Derivative/tangency problem?

**lysserloo** · October 12th 2009, 09:25 PM

I really don't understand how to even begin this problem...otherwise I'd show my attempt. But I've just been staring at it unable to figure it out at all.

All I have figured out thus far is that the equation for intersection is just sinx = ke^-x

Problem: If $k \geq 1$ , the graphs of y = sinx and y = ke^-x intersect for $x \geq 0$ . Find the smallest value for k for which the graphs are tangent. What are the exact coordinates of the point of tangency? [Hint: you should have two equations involving x and k, one for intersection, and one for tangency. It will be difficult to deal with either of these equations separately, but you should be able to figure out the answer by combining them.]

Thank you in advance!

**Chris L T521** · October 12th 2009, 10:02 PM

Originally Posted by lysserloo

I really don't understand how to even begin this problem...otherwise I'd show my attempt. But I've just been staring at it unable to figure it out at all.

All I have figured out thus far is that the equation for intersection is just sinx = ke^-x

Problem: If $k \geq 1$ , the graphs of y = sinx and y = ke^-x intersect for $x \geq 0$ . Find the smallest value for k for which the graphs are tangent. What are the exact coordinates of the point of tangency? [Hint: you should have two equations involving x and k, one for intersection, and one for tangency. It will be difficult to deal with either of these equations separately, but you should be able to figure out the answer by combining them.]

Thank you in advance!

You need the equation for point of tangency, i.e. when $\cos x=-ke^{-x}$

To combine them, note that $\sin^2x+\cos^2x=1\implies \left(ke^{-x}\right)^2+\left(-ke^{-x}\right)^2=1$ $\implies 2k^2e^{-2x}=1\implies x=\tfrac{1}{2}\ln\left(2k^2\right)=\tfrac{1}{2}\ln 2+\ln k$

Also note that $\frac{\sin x}{\cos x}=\frac{ke^{-x}}{-ke^{-x}}=-1\implies \tan x=-1\implies x=\tfrac{3\pi}{4}$ .

Thus, $x=\tfrac{3\pi}{4}=\tfrac{1}{2}\ln 2+\ln k\implies\frac{3\pi}{2}=\ln\!\left(2k^2\right)\imp lies k^2=\tfrac{1}{2}e^{\frac{3\pi}{2}}\implies k=\tfrac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}$

Thus, the point of tangency is $\left(\tfrac{3\pi}{4},\tfrac{\sqrt{2}}{2}\right)$

Does this make sense? If not, please post back!

**lysserloo** · October 12th 2009, 10:51 PM

I understand how you got the equation for tangency, and I sort of follow the concept of sin^2x + cos^2x = 1 and how you applied that.

So in that first line, did you just solve for x?

The rest of it... I have no idea what you're doing or why. I'm sorry, I just don't follow at all.

In particular, I want to know the why so that I can apply it to other, similar problems.

Can you explain your steps a little bit? And can you explain how you knew to do what you did?

**Chris L T521** · October 13th 2009, 06:50 AM

Originally Posted by lysserloo

I understand how you got the equation for tangency, and I sort of follow the concept of sin^2x + cos^2x = 1 and how you applied that.

So in that first line, did you just solve for x?

The rest of it... I have no idea what you're doing or why. I'm sorry, I just don't follow at all.

In particular, I want to know the why so that I can apply it to other, similar problems.

Can you explain your steps a little bit? And can you explain how you knew to do what you did?

The key idea to this is to come up with two relationships between sine and cosine. This required some thinking and ingeniuity, but it was doable.

Here, the first one I thought of was $\sin^2x+\cos^2x=1$ . I noticed that this would give me a value of x in terms of k.

However, this isn't enough. We need to find a number k. So how do we do that?

It took me a little bit to noticed the nice cancellation that happened when we applied the identity $\frac{\sin x}{\cos x}=\tan x$ .

In this case, we happen to get rid of the $ke^{-x}$ term. So we were left with an equation that could easily be solved: $\frac{\sin x}{\cos x}=\frac{ke^{-x}}{-ke^{-x}}\implies\tan x=-1\implies x=\tfrac{3}{4}\pi$

Now that we know what x is, we can find k. (Substitute x into $x=\tfrac{1}{2}\ln\!\left(2k^2\right)$ then solve for k).

To find the point of tangency, plug in $x=\tfrac{3}{4}\pi$ into $\sin x$ .

Does this clarify anything?

**lysserloo** · October 13th 2009, 08:38 AM

Ah yes thank you so much for explaining! I completely understand now.

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