Question

What is the slope of the tangent line of `r=theta/3+sin((3theta)/8-(5pi)/3)` at `theta=(7pi)/6`?

Answer 1

`-1.847`

Explanation:

First, let's go ahead and find the values of `r` and `(dr)/(d theta)` at `theta = (7pi)/6`, just to make things easier later.

`r = theta/3 + sin((3theta)/8-(5pi)/3)`

`= (7pi)/18 + sin((21pi)/48 - (5pi)/3)`

`~~ 1.881`

`(dr)/(d theta) = 1/3 + 3/8 cos((3theta)/8 - (5pi)/3)`

`= 1/3 + 3/8cos((21pi)/48 - (5pi)/3)`

`~~ 0.0514`

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Now, the slope of the tangent line at any point is `dy/dx`, but the problem is that we don't have `y` and `x`, we have `r` and `theta`.

Luckily, we can apply a version of the chain rule which states that

`dy/dx = (dy"/"d theta)/(dx"/"d theta)`

We will also have to use the rectangular --> polar coordinate formulas:

`x = rcostheta`
`y = rsintheta`

Since we have expressions for `x` and `y`, we can use the product rule to derive them with respect to `theta`.

`dy/(d theta) = d/(d theta) rsintheta = (dr)/(d theta)sintheta + rcostheta`

`dx/(d theta) = d/(d theta) rcostheta = (dr)/(d theta)costheta -rsintheta`

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With these formulas, we can find the slope of the line at `theta = (7pi)/6`.

`dy/dx = (dy"/"d theta)/(dx"/"d theta) = ((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta)`

We already know that at `theta = (7pi)/6`, `r = 1.881` and `(dr)/(d theta) = 0.0514`. All we need to do now is find the values of `sintheta` and `costheta`, and then plug everything in.

`sin((7pi)/6)=-0.5`
`cos((7pi)/6) = -0.866`

Now we can find our slope:

`((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta) = (0.0514 * (-0.5) + 1.881 * (-0.866))/(0.0514 * (-0.866) - 1.881 * (-0.5))`

`= -1.847`

Final Answer