# Find the acute angle of intersection of the curves $y=\cos x$ and $y=e^{-x}$ at the point $(0,1)$.

Find the acute angle of intersection of the curves $y=\cos x$ and $y=e^{-x}$ at the point $(0,1)$.

My method:
$y=\cos(x)$ $(0,1)$
$1=\cos(0)$
$=0$

$\frac{dy}{dx}=-\sin(x)$
$=-\sin(0)$ $=0$

I did the above step exactly from the example given in the text book, but I can't get the answer.
The answer is $45^{\circ}$

Help me out by step by step solution. thanks

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$\sin 0$ is not 1! –  Host-website-on-iPage Jul 15 '12 at 11:36

$$f(x)=\cos x\Longrightarrow f'(x)=-\sin x\Longrightarrow f'(0)=0=:m_1$$

$$g(x)=e^{-x}\Longrightarrow g'(x)=-e^{-x}\Longrightarrow g'(0)=-1=:m_2$$

So you have that the functions' tangent lines at $\,(0,1)\,$ have slopes $\,0\,$ (this means the tangent line of $\,f\,$ at this point is horizontal) and $\,-1\,$ , so what's the acute angle between two lines with these slopes?

Yup, it is $\,45^\circ\,$ , as you can readily check. Of course, you can use the formula

$$\tan\alpha = \arctan\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\arctan\frac{1}{1}=\frac{\pi}{4}\text{radians}=45^\circ$$with $\,\alpha\,=$ the angle between the curves.

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why is $45^{\circ}$, it is because $tan(1)=45^{\circ}$? thx –  Sb Sangpi Jul 15 '12 at 11:56
Yes, of course. Rememer that $$\sin 45^\circ=\cos 45^\circ=\frac{\sqrt 2}{2}\Longrightarrow \tan 45^\circ=1$$ –  DonAntonio Jul 15 '12 at 12:00
sorry, how come $\frac{\pi}{4}$ ? thx –  Sb Sangpi Jul 15 '12 at 12:22
because If i have $-2$ how about that? –  Sb Sangpi Jul 15 '12 at 12:31
$\,\pi/4\,$ is the measure of an angle of $\,45^\circ\,$ in radians. If you haven't studied yet this forget it. About your question "if I have -2 how about that?" I'm not sure I understand, but you can always evaluate $\,\arctan -2\,$... –  DonAntonio Jul 15 '12 at 12:35

$\frac{dy}{dx}$ at $P$, where is the slope of the tangent to the curve in concern, at $P$.

For $y=\cos x$ you were almost there. $\sin 0$ is correct, but it's equal to 0.

For the other one, however, $\frac{dy}{dx}=-e^{-x}$ and that's $-e^0=-1$ so take tan inverse of the answers ($0$ and $-1$) and take the difference. i.e., $0^o$ and $-45^o$. It gives you the answer!

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