# Thread: Tangent to each other

1. ## Tangent to each other

Im working on a problem and cant seem to figure out how to do it.

The problem is:
Prove the curves y = e^(-x) and y = (e^(-x))(cosx) are tangents to each other at all intersection points.

I did some work and found they both intersect when x = 0 but im not sure how to prove they are tangents to each other at this point.

2. Hello

In fact, there are (much) more than one point of intersection :
if $(x,y)$ is a point of intersection, $\exp -x = \cos x \exp -x \Leftrightarrow (1-\cos x)\exp-x=0\Leftrightarrow x=\ldots$

(you should sketch the graph of the functions, it'd help you to see this)

To show that two tangents are parallel, you only need to show that they have the same slope.

3. Differentiate both then prove that they are the same when they intersect at any given point $x$.

Let's find where they intersect;

$y~=~e^{-x}~,~y~=~(e^{-x})(cos~x)$

$\implies e^{-x}~=~(e^{-x})(cos~x)$
$\implies (cos~x)~=~1$

Now differentiate

$y~=~e^{-x}$

$\frac{dy}{dx}~=~-e^{-x}$

$y~=~(e^{-x})(cos~x)$

Let $u~=~e^{-x}~,~v~=~cos~x$

$\frac{du}{dx}~=~-e^{-x}~,~\frac{dv}{dx}~=~-sin~x$

$\implies \frac{dy}{dx}~=~-(e^{-x})(sin~x)-(e^{-x})(cos~x)$

$=~-(e^{-x})(sin~x+cos~x)$

If $~~(cos~x)~=~1~~$ then $(cos~x+sin~x)~=~1$

hence $~\frac{dy}{dx}~=~-(e^{-x})~$ which is the same as the gradient of the other curve therefore the 2 are tangents at each other when they intersect at any given point $x$

4. Thanks for the quick replies! I was up to you until you said:

if cosx = 1
then cosx+sinx = 1

Math isnt my strength so am I missing an identity or just not seeing something.

5. Oh sorry I forgot that x = 0 therefore sinx = 0.

I do have a question though. flyingsquirrel said that there are more than point of intersection, im not too sure on how I can algebraically find out more than one point?

6. If $\cos x=1,\,x=2k\pi$ with $k\in \mathbb{Z}$ and as $\sin(2k\pi)=0,\,\cos(2k\pi)+\sin(2k\pi)=1$

It seem you have forgotten that sine and cosine are $2\pi$-periodic ?