Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

2 Answers 2

up vote 2 down vote accepted

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

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

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.

share|improve this answer
    
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!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.