# What is the slope of the tangent line of #r=(sin^2theta)/(theta-thetacos^2theta)# at #theta=(pi)/4#?

##### 1 Answer

#### Explanation:

First, note that we can simplify the equation:

#r=sin^2theta/(theta-thetacos^2theta)=sin^2theta/(theta(1-cos^2theta))=sin^2theta/(thetasin^2theta)=1/theta#

The slope of the equation is given by

To do this, we need to find the use the relationship between

Thus, we can do:

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

Since we have

#dy/dx=(d/(d theta)theta^-1sintheta)/(d/(d theta)theta^-1costheta)#

Using the product rule:

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

Multiply through by

#dy/dx=(-sintheta+thetacostheta)/(-costheta-thetasintheta)#

Recalling that

#dy/dx=(-1/sqrt2+pi/(4sqrt2))/(-1/sqrt2-pi/(4sqrt2))#

Multiply through by

#dy/dx=(-4+pi)/(-4-pi)=(pi-4)/(pi+4)#