Nov 2017, 06:36
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?

A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3

Find $$s$$.

Area of square = $$s^2 = \frac{9\pi^2}{16}$$

$$\sqrt{s^2} = \frac{(\sqrt{9})*(\sqrt{\pi^2})}{\sqrt{16}}$$

$$s= \frac{3\pi}{4}$$

Perimeter = $$4s=(\frac{3\pi}{4})*4 = 3\pi$$

_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

Intern
Joined: 13 Oct 2017
Posts: 3
GMAT 1: 530 Q43 V20
GPA: 3.99
Re: What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

30 Nov 2017, 07:04
Area = 9p^2/16
So a^2 = 9p^2/16 ( assumed side of a square as a)

So a = 3p/4
And perimeter of square is sum of all sides
So perimeter = 4a

So ans is 4(3p/4) = 3p

Sent from my iPhone using GMAT Club Forum
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 1994
Re: What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

30 Nov 2017, 12:15
1
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?

A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3

We know that the area of square = (side)^2

Thus, we can write :
$$(Side)^2 = \frac{9p^2}{16}$$

$$(Side)^2 = (\frac{3p}{4})^2$$

Since the side of square is always positive, we can conclude that the length of each side $$= \frac{3p}{4}$$

And the perimeter of the square $$= 4 * side = 4 * \frac{3p}{4} = 3p$$

The correct answer is Option C.
_________________

Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3516
Location: United States (CA)
Re: What is the perimeter of a square with area 9p^2/16 ?  [#permalink]

### Show Tags

03 Dec 2017, 18:47
1
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?

A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3

Since a side of the square is equal to √(9p^2/16) = 3p/4, then the perimeter of the square is 4 x 3p/4 = 12p/4 = 3p.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: What is the perimeter of a square with area 9p^2/16 ? &nbs [#permalink] 03 Dec 2017, 18:47
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


What's new

# Miller - Chapter 4 - Distributions - Study Notes, Pg 75, question #2

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi David,

In question (2) referenced as above,

(2) At the start of the year, a stock price is $100. A twelve-step binomial model describes the stock price evolution such that each month the extreme volatility price with either jump from S(t) to S(t)*u with 60.0% probability or down to S(t)*d with 40.0% probability. The up jump (u) = 1.1 and the down jump (d) = 1/1.1; note these (u) and (d) parameters correspond to an annual volatility of about 33% as exp[33%*SQRT(1/12)] ~= 1.10. At the end of the year, which is nearest to the probability that the stock price will be exactly$121?

(a) 0.33%
(b) 3.49%
(c) 12.25%
(d) 22.70%

There are 13 outcomes at the end of the 12-step binomial, with $100 as the outcome that must correspond to six up jumps and six down jumps. Therefore,$121 must be the outcome due to seven up jumps and five down jumps: $100*1.1^7*(1/1.1)^5 =$121
Such that we want the binomial probability given by:
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%

Clarification

Although, intuitively, 13 outcomes at the end of the 12-step binomial seems right (ie expected value of stock price at the end of 12 steps ??), how do you figure out six up jumps and six down jumps from the Binomial model to get \$100 as the outcome? Also how do you get :
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
I guess I am missing something that I should understand.

Thanks!
Jayanthi

#### ShaktiRathore

##### Well-Known Member
Subscriber
Hi
Consider n up jumps and 12-n down jumps to get 121 in the end
Therefore,100*(1.1)^n*(1/1.1)^12-n=121=>1.1^(2n-12)=1.1^2=>2n-12=2=>2n=14=>n=7 therefore there are 7 up and 12-7=5 down jumps.
Binomial probability is 12C7(.6)^7*(.4)^5=22.69%
Thanks

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi Shakti,

Yes - that is indeed great! Thanks a tonne for making it so simple to understand!

Jayanthi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @Jayanthi Sankaran the source question thread contains a visual that should help clarify, too. Please see https://www.bionicturtle.com/forum/...ributions-i-miller-chapter-4.7025/#post-28837
ie..,

#### Dr. Jayanthi Sankaran

##### Well-Known Member
This is neat, David

Thanks!
Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi David,

Does the above visual come with its own spreadsheet?

Thanks
Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Thanks David - will go through it, at leisure

Warm regards
Jayanthi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!

#### eva maria

##### New Member
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!

Many thanks David

Last edited by a moderator:


# Characterizing discontinuous derivatives

Apparently the set of discontinuity of derivatives is weird in its own sense. Following are the examples that I know so far:

$1.$ $$g(x)=\left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & x \in (0,1] \\ 0 & x=0 \end{array}\right.$$ $g'$ is discontinuous at $x=0$.

$2.$ The Volterra function defined on the ternary Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Cantor set ,that is on a nowhere dense set of measure zero.

$3.$ The Volterra function defined on the Fat-Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Fat-Cantor set ,that is on a set of positive measure, but not full measure.

$4.$ I am yet to find a derivative which is discontinuous on a set of full measure.

Questions:

1.What are some examples of functions whose derivative is discontinuous on a dense set of zero measure , say on the rationals?

2.What are some examples of functions whose derivative is discontinuous on a dense set of positive measure , say on the irrationals?

Update: One can find a function which is differentiable everywhere but whose derivative is discontinuous on a dense set of zero measure here.

• The Volterra function is not defined just on the Cantor set, but the whole interval. It equals $0$ on the Cantor set. I know you know this. But perhaps "defined relative to the Cantor set" might be more accurate. – zhw. Feb 24 '18 at 17:32
• @zhw Yeah, but I think its pretty obvious though. :P – yasir Feb 24 '18 at 17:44
• "Some good discussion about this can be found here and here." The first link you gave gives several references to the fact that any $F_{\sigma}$ first category set (equivalently, any countable union of closed nowhere dense sets) is the discontinuity set of some derivative. Those references that give a proof of this result do so in a constructive way, in the sense that given any countable union of closed nowhere dense sets (note that $\mathbb Q$ is a countable union of singleton sets), a differentiable function is constructed whose derivative is discontinuous exactly on that countable union. – Dave L. Renfro Feb 24 '18 at 21:08

There is no everywhere differentiable function $f$ on $[0,1]$ such that $f'$ is discontinuous at each irrational there. That's because $f',$ being the everywhere pointwise limit of continuous functions, is continuous on a dense $G_\delta$ subset of $[0,1].$ This is a result of Baire. Thus $f'$ can't be continuous only on a subset of the rationals, a set of the first category.

But there is a differentiable function whose derivative is discontinuous on a set of full measure.

Proof: For every Cantor set $K\subset [0,1]$ there is a "Volterra function" $f$ relative to $K,$ which for the purpose at hand means a differentiable function $f$ on $[0,1]$ such that i)$|f|\le 1$ on $[0,1],$ ii) $|f'|\le 1$ on $[0,1],$ iii) $f'$ is continuous on $[0,1]\setminus K,$ iv) $f'$ is discontinuous at each point of $K.$

Now we can choose disjoint Cantor sets $K_n \subset [0,1]$ such that $\sum_n m(K_n) = 1.$ For each $n$ we choose a Volterra function $f_n$ as above. Then define

$$F=\sum_{n=1}^{\infty} \frac{f_n}{2^n}.$$

$F$ is well defined by this series, and is differentiable on $[0,1].$ That's because each summand above is differentiable there, and the sum of derivatives converges uniformly on $[0,1].$ So we have

$$F'(x) = \sum_{n=1}^{\infty} \frac{f_n'(x)}{2^n}\,\, \text { for each } x\in [0,1].$$

Let $x_0\in \cup K_n.$ Then $x_0$ is in some $K_{n_0}.$ We can write

$$F'(x) = \frac{f_{n_0}'(x)}{2^{n_0}} + \sum_{n\ne n_0}\frac{f_n'(x)}{2^n}.$$

Now the sum on the right is continuous at $x_0,$ being the uniform limit of functions continuous at $x_0.$ But $f_{n_0}'/2^{n_0}$ is not continuous at $x_0.$ This shows $F'$ is not continuous at $x_0.$ Since $x_0$ was an aribtrary point in $\cup K_n,$ $F'$ is discontinuous on a set of full measure as desired.

• @ zhw So, if $f'$ is discontinuous on a set $D$, then $D$ can be dense only if its Lebesgue measure is zero? – yasir Feb 25 '18 at 3:04
• No, certainly not. A set of full measure is dense. – zhw. Feb 25 '18 at 7:09
• @ zhw. Is the set on which $F'$ is continuous dense, as per your claim in first paragraph? – yasir Feb 25 '18 at 9:26
• @yasir Yes, it has to be. – zhw. Feb 25 '18 at 17:14


Small Solver Program
02-14-2019, 05:25 AM
Post: #1
 Gamo Senior Member Posts: 518 Joined: Dec 2016
Small Solver Program
Recently I try to make a small but versatile solver program specifically
for HP Voyager Series calculator.

This solver solve for f(x) = 0 and use some type of counter for iteration.
Due to the iteration process this program will run slow on Original Calculator.
With modern 12C, 15C, and all other emulator this will run extremely fast.

This program is intended for HP-11C but this example will use
HP-15C emulator so that this solver program result can be compare
with the 15C build-in SOLVE function.

---------------------------------------------------------
To run program.

Enter 2 guesses closest to the root.

X1 [ENTER] X2 [A] display answer // Use this Solver

X1 [ENTER] X2 f [SOLVE] [E] display answer // Use 15C build-in SOLVE
---------------------------------------------------------
Example for X^X = Y // X to the power of X equal to Y

X on R1 // Register Storage 1 // For this Sover
Y on R5 // Register Storage 5

X on R2 // Register Storage 2 // For 15C SOLVE
Y on R5 // Register Storage 5
--------------------------------------------------------
Program:
Code:
 LBL A   // This Solver STO 0 Rv STO 1 90    STO I CLx LBL 1 RCL 0 / CHS RCL 1 + STO 1 DSE I GTO B RCL 1 RTN ------------------------------ LBL B  // f(x) program for X^X = Y RCL 1 LN RCL 1 x RCL 5 LN - GTO 1 ------------------------------- LBL E  // 15C SOLVE STO 2 // f(x) program for X^X = Y LN RCL 2 x RCL 5 LN - RTN
-------------------------------------------
Example: X^X = 1000

STORE 1000 to R5 // [STO] 5

This Solver

4 [ENTER] 6 f [A] display 4.555535704
-------------------------------------------------------
15C SOLVE

4 [ENTER] 6 f [SOLVE] [E] display 4.555535705
-------------------------------------------------------
Remark:
For this solver answer is in R1
For 15C SOLVE answer is in R2

Gamo
02-14-2019, 07:06 AM
Post: #2
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote:  Example: X^X = 1000

You can rewrite this equation such that $$x$$ is a fixed point of $$f(x)$$, that is $$x=f(x)$$:

$$x=\frac{3}{LOG(x)}$$

Then you can iterate the following program with a starting value, say $$x=4$$:
Code:
001-  43  13 :    LOG 002-       3 :    3 003-      34 :    x<>Y 004-      10 :    ÷

Example:

4
R/S
4.982892143
R/S
4.301189432
R/S
4.734933901
R/S
4.442378438
R/S
4.632377942
R/S
4.505830646
R/S
4.588735607
R/S
4.533824357
R/S
4.569933524
R/S
4.546075273
R/S
4.561789745
R/S
4.551417837
R/S
4.558254212
R/S
4.553744141
R/S
4.556717747
R/S
4.554756405
R/S
4.556049741
R/S
4.555196752
R/S
4.555759258
R/S
4.555388285
R/S
4.555632930
R/S
4.555471588
R/S
4.555577990
R/S
4.555507820
R/S
4.555554095
R/S
4.555523578
R/S
4.555543703
R/S
4.555530431
R/S
4.555539183
R/S
4.555533412
R/S
4.555537217
R/S
4.555534708

4.555535704
R/S
4.555535706
R/S
4.555535704
R/S
4.555535706

From then on the result flips between these last two values.

Cheers
Thomas
02-14-2019, 07:15 AM (This post was last modified: 02-14-2019 07:16 AM by Thomas Klemm.)
Post: #3
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
If you don't want to press R/S repeatedly:
Code:
001-  43  13 :    LOG 002-       3 :    3 003-      34 :    x<>Y 004-      10 :    ÷ 005-  42  31 :    PSE 006-  43  40 :    x=0
02-15-2019, 12:07 AM (This post was last modified: 02-17-2019 12:16 PM by Albert Chan.)
Post: #4
 Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-14-2019 07:06 AM)Thomas Klemm Wrote:  $$x=\frac{3}{LOG(x)}$$

4
R/S
4.982892143
R/S
4.301189432
...

It would be nice if we can temper the oscillation, or slow convergence.
Let x0 = 4, x1, x2 = the first two iterated values.

Rate = (x2-x1)/(x1-x0) ~ (4.30 - 4.98) / (4.98 - 4) = -0.694

If the same trend continued, we expect final % = 1/(1-r) ~ 60%
Use weighted fixed-point equation x = 0.4x + 0.6 * 3/log10(x):

4.6
4.555924149
4.555537395
4.555535712
4.555535705 (converged)

Edit: compare with Newton's method, x = (ln(1000) + x) / (ln(x) + 1)
4
4.571001573
4.555546101
4.555535705 (converged)
02-15-2019, 06:26 PM
Post: #5
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 12:07 AM)Albert Chan Wrote:  It would be nice if we can temper the oscillation, or slow convergence.

We can also use Aitken's delta-squared process to accelerate the speed of convergence:

$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$

This leads to the following program for the HP-11C:
Code:
001-42,21, 1 :    ▸LBL 01 002-      36 :     ENTER 003-    32 0 :     GSB 00 004-      30 :     - 005-  43  36 :     LSTx 006-      36 :     ENTER 007-    32 0 :     GSB 00 008-      30 :     - 009-  43  36 :     LSTx 010-      34 :     x<>y 011-  43  33 :     R↑ 012-      34 :     x<>y 013-      30 :     - 014-  43  36 :     LSTx 015-  43  11 :     x² 016-      34 :     x<>y 017-      10 :     ÷ 018-      30 :     - 019-  43  32 :     RTN 020-42,21, 0 :    ▸LBL 00 021-  43  12 :     LN 022-    45 0 :     RCL 00 023-      34 :     x<>y 024-      10 :     ÷ 025-  43  32 :     RTN

Example:

Solve: $$x^x=1000$$

1000 LN
STO 0

4.6
GSB 1
4.555665779
R/S
4.555535706
R/S
4.555535705
R/S
Error 0

Feel free to add a check for 0 after line 13 to prevent the error and loop back to label 1 instead of returning in line 19.

Cheers
Thomas
02-15-2019, 09:16 PM (This post was last modified: 02-15-2019 09:31 PM by Albert Chan.)
Post: #6
 Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-15-2019 06:26 PM)Thomas Klemm Wrote:
(02-15-2019 12:07 AM)Albert Chan Wrote:  It would be nice if we can temper the oscillation, or slow convergence.

We can also use Aitken's delta-squared process to accelerate the speed of convergence:

$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$

Amazingly, my rate formula is same as Aitken extrapolation formula !
Assuming we have 3 values, x0,x1,x2 and tried to guess where it should end up.

My rate analysis: r = (x2-x1)/(x1-x0) = Δx1 / Δx0

We need this for the proof:
(Δx0)²
= ((Δx0 - Δx1) + Δx1)²
= (Δx0 - Δx1)² + 2 * Δx1 (Δx0 - Δx1) + (Δx1)²

If same trend continue, where it will ends up
= x0 + Δx0 * 1/(1-r)
= x0 + (Δx0)² / (Δx0 - Δx1)
= x0 + (Δx0 - Δx1) + 2 * Δx1 + (Δx1)² / (Δx0 - Δx1)
= x0 + (x1-x0) + (x1-x2) + 2*(x2-x1) − (Δx1)² / (Δx1 - Δx0)
= x2 − (Δx1)² / (Δx1 - Δx0)
02-16-2019, 03:58 AM
Post: #7
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 09:16 PM)Albert Chan Wrote:  Amazingly, my rate formula is same as Aitken extrapolation formula !

That's not really a surprise, is it?
It's essentially the same idea: geometric series

$$1\,+\,r\,+\,r^{2}\,+\,r^{3}\,+\,\cdots \;=\;{\frac {1}{1-r}}$$
02-16-2019, 12:24 PM (This post was last modified: 02-16-2019 12:34 PM by Csaba Tizedes.)
Post: #8
 Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote:  a small but versatile solver program specifically for HP Voyager Series calculator.
Example for X^X = Y // X to the power of X equal to Y

Yes, the classic problem: which number you can power to itself to get the largest number on your calculator: x^x=10^100. Do LOG() of both sides: x×LOG(x)=100, then do the iteration: x_0:=50, then x_i+1:=100/log(x_i). The simple fixed point iteration and converges.

By the way, if you interested, here is my SOLVE(i) for HP-12C, which can solve any equation for ANY variable like on 15C, without rewriting the equation. Looks like indirect addressing without indirect addressing. 28 steps only.

Yes, my English is terrible, but more convenient than read the full material in Hungarian.
At the end of pdf you can find lots of applications which solved with this little solver (like ODE solve with EULER+SOLVE(i) on 12C (!!!), pressure vessel pneumatic conveying pipe sizing program, humidity and dew point solving, outer temperature calculation of a concrete silo wall or comparison of present values of educated and not educated people salaries with the little SOLVE combined with 12C's cash flow calculation capabilities - I guess this is shows clearly the power of this little routine possibilities and why I say everywhere, the SOLVER is MUST to implement ALL current calculators, I hope somebody read and understand what I want to say).

The small secant method in this paper for CASIO fx-50F meantime reduced to 13 steps, I guess the smallest solver for these blind machines. You can find attached.

Csaba

Attached File(s)
02-16-2019, 01:42 PM
Post: #9
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-16-2019 12:24 PM)Csaba Tizedes Wrote:  By the way, if you interested, here is my SOLVE(i) for HP-12C

Steps of secant method from 03 to 18:
Code:
03-   45 2 :    RCL 2 04-   45 0 :    RCL 0 05-     30 :    - 06-  43 36 :    LSTx 07-   44 2 :    STO 2 08-     35 :    CLx 09-   45 3 :    RCL 3 10-   45 1 :    RCL 1 11-     30 :    - 12-  43 36 :    LSTx 13-   44 3 :    STO 3 14-     33 :    R↓ 15-     10 :    ÷ 16-   45 1 :    RCL 1 17-     20 :    × 18-44 30 0 :    STO+ 0

This can be shortened to:
Code:
03-   45 2 :    RCL 2 04-   45 0 :    RCL 0 05-   44 2 :    STO 2 06-     30 :    - 07-   45 3 :    RCL 3 08-   45 1 :    RCL 1 09-   44 3 :    STO 3 10-     30 :    - 11-     10 :    ÷ 12-   45 1 :    RCL 1 13-     20 :    × 14-44 30 0 :    STO+ 0

Don't hesitate to publish your programs in this forum. You can use Google to translate to English.

Cheers
Thomas
02-16-2019, 03:24 PM
Post: #10
 Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-16-2019 01:42 PM)Thomas Klemm Wrote:  Steps of secant method from 03 to 18 can be shortened to.
Thanks, this was eons before, I am sure this is only the first iteration and required to fine tuning.

(02-16-2019 01:42 PM)Thomas Klemm Wrote:  Don't hesitate to publish your programs in this forum. You can use Google to translate to English.
OK, sometimes I feel myself as Don Quixote... Frankly speaking, who want to discuss about calculators and calculator programming now? 15 years before I wrote something useful, today I am only (an angry) user...

Csaba
02-17-2019, 02:57 AM (This post was last modified: 02-17-2019 03:20 AM by Gamo.)
Post: #11
 Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
I just fine tune this solver program for HP-11C

Added the iterations limit to 28 loops count
Added the condition test to end program when root is found.

If iteration is over 28 loops count this indicate that user must use a closer guess.

-------------------------------------------

Procedure:

Use R1 for unknown X in your equation. End your formula with [RTN]

X1 ENTER X2 [A] display Answer

------------------------------------------------------------------------
Program:
Code:
 LBL A STO 0 Rv STO 1 28 STO I  // Store Loop Count Limit CLx --------------------- LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y  // End if Root is found GTO 2 DSE  // Loop Limit Counter GTO 0 CLx FIX 9  // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4  GTO 2 --------------------- LBL 1  // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN -------------------- LBL 2 RCL 1  // Answer  RTN -------------------- LBL B .  // f(x) equation start here . .  // X = Register 1 [R1] . . RTN

Gamo
02-17-2019, 09:06 AM
Post: #12
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:57 AM)Gamo Wrote:  I just fine tune this solver program for HP-11C

Your program doesn't work for this simple example: $$x^2+x=x(x+1)=0$$
I've entered the following program:
Code:
LBL B 1 RCL 1 + LSTx × RTN

Example:

-2
ENTER
-0.5
A

9.999999 99

The reason is that the function value is divided by -0.5, making it larger and larger:
Code:
LBL 1  // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN

Another thing is this in this loop:
Code:
LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y  // End if Root is found GTO 2 DSE  // Loop Limit Counter GTO 0

When you return to label 0, the same value in register 1 will be used as before.
This function evaluation can be avoided if you save the result of the previous evaluation.

You could move this block:
Code:
LBL 2 RCL 1  // Answer  RTN

Over here and avoid the jump GTO 2:
Code:
CLx FIX 9  // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4  LBL 2 RCL 1  // Answer  RTN

Take a look at Csaba's solution for the HP-12C using the secant method.

Cheers
Thomas
02-17-2019, 02:33 PM
Post: #13
 Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Thanks Thomas Klemm

I try single guess and it work.

CHS 2 [A] display -1

If using two negative guesses and result is
Overflow then use Single guess instead.

Formula Routine is

RCL 1
1
+
RCL 1
X

Gamo
02-17-2019, 04:57 PM
Post: #14
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:33 PM)Gamo Wrote:  If using two negative guesses and result is Overflow then use Single guess instead.

Then try this function $$f(x)=4x(x+1)$$:
Code:
LBL B RCL 1 1 + RCL 1 × 4 × RTN

Examples:

-0.5
ENTER
0.5
A

9.999999 99

-0.5
ENTER
A

9.999999 99

0.5
ENTER
A

9.999999 99

Even initial guesses very close to the solutions lead to the same result:

-1.00001
ENTER
-0.99999
A

9.999999 99

-0.00001
ENTER
0.00001
A

9.999999 99

Cheers
Thomas
02-18-2019, 03:49 AM
Post: #15
 Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Your're right that only work for certain situations.

Gamo
02-18-2019, 05:20 AM
Post: #16
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
If you compare your algorithm with Newton's method:

$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$

you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.

And with this in mind indeed the solution can now be found:

$$f'(x) = \frac{d}{dx}4 x (x + 1) = 8 x + 4$$

$$f'(-1) = -4$$

$$f'(0) = 4$$

Examples:

-1.5
ENTER
-4
A

-1.00000

0.5
ENTER
4
A

0.00000

Thus your 2nd parameter is an estimate of the slope at the root.

And then you don't really have to call the function twice with the same value:
Code:
LBL A STO 0 Rv STO 1 28 STO I  // Store Loop Count Limit LBL 0 GSB B RCL 0 ÷ STO - 1 X=0  // End if Root is found GTO 1 DSE  // Loop Limit Counter GTO 0 CLx FIX 9  // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 LBL 1 RCL 1  // Answer  RTN LBL B .  // f(x) equation start here . .  // X = Register 1 [R1] . . RTN

So the question remains how to approximate the derivation at the roots.
If you keep record of the previous function evaluation you can get an estimate of the slope using the secant method.

Cheers
Thomas
02-18-2019, 07:46 PM
Post: #17
 Dieter Senior Member Posts: 2,398 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 05:20 AM)Thomas Klemm Wrote:  If you compare your algorithm with Newton's method:

$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$

you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.

Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.

For more details take a look at this post: http://www.hpmuseum.org/forum/thread-123...#pid111680

Gamo, you really should correct and update the instructions. Here it still says:
Quote:Enter 2 guesses closest to the root.

But again: that's NOT true. The first input is NOT a guess for the root. You now have seen a variety of cases that prove it.

Dieter
02-18-2019, 10:22 PM
Post: #18
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 07:46 PM)Dieter Wrote:  Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.

You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.

So all in all I'm happy with the outcome.

Cheers
Thomas
02-19-2019, 01:10 AM (This post was last modified: 02-19-2019 01:22 AM by Albert Chan.)
Post: #19
 Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG

Linear regression on slope had another benefit, which my Casio cannot do.
Doing Quadratic Regression on Casio, you don't really know if it is any good.
There is no "correlation coefficient" to lookup. (Why?)

Linear regression on slope can show whether a Quadratic curve is warranted.

I think the code add slope data, like this:

Data points (0,90) and (30, 90.2), add to linear regression: (30+0)/2, (90.2-90)/(30-0)
Next point (60, 86.4), add to linear regression: (60+30)/2, (86.4-90.2)/(60-30)
...
N points (sorted) thus generate N-1 slopes to regress.
02-19-2019, 08:39 AM
Post: #20
 Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-18-2019 10:22 PM)Thomas Klemm Wrote:
(02-18-2019 07:46 PM)Dieter Wrote:  Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.

You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.

So all in all I'm happy with the outcome.

Cheers
Thomas

I feel myself like a star (at least for 15 minutes...)

Yes, this is my "linear regression on slope", as Chan wrote. When I was in my "Summer Practice" at a company I have modelled lots of water network flows between pump houses (sources), consumers and reservoirs, and sometimes the characteristics curves given only in paper form. Thats why I wrote this little routine.

As you can see, if the measured points has significant error, the slope will be very inaccurate then the regression also inaccurate and the coefficients can be meaningless.

But as a programming question this was an interesting "game" for me.

To the original question from Gamo: I will check carefully the lists here, but I have another idea: what if, if we just simple walking along on the function's curve until the sign is changing, then we turning back, decreasing the stepsize and walking again. If the sign of the function changed, we are turning back again, decreasing the size of the steps and so on... If the stepsize is less than a small value, we found the root.

This "marching method" not so elegant, not so fast, but good to play with it, just for a try - maybe it can be shorter than the other methods. (I have an exactly same method for finding a minimum point of a function for CASIO fx-3650P/Sencor SEC103; perfectly works, like a ball at the bottom of a hole, slowly find the deepest point, when swings around the extremum).

Csaba
 « Next Oldest | Next Newest »

User(s) browsing this thread: 1 Guest(s)


# Representing displacement vectors in cylindrical coordinates and finding the distance in cylindrical coordinates?

In cartesian coordinates, we can derive the vector $\vec v_3$ by vector subtraction $\vec v_2-\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\vec v_3$ which then is: $$\lvert \vec v_3\rvert = \lvert \vec v_2-\vec v_1 \rvert= \lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \rvert = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

But how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\rho_1$ and $\theta_1$ for $\vec v_1$ as well as $\rho_2$ and $\theta_2$ for $\vec v_2$ in cylindrical coordinates $(\rho,\theta,z)$, where $z=0$ in this example.

Your problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\vec v_3|^2=(\vec v_2-\vec v_1)\cdot(\vec v_2-\vec v_1)=|\vec v_1|^2+|\vec v_2|^2-2\vec v_2\cdot \vec v_1$$ Now using cylindrical coordinates, $|\vec v_i|^2=r_i^2+z_i^2$ and $\vec v_2\cdot \vec v_1=r_1r_2\cos(\theta_2-\theta_1)+z_1z_2$. Then the final answer will be $$|\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)+(z_2-z_1)^2$$
• Yes, I've seen this solution, but isn't it based in terms of cartesian coordinates/base vectors, $\hat e_x$, $\hat e_y$ and $\hat e_z$ after all? This since, I guess, you must express a distance in constant base vectors? I'm a bit confused about how to interpret the problem I have to admit. How would it look if I want to express the solution completely in cylindrical coordinates with $\vec v_1=\rho_1 \hat e_\rho (\theta_1)$ and base vectors $\hat e_\rho$, $\hat e_\theta$, and $\hat e_z$ etc? Would that be possible? Sep 6, 2018 at 5:46
• I did not use Cartesian coordinates. The $z$ coordinate is there in cylindrical system. The only other reference to the Cartesian system is the angle $\theta$. The issue that you have is that the basis of the cylindrical coordinate system changes with the vector, therefore equations will be more complicated. Sep 6, 2018 at 6:38
• If you choose one of the vectors as the reference, the cylindrical coordinate system is just a Cartesian system rotated with respect to the original. $\hat e_\rho$ is the new $\hat e_x$, and $\hat e_\theta$ is the new $\hat e_y$. Sep 6, 2018 at 7:19


# How to determine the curve?

In the figure above, segment $PQ$ is determined by two points: $P: (t,0)$ and $Q: (1,t)$, where $t\in [0,1]$ continuously increases and decreases between $0$ and $1$.

Then this gives a close region swept by $PQ$, the upper edge of which is a curve.

How to determine the equation of the curve (maybe implicit form)?

My own method

Suppose the curve is $y=y(x)$, then for any fixed $x_0\in (0,1)$, there is a vertical line $x=x_0$, which intersects a bundle of such $PQ(t)$ segments:

$$y=\frac{t(x-t)}{1-t}$$

Easy to conclude that, the desired $y_0$ on the curve corresponding to $x_0\left(\in(0,1)\right)$ is the maximum of:

$$y_0= \max\limits_{t\in (0,1)}\frac{t(x_0-t)}{1-t}=2-2\sqrt{1-x_0}-x_0$$

How to use the envelope concept? Is there any elementary method since the area is $\frac{1}{6}$?

• Do you know derivatives, tangent lines and integrals? – Tomas Feb 6 '15 at 9:47
• Yes, I know; I am not sure whether calculus is necessary in order to determine the area, but I am pretty sure in order to determine the curve, calculus would be a help – LCFactorization Feb 6 '15 at 9:53
• Look at the wiki entry for Envelope, it teach you how to determine the curve. – achille hui Feb 6 '15 at 10:29
• I use optimization method and obtains $y=2-2\sqrt{1-x}-x$, then the area is $1/6$. Is there any elementary method in order to determine the area? – LCFactorization Feb 6 '15 at 11:12

I know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.

$$\begin{pmatrix}t\\0\\1\end{pmatrix}\times \begin{pmatrix}1\\t\\1\end{pmatrix}= \begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}$$

This could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:

\begin{align*} a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\ ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4 \end{align*}

The $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:

$$a^2 + ac - bc = 0$$

Written as a matrix:

$$(a,b,c)\cdot\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = 0$$

Now, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:

$$(x,y,1)\cdot\begin{pmatrix}1&1&0\\1&1&-2\\0&-2&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix} = 0$$

Since the determinant of the upper left $2\times2$ matrix is zero, this is a parabola. It can also be described by the equation

\begin{align*} x^2 + 2xy + y^2 &= 4y \\ (x+y)^2 &= 4y \end{align*}

Since this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as

$$\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix} =-\begin{pmatrix}2t-t^2\\t^2\\1\end{pmatrix}$$

So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.

HINT

I would say

$F(x,y,t)=0\equiv t(x-t)-y(1-t)=0,\quad t\in\langle 0,1 \rangle$

$F'_t(x,y,t)=0\equiv x-2t+y=0$

Solving this system of equations for the x, y:

$x=t(2-t), y=t^2, \quad t\in\langle 0,1 \rangle$ is a parametric curve of equation to search.

Plot

• Very simple and neat answer. All three are acceptable. I choose the first one that also uses projective geometry. Thank you very much! @MvG – LCFactorization Feb 7 '15 at 2:40

I have a simpler solution.

1. Find the equation of the line with $(t,0,1)\times(1,t,1)=(-t,1-t,t^2)$ or $$(-t)x+(1-t)*y+(t^2)=0$$
2. Find the extema $t$ with $$\frac{{\rm d}}{{\rm d}t}\left(-t\,x+(1-t)\,y+t^2\right)=0$$ $$\left. -x-y+2 t=0 \right\}\; t=\frac{x+y}{2}$$
3. Plug $t$ into line equation for $$-\frac{(x+y)^2}{4}+y=0$$ which is the resulting curve

Hint: The envelope of an implicit curve $f(x,y,t)=0$ is found by solving for $t$ in $\frac{\partial f}{\partial t}=0$ and using it into the original equation.

My other answer describes how I myself think about this. But I realize that there is a lot of background knowledge in there, which might make this less accessible to members of other communities than projective geometry. So here is my attempt at a more generic solution.

Suppose you have found the equation of the line to be $tx + (t-1)y = t^2$. Take a second line from your family, using $u$ instead of $t$ as the parameter, i.e. $ux + (u-1)y = u^2$. These two intersect in a point $(t+u-tu, tu)$. Now if you consider $\lim_{u\to t}$ then strictly speaking the point of intersection becomes undefined, but the above will simply become $(2t-t^2,t^2)$. So that's the one point that your line does not have in common with any other line of the family (extended from $t\in[0,1]$ to $t\in\mathbb R$). So at that point this line is the one defining the curve. Therefore that's your parametric equation of the envelope curve.


# Question Solved1 Answer1. 2. 3. Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).

CW7XUD The Asker · Calculus

1.

2.

3.

Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
More
Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
See Answer
Add Answer +20 Points
Community Answer
CXSSZN The First Answerer
See all the answers with 1 Unlock
Get 4 Free Unlocks by registration

1)the average velocity,Average Velocity =(" Total Displacement ")/(" Total Time ")Average Velocity =(d(b)-d(a))/(b-a)2)we ... See the full answer