Question

What is the slope of the polar curve `f(theta) = theta^2-theta - cos^3theta + tan^2theta` at `theta = pi/4`?

Answer 1

The slope is: `(pi^2+4pi+8sqrt2+64)/(pi^2+12pi+16sqrt2+32)approx0.9564910`

Explanation:

The slope will be given by `dy/dx` of the curve evaluated at `theta=pi/4`.

To get the equation in terms of `x` and `y`, we should use the polar-to-rectangular identities: `{(x=rcostheta),(y=rsintheta):}`

So here, where `r=f(theta)`, we have:

`x=costheta(theta^2-theta-cos^3theta+tan^2theta)`

`color(white)x=theta^2costheta-thetacostheta-cos^4theta+tanthetasintheta`

`y=sintheta(theta^2-theta-cos^3theta+tan^2theta)`

`color(white)y=theta^2sintheta-thetasintheta-cos^3thetasintheta+sinthetatan^2theta`

To find `dy/dx`, we first need to find `dx/(d theta)` and `dy/(d theta)`. This will take a fair amount of product and chain rule, so be careful.

`dx/(d theta)=2thetacostheta-theta^2sintheta-costheta+thetasintheta+4cos^3thetasintheta+sec^2thetasintheta+tanthetacostheta`

`color(white)(dx/(d theta))=2thetacostheta-theta^2sintheta-costheta+thetasintheta+4cos^3thetasintheta+tanthetasectheta+sintheta`

`dy/(d theta)=2thetasintheta+theta^2costheta-sintheta-thetacostheta+3cos^2thetasin^2theta-cos^4theta+costhetatan^2theta+2sinthetatanthetasec^2theta`

`color(white)(dy/(d theta))=2thetasintheta+theta^2costheta-sintheta-thetacostheta+3cos^2thetasin^2theta-cos^4theta+sinthetatantheta+2tan^2thetasectheta`

Now we can find `dy/dx` using `dy/dx=(dy//d theta)/(dx//d theta)`.

Just so we don't have to write out a giant fraction, let's evaluate `dx/(d theta)` and `dy/(d theta)` at `theta=pi/4` separately.

`dx/(d theta)=2 pi/4 1/sqrt2+pi^2/16 1/sqrt2-1/sqrt2+pi/4 1/sqrt2+4 1/(2sqrt2)1/sqrt2+2 1/sqrt2+1 1/sqrt2`

`color(white)(dx/(d theta))=pi/(2sqrt2)+pi^2/(16sqrt2)+pi/(4sqrt2)+1+sqrt2`

`color(white)(dx/(d theta))=1/(16sqrt2)(pi^2+12pi+16sqrt2+32)`

`dy/(d theta)=2 pi/4 1/sqrt2+pi^2/16 1/sqrt2-1/sqrt2-pi/4 1/sqrt2+3 1/2 1/2-1/4+1/sqrt2 1+2*1sqrt2`

`color(white)(dy/(d theta))=pi/(2sqrt2)+pi^2/(16sqrt2)-pi/(4sqrt2)+1/2+2sqrt2`

`color(white)(dy/(d theta))=1/(16sqrt2)(pi^2+4pi+8sqrt2+64)`

Thus, at `theta=pi/4`, the slope is:

`dy/dx=(dy//d theta)/(dx//d theta)=(pi^2+4pi+8sqrt2+64)/(pi^2+12pi+16sqrt2+32)approx0.9564910`