Question

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

Answer 1

`(pi-4)/(pi+4)`

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 `dy/dx`.

To do this, we need to find the use the relationship between `r,theta` and `x,y`. Recall that `x=rcostheta` and `y=rsintheta`.

Thus, we can do:

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

Since we have `r=theta^-1`, we get:

`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 `theta^2/theta^2`:

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

Recalling that `sin(pi/4)=cos(pi/4)=1/sqrt2`, we see that the slope at `theta=pi/4` the slope is:

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

Multiply through by `(4sqrt2)/(4sqrt2)`:

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