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Given $y\sin(8x) = x\cos(2y)$ find the tangent line at the point ($\pi\over2$, $\pi\over4$).

I got $y = 2x - 2.36$, but my teacher wants a fraction. Can somebody help me get the answer in fraction form?

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$2.36=236/100 $. Is that good enough? –  Pedro Tamaroff Oct 19 '12 at 22:40
    
I'm doing the problem in WebAssign and it says I got it wrong when I entered y = 2x - 2.36 and I don't know why. I don't think she wants 236/100. –  dsta Oct 19 '12 at 22:43
    
$2.36 \approx 3\pi/4$. This is most likely the fraction they're after. –  Arthur Oct 19 '12 at 22:46
    
Ok I will try that, thank you. –  dsta Oct 19 '12 at 22:46
    
Nope, I still got it wrong. I am locked out now so I'll ask my teacher on Monday. Thanks everyone for your help. –  dsta Oct 19 '12 at 22:47

4 Answers 4

Differentiate (implicitly). We get $$8y\cos(8x)+\sin(8x)y'=-2x\sin(2y)y'+\cos(2y).$$ When we substitute the given values of $x$ and $y$, the numbers become very simple, since $\cos 2y=0$ and $\sin 8x=0$. We get that at our target point, $$2\pi =-\pi y',$$ and our slope is $-2$. The tangent line therefore has equation $$y-\frac{\pi}{4}=-2\left(x-\frac{\pi}{2}\right).$$ This simplifies to $y=-2x +\dfrac{5\pi}{4}.$

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Differentiating implicitly the equation $\,y\sin 8x=x\cos 2y\,$ ,we get

$$\sin 8x\,dy+8y\cos 8x\,dx=\cos 2y\,dx-2x\sin 2y\,dy\Longrightarrow$$

$$(\sin 8x+2x\sin 2y)dx=(\cos 2y-8y\cos 8x)dx\Longrightarrow$$

$$\frac{dy}{dx}=\frac{\cos 2y-8y\cos 8x}{\sin 8x+2x\sin 2y}\Longrightarrow$$

$$\left.\frac{dy}{dx}\right|_{\left(\frac{\pi}{2},\frac{\pi}{4}\right)}=\frac{0-2\pi}{0+\pi}=-2$$

Thus, the tangent line's given by

$$y-\frac{\pi}{4}=-2\left(x-\frac{\pi}{2}\right)\Longleftrightarrow y=-2x+\frac{5\pi}{4}$$

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You have $$y\sin(8x)=x\cos(2y)$$

Then

$$y'\sin(8x)+8y \cos(8x)=\cos(2y)-2y'x\sin(2y)$$

Thus

$$y'(\sin(8x)+2x\sin(2y))=\cos(2y)-8y\cos(8x)$$

$$y'=\frac{\cos(2y)-8y\cos(8x)}{\sin(8x)+2x\sin(2y)}$$

We look at $$(\pi/2,\pi/4)$$

$$y'_{(\pi/2,\pi/4)}=\frac{\cos(2 \pi/4)-8 \pi/4\cos(8 \pi/2)}{\sin(8 \pi/2)+2 \pi/2\sin(2 \pi/4)}$$

$$\eqalign{ & {{y'}_{(\pi /2,\pi /4)}} = \frac{{\cos (\pi /2) - 2\pi \cos (4\pi )}}{{\sin (4\pi ) + \pi \sin (\pi /2)}} \cr & {{y'}_{(\pi /2,\pi /4)}} = \frac{{0 - 2\pi }}{{0 + \pi }} = - 2 \cr} $$

Since $y(\pi/2)=\pi/4$, we get

$$y_T=-2(x-\pi/2)+\pi/4$$

$$y_T=-2x+\pi+\pi/4=-2x+5\pi/4$$

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General and short formula:

If $F(x,y)=0$ defines $y$ as a function of $x$ implicity, then we always have $y'=\frac{-F_x}{F_y}$.

Here we have $y\sin(8x) = x\cos(2y)$ so $$F(x,y)=y\sin(8x) - x\cos(2y)=0$$ and then $$F_x=8y\cos(8x)-\cos(2y),~~~F_y=\sin(8x)+x\sin(2y)$$

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