Question

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

Answer 1

The slope, `m ~~ 2.7`

Explanation:

Slope of a tangent to a polar curve

From the reference we have the equation:

`dy/dx = ((dr(theta))/(d theta)sin(theta) + r(theta)cos(theta))/((dr(theta))/(d theta)cos(theta) - r(theta)sin(theta))`

We are given `r(theta)`:

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

Use WolframAlpha to compute `(dr(theta))/(d theta)`:

`(dr(theta))/(d theta) = 2 θ+3 sin(θ) cos^2(θ)+2 tan(θ) sec^2(θ)-1`

Use WolframAlpha to evaluate `(dr(pi/3))/(d theta) ~~ 15.6`

Evaluate `r(pi/3) ~~ 2.9`

When we evaluate the remaining parts of `dy/dx` at `pi/3`, we obtain the slope, m, of the tangent line.

`m = (15.6sin(pi/3) + 2.9cos(pi/3))/(15.6cos(pi/3) - 2.9sin(pi/3))`

`m ~~ 2.7`