Find the locations where the tangent line to r=e^theta is vertical on [0,2pi]

Question

Find the locations where the tangent line to r=e^theta  is vertical on [0,2pi]

zarkon · Answer

$$x=r\cos(\theta), y=r\sin(\theta)$$

$$\Rightarrow x=e^\theta\cos(\theta), y=e^\theta\sin(\theta)$$


$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$


$$\ldots$$

anonymous · Answer

dr/dt cos(t)-rsin(t) = 0

anonymous · Answer

Sorry, i meant:  dr/dt= f '(t)cos(t) -rsin(t)
                         -sin(t)cos(t)-(e^t)sin(t)=0
Now i need to knwo how to solve.

zarkon · Answer

you should end up with $$\frac{dy}{dx}=\frac{\cos(\theta)+\sin(\theta)}{\cos(\theta)-\sin(\theta)}$$

zarkon · Answer

solve $$\cos(\theta)-\sin(\theta)=0$$

anonymous · Answer

t=0 --> 1-0=0 -->1=0
t=2pi --> 1-0=0 --> 1=0???? Wouldnt it just be 0??

zarkon · Answer

no...you should get $$\theta=\frac{\pi}{4},\frac{5\pi}{4}$$

anonymous · Answer

whoa!? Im not supposed to plug in 0 & 2pi??

zarkon · Answer

$$[0,2\pi]$$ is the interval in which $\theta$ takes its values

anonymous · Answer

But how are u getting those specific values?? Or better yet, im supposed to solve the equation but not plug in the values??

zarkon · Answer

you need to do what I typed above in my first post

zarkon · Answer

then you can find the derivative (dy/dx) like I have in my second post 

then you find where you have vertical slope based on the equation for dy/dx