Parallel Tangent Lines

[Jason76]

At what value of \(\displaystyle x\) do the graphs of \(\displaystyle a\) and \(\displaystyle b\) have parallel tangent lines? Answer: \(\displaystyle -0.484\)

Graph a

\(\displaystyle a(x) = e^{3x}\)

Graph b

\(\displaystyle b(x) = x^{3}\)

In order for them to have parallel tangent lines, they must have the same slope (derivative). So how to find that?

 

[lookagain]

At what value of \(\displaystyle x\) do the graphs of \(\displaystyle a\) and \(\displaystyle b\) have parallel tangent lines? Answer: \(\displaystyle -0.484\)

Graph a

\(\displaystyle a(x) = e^{3x}\)

Graph b

\(\displaystyle b(x) = x^{3}\)

In order for them to have parallel tangent lines, they must have the same slope (derivative). So how to find that?

Jason76, you have been taking derivatives of polynomial functions, so you should be able to find \(\displaystyle \ b \ '(x).\)

The derivative of the e-based exponential function is itself multiplied by the derivative of the exponent.

After you work out what the derivatives are, set them equal.

To solve for a decimal approximation for x, expect to use a numerical method, a graphics calculator, or an online graphing system, for instance.

 

[Jason76]

Jason76, you have been taking derivatives of polynomial functions, so you should be able to find \(\displaystyle \ b \ '(x).\)

The derivative of the e-based exponential function is itself multiplied by the derivative of the exponent.

After you work out what the derivatives are, set them equal.

To solve for a decimal approximation for x, expect to use a numerical method, a graphics calculator, or an online graphing system, for instance.

Right, I was thinking along the same lines. If they both have the same derivative at a certain \(\displaystyle x\), then set those derivatives equal to each other, and solve for \(\displaystyle x\).

1. Take the derivative of both equations.

2. Set them equal to each other.

3. Find x.

1.

At what value of \(\displaystyle x\) do these two equations have parallel tangent lines (same derivative)?

\(\displaystyle y = e^{3x}\) - graph a

\(\displaystyle \dfrac{dy}{dx} = 3e^{3x} \) - derivative of graph a

\(\displaystyle y = x^{3}\) - graph b

\(\displaystyle \dfrac{dy}{dx} = 3x^{2} \) - derivative of graph b

\(\displaystyle 3e^{3x} = 3x^{2} \) - Set derivatives equal to each other and solve for \(\displaystyle x\)

\(\displaystyle \dfrac{3e^{3x}}{3} = \dfrac{3x^{2}}{3} \)

\(\displaystyle e^{3x} = x^{2} \)

\(\displaystyle \ln(e^{3x}) = \ln(x^{2})\)

\(\displaystyle 3x = \ln(x^{2})\) - What trick could I do here?

 

[Bob Brown MSEE]

Put: 1 in cell A1 of spreadsheet
Put: =-EXP(A1*3/2) in cell A2 of spreadsheet
Duplicate cell A2 downward about 50 times

EDIT: This is a numerical method called "Fixed Point Iteration", it is by far the simplest/quickest way to an answer without a graphing calculator. Warning, this method often does not converge or does so slowly. Another approach is the "bi-section method" that is more likely to converge, but often even more slowly. The fastest converging and relative robust numerical solution presented is Newton's Method. It is worth investing the time to learn how to use a graphing calculator, there are several on line. (click on the link in this edit -- hover over the intersection point on the two graphs to read the numerical solution)

 

[HallsofIvy]

Right, I was thinking along the same lines. If they both have the same derivative at a certain \(\displaystyle x\), then set those derivatives equal to each other, and solve for \(\displaystyle x\).

1. Take the derivative of both equations.

2. Set them equal to each other.

3. Find x.

1.

At what value of \(\displaystyle x\) do these two equations have parallel tangent lines (same derivative)?

\(\displaystyle y = e^{3x}\) - graph a

\(\displaystyle \dfrac{dy}{dx} = 3e^{3x} \) - derivative of graph a

\(\displaystyle y = x^{3}\) - graph b

\(\displaystyle \dfrac{dy}{dx} = 3x^{2} \) - derivative of graph b

\(\displaystyle 3e^{3x} = 3x^{2} \) - Set derivatives equal to each other and solve for \(\displaystyle x\)

\(\displaystyle \dfrac{3e^{3x}}{3} = \dfrac{3x^{2}}{3} \)

\(\displaystyle e^{3x} = x^{2} \)

\(\displaystyle \ln(e^{3x}) = \ln(x^{2})\)

\(\displaystyle 3x = \ln(x^{2})\) - What trick could I do here?

This can be simplified slightly: 3x= 2ln(x) or ln(x)= (3/2)x- although, in fact, it is better to stay with \(\displaystyle e^{3x}= x^2\). That, involving x both "inside" an transcendental function (ln(x) or \(\displaystyle e^x\)) and "outside" cannot be solved using "standard algebraic" means. It could be solved numerically, perhaps by graphing y= 2ln(x) and y= 3x and seeing where they intersect (I get x about -.484).

 

[Jason76]

This can be simplified slightly: 3x= 2ln(x) or ln(x)= (3/2)x- although, in fact, it is better to stay with \(\displaystyle e^{3x}= x^2\). That, involving x both "inside" an transcendental function (ln(x) or \(\displaystyle e^x\)) and "outside" cannot be solved using "standard algebraic" means. It could be solved numerically, perhaps by graphing y= 2ln(x) and y= 3x and seeing where they intersect (I get x about -.484).

I suppose on a test this would require a graphing calculator. Are you sure there is NO way to get out of this "transcendental situation" by using algebra?

 

[JeffM]

Jason

Yesterday, about a different problem, I asked what you are currently studying. Context matters, but you refuse to tell us in what context problems are posed. I also asked whether the problem asks for an exact or an approximate answer. Again, you did not respond. For this problem, all you can get is an approximation.

\(\displaystyle a(x) = e^{3x} = e^u \implies \dfrac{du}{dx} = 3\ and\ a'(u) = e^u \implies a'(x) = 3 * e^u = 3e^{3x}.\) Easy, right?

\(\displaystyle b(x) = x^3 \implies b'(x) = 3x^2.\) Basic stuff.

\(\displaystyle a'(x) = b'(x) \implies a'(x) - b'(x) = 0 \implies 3e^{3x} - 3x^2 = 0 \implies e^{3x} - x^2 = 0.\)

\(\displaystyle Let\ c(x) = e^{3x} - x^2.\)

What techniques do you know for approximating roots.

Hint: What is the sign of c(0)? What is the sign of of c(- 1)? What does that tell you?

 

[Jason76]

Jason

Yesterday, about a different problem, I asked what you are currently studying. Context matters, but you refuse to tell us in what context problems are posed. I also asked whether the problem asks for an exact or an approximate answer. Again, you did not respond. For this problem, all you can get is an approximation.

\(\displaystyle a(x) = e^{3x} = e^u \implies \dfrac{du}{dx} = 3\ and\ a'(u) = e^u \implies a'(x) = 3 * e^u = 3e^{3x}.\) Easy, right?

\(\displaystyle b(x) = x^3 \implies b'(x) = 3x^2.\) Basic stuff.

\(\displaystyle a'(x) = b'(x) \implies a'(x) - b'(x) = 0 \implies 3e^{3x} - 3x^2 = 0 \implies e^{3x} - x^2 = 0.\)

\(\displaystyle Let\ c(x) = e^{3x} - x^2.\)

What techniques do you know for approximating roots.

Hint: What is the sign of c(0)? What is the sign of of c(- 1)? What does that tell you?

The answer given (by HallsOfIvy) of -.484 (from observing the graph, not doing algebra (at least, after the ln stage) was the practice test answer.

 

[JeffM]

The answer given (by HallsOfIvy) of -.484 (from observing the graph, not doing algebra (at least, after the ln stage) was the practice test answer.

Oh good. You have the answer.

 

[Deleted member 4993]

I suppose on a test this would require a graphing calculator. Are you sure there is NO way to get out of this "transcendental situation" by using algebra?

Not really - there is an approximation process called "bi-section method" and you can dothat with calculator>

[

x	e^(3x)-x^2
0	1
-1	-0.950212932
-0.5	-0.02686984
-0.25	0.409866553
-0.375	0.184027467
-0.4375	0.077740099
-0.46875	0.025333977
-0.484375	-0.000780741
-0.4765625	0.012271902
-0.4804688	0.005744593
-0.4785156	0.009007977
-0.4814453	0.00411309
-0.4829102	0.001666053
-0.4836426	0.000442627
-0.4840088	-0.000169064
-0.4838257	0.00013678
-0.4839172	-1.61427E-05
-0.4838715	6.03183E-05

/QUOTE]

 

[Jason76]

No Thanks I think I'd rather graph it, and look at the answer. :mrgreen:

Not really - there is an approximation process called "bi-section method" and you can dothat with calculator>

x	e^(3x)-x^2
	1
-1	-0.950212932
-0.5	-0.02686984
-0.25	0.409866553
-0.375	0.184027467
-0.4375	0.077740099
-0.46875	0.025333977
-0.484375	-0.000780741
-0.4765625	0.012271902
-0.4804688	0.005744593
-0.4785156	0.009007977
-0.4814453	0.00411309
-0.4829102	0.001666053
-0.4836426	0.000442627
-0.4840088	-0.000169064
-0.4838257	0.00013678
-0.4839172	-1.61427E-05
-0.4838715	6.03183E-05

/QUOTE]

 

[MarkFL]

One of the simplest of numeric root finding methods is Newton's method, aka the Newton-Raphson method. To begin, we define the function:

\(\displaystyle f(x)=e^{3x}-x^2=0\)

\(\displaystyle f'(x)=3e^{3x}-2x=0\)

and then the method gives us the recursion:

\(\displaystyle x_{n+1}=x_n-\dfrac{f(x)}{f'(x)}=x_n-\dfrac{e^{3x_n}-x_n^2}{3e^{3x_n}-2x_n}=\dfrac{x_n\left(3e^{3x_n}-2x_n \right)-e^{3x_n}+x_n^2}{3e^{3x_n}-2x_n}=\dfrac{e^{3x_n}\left(3x_n-1 \right)-x_n^2}{3e^{3x_n}-2x_n}\)

Using \(\displaystyle x_0=0\) as our initial guess, we then find:

\(\displaystyle x_1\approx-0.33333333333333\)

\(\displaystyle x_2\approx-0.47837519419036\)

\(\displaystyle x_3\approx-0.4839063724388\)

\(\displaystyle x_4\approx-0.483907571844\)

\(\displaystyle x_5\approx-0.48390757184\)

 

[Bob Brown MSEE]

Put: 1 in cell A1 of spreadsheet
Put: =-EXP(A1*3/2) in cell A2 of spreadsheet
Duplicate cell A2 downward about 50 times

EDIT: This is a numerical method called "Fixed Point Iteration", it is by far the simplest/quickest way to an answer without a graphing calculator. Warning, this method often does not converge or does so slowly. Another approach is the "bi-section method" that is more likely to converge, but often even more slowly. The fastest converging and relative robust numerical solution presented is Newton's Method. It is worth investing the time to learn how to use a graphing calculator, there are several on line. (click on the link in this edit -- hover over the intersection point on the two graphs to read the numerical solution)

Click Here

I got the above link by simply speaking the request to WolframAlpha.

 

[Bob Brown MSEE]

A comment about numerical methods

WolframAlpha.

It is a good idea to learn how to get numerical solutions. However, in a class-room environment it is important to use the methods taught in the class. (numerical solutions can get you zero credit, when they are not required) You may be frustrated as a student, to know when they are required. It will take some experience before you can spot a problem for which there is no "closed form" or clever substitution. Do not feel like you should know, at least not at this point in your studies.

Rules:
1) Never use numerical methods on tests (unless specifically required)
2) Always use numerical methods on homework to check answers.
3) Always use numerical methods on homework to obtain insight.