Question

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

Answer 1

Slope is given by `dy/dx`. However, this is not explicitly defined when we have a polar function. We have to use the identities for `x` and `y`:

• `x=rcostheta`
• `y=rsintheta`

And to find `dy/dx`, we need to see that:

`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 `r=tan^2theta-theta`, we see that:

`(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 `theta=(3pi)/8` for the slope.