Question

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

Answer 1

\[x=r\cos(\theta), y=r\sin(\theta)\]

\[\Rightarrow x=e^\theta\cos(\theta), y=e^\theta\sin(\theta)\]


\[\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]


\[\ldots\]

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

solve \[\cos(\theta)-\sin(\theta)=0\]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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