Find equation of tangent to curve at $t=-\pi$

[ineedhelpnow]

find an equation of the tangent to the curve at the point corresponding to the given value of the parameter $x=t\cos\left({t}\right)$, $y=t\sin\left({t}\right)$, $t=-\pi$

help me

 

[ineedhelpnow]

here's what i have so far:

$dx=cos(t)+tsin(t) dt$
$dy=sin(t)-tcos(t) dt$
$d\pi=-\pi t dt$$\frac{dy}{dx}=\frac{sin(t)-tcos(t)}{cos(t)+tsin(t)}$

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now i plug in $-\pi$ in for t, correct?

 

[MarkFL]

In order to find the tangent line, you need a slope and a point. Let's begin with the point:

\(\displaystyle (x,y)=\left(-\pi\cos(-\pi),-\pi\sin(-\pi)\right)=?\)

Next, compute the slope, using the chain rule:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\)

This will be a function of $t$. Then find:

\(\displaystyle m=\left.\frac{dy}{dx}\right|_{t=-\pi}\)

Now you have your point and your slope, so plug into the point-slope formula...what do you find?

 

[ineedhelpnow]

here's what i did:

so $(x,y)=(-\pi cos(-\pi), -\pi sin(-\pi))=(\pi,0)$

$\frac{dy}{dx}=\pi$

$y=mx+b$
$0=(\pi)(\pi)+b$
$0=\pi^2+b$
$b=-\pi^2$

$y=\pi x-\pi^2$

 

[MarkFL]

You have the correct point, but the wrong slope. Check to make sure you differentiated correctly.

 

[ineedhelpnow]

here's what i have so far:

$dx=cos(t)+tsin(t) dt$
$dy=sin(t)-tcos(t) dt$
$d\pi=-\pi t dt$$\frac{dy}{dx}=\frac{sin(t)-tcos(t)}{cos(t)+tsin(t)}$

- - - Updated - - -

now i plug in $-\pi$ in for t, correct?

is the dy/dx in this step correct?

 

[MarkFL]

is the dy/dx in this step correct?

No, you have not differentiated correctly. Apply the product rule again, and be sure you differentiate the trig. functions correctly. :D

 

[ineedhelpnow]

got the signs wrong

 

[MarkFL]

got the signs wrong

Yes, so what is the slope and then the line?

 

[ineedhelpnow]

$y=-\pi x+ \pi^2$

 

[MarkFL]

$y=-\pi x+ \pi^2$

Yes, that's the one! (Yes)