Tangency problem

[malonmad]

I've been having some problems trying to solve this problem.

If k > 0, the graphs of y = 0.9sinx and zk(x) = ke−x intersect for some x ≥ 0. Find the smallest value of k for which the graphs of the two functions are tangent. What are the coordinates of the point of tangency?

If anyone could help that would be awesome.

 

[DrPhil]

I've been having some problems trying to solve this problem.

If k > 0, the graphs of y = 0.9sinx and zk(x) = ke−x intersect for some x ≥ 0. Find the smallest value of k for which the graphs of the two functions are tangent. What are the coordinates of the point of tangency?

If anyone could help that would be awesome.

Takes some efort to type math expressions inline. Like lots of extra parentheses to show the order of operations, and also conventions like ^ for superscript and _ for subscript. May I read the 2nd graph to say
z_k(x) = k e^(-x) ?
If we use LaTeX, we can make that look better:

$\displaystyle \displaystyle z_k(x) = k\ e^{-x}$

If that is ok, then you can see that the slope dz/dx is always negative, so any point of tangency must be on a downslope of the sine function. "Smallest" k suggests that we have to deal with just the first hump of the sine function. I would therefore limit the domain of x to pi/2 < x < pi.

You have two unknowns, k and x, so you need two independent equations. Those would be that y(x) = z_k(x), and also y'(x) = z'_k(x). What do you know about solving a system of non-linear equations?

 

[malonmad]

I know a little but I'm rally just confused by this problem because I have no idea how to approach solving for the value of k. I know that they are tangent when the slopes are equal but so I would have to put the derivitives equal to each other but that's as far as i've gotten.

 

[DrPhil]

I know a little but I'm rally just confused by this problem because I have no idea how to approach solving for the value of k. I know that they are tangent when the slopes are equal but so I would have to put the derivitives equal to each other but that's as far as i've gotten.

Derivatives equal -->
..........$\displaystyle 0.9\ \cos(x) = - k\ \mathrm e^{-x}$
Functions equal -->
..........$\displaystyle 0.9\ \sin(x) = k\ \mathrm e^{-x}$

That is a system of two equations in two unknowns.
Think about that for a bit. How might you combine the equations to eliminate k ?
You could solve for x, then go back for k.

 

[malonmad]

I found that x= -ln(-.9cos(x)/k) is that close to what I should get, and if it is how do I use that to solve for K.

 

[DrPhil]

I found that x= -ln(-.9cos(x)/k) is that close to what I should get, and if it is how do I use that to solve for K.

While true, that statement is not useful because it is still a single equation in two unknowns. You can't solve either equation by itself - you have a system of simultaneous equations.

PLEASE LOOK AT MY PREVIOUS POST
Think about it for a bit. How might you combine the equations to eliminate k ?
You could then solve for x, then go back for k.

 

[malonmad]

would I use .9sin(x) = ke^-x and substitute it into the derivitive equation

like this? .9cos(x) = -(.9sin(x) ) then solve for x?

 

[DrPhil]

would I use .9sin(x) = ke^-x and substitute it into the derivitive equation

like this? .9cos(x) = -(.9sin(x) ) then solve for x?

YES.

For what angle(s) are the magnitudes of sine and cosine equal?

Remember the original qualitative statement that the tangency point will be on the downslope of the first maximum of the sine function. Look for positive sine and negative cosine - that is, in the second quadrant.