## Calculus 8th Edition

$y=\pi x +\pi ^2$
Given: $$x=t\cos t$$ and $$y=t\sin t.$$ As we know the slope of this line is the derivative of the curve at the point where $t=\pi$, and because this line is tangent to the curve at the point where $t=\pi$, we know the point where $t=\pi$ lies on this line. So, let us first find the derivative at $t=\pi$. The curve is given by the parametric equations $$x=t\cos t$$ and $$y=t\sin t.$$ So, the derivative at $t=\pi$ is $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}},$$ evaluated at $t=\pi$ So, we find the derivatives of each of the parametric equations with respect to $t$. $$\frac{dx}{dt}=\frac{dt}{dt} \cos t+ t\frac{d}{dt}(\cos t)=1(\cos t) +t(-\sin t)=\cos t-t\sin t$$ and $$\frac{dy}{dt}=\frac{dt}{dt} \sin t+ t\frac{d}{dt}(\sin t)=1(\sin t) +t(\cos t)=\sin t+t\cos t.$$ Then $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\sin t+t\cos t}{\cos t-t\sin t}.$$ Next, we plug in $t=\pi$ to get $$\frac{\sin \pi+\pi\cos \pi}{\cos \pi-\pi\sin \pi}=\frac{0+\pi (-1)}{-1-\pi (0)}=\frac{-\pi }{-1}=\pi .$$ Thus, the slope of our line is $\pi$. Now, to find the point that lies on the line, we plug $t=\pi$ into our parametric equations. We get $$x=t\cos t=\pi \cos \pi=\pi (-1)=-\pi$$ and $$y=t\sin t=\pi sin \pi =\pi (0)=0.$$ Thus, our point is $(-\pi , 0)$. Finally, we are ready to find the equation of our line. We will use the formula $$y-y_{1}=m(x-x_{1}),$$ where $m$ is the slope and $(x_{1},y_{1})$ is the point that lies on the line. For us, $m=\pi$ and $(x_{1},y_{1})=(-\pi , 0)$ We plug these into the equation to get $$y-0=\pi (x+\pi ).$$ Thus, we have $$y=\pi (x+\pi ).$$ Distributing the $\pi$ on the right, we get $y=\pi x +\pi ^2$