# What is the slope of #r=tantheta^2-theta# at #theta=(3pi)/8#?

##### 1 Answer

May 11, 2017

Slope is given by

#x=rcostheta# #y=rsintheta#

And to find

#dy/dx=(dy//d theta)/(dx//d theta)=(d/(d theta)(rsintheta))/(d/(d theta)(rcostheta))#

Using the product rule, we can say that:

#dy/dx=((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)#

Since

#(dr)/(d theta)=2tantheta(sec^2theta)-1#

So:

#dy/dx=((2tanthetasec^2theta-1)sintheta+(tan^2theta-theta)costheta)/((2tanthetasec^2theta-1)costheta-(tan^2theta-theta)sin theta)#

Evaluate this at