Teresa Manning

Answered question

2023-03-14

What is the slope of the tangent line of $r=-2\mathrm{sin}\left(3\theta \right)-12\mathrm{cos}\left(\frac{\theta }{2}\right)$ at $\theta =\frac{-\pi }{3}$?

Answer & Explanation

Gregory Ferguson

Beginner2023-03-15Added 3 answers

The slope of the tangent line of r at $\theta =\frac{-\pi }{3}$, is the same as the derivative of the function for that exact x-value. Therefore, we need to differentiate both sides of the function, so we get an expression for $\frac{d}{d\theta }$ r.
$\frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }\left[-2\mathrm{sin}\left(3\theta \right)-12\mathrm{cos}\left(\frac{\theta }{2}\right)\right]$
$\frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }\left[-2\mathrm{sin}\left(3\theta \right)\right]-\frac{d}{d\theta }\left[-12\mathrm{cos}\left(\frac{\theta }{2}\right)\right]$
From this point, we need to know how to differentiate $\mathrm{sin}\theta$ and $\mathrm{cos}\theta$:
$\frac{d}{d\theta }\mathrm{sin}\theta =\mathrm{cos}\theta$
$\frac{d}{d\theta }\mathrm{cos}\theta =-\mathrm{sin}\theta$
We need to know how the chain rule works as well. In this case, we have to substitute $3\theta$ with u and $\frac{\theta }{2}$ with v. When we then differentiate, we have to also differentiate our substitute.
We then have:
$\frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }\left[-2\mathrm{sin}\left(u\right)\right]-\frac{d}{d\theta }\left[12\mathrm{cos}\left(v\right)\right]$
$\frac{d}{d\theta }\left[r\right]=-2\mathrm{cos}\left(u\right)\cdot \frac{d}{d\theta }\left[u\right]-\left(-12\mathrm{sin}\left(v\right)\cdot \frac{d}{d\theta }\left[v\right]\right)$
Let's now substitute u and v back to our original functions.
$\frac{d}{d\theta }\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot \frac{d}{d\theta }\left[3\theta \right]-\left(-12\mathrm{sin}\left(\frac{\theta }{2}\right)\cdot \frac{d}{d\theta }\left[\frac{\theta }{2}\right]\right)$
$\frac{d}{d\theta }\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot 3+12\mathrm{sin}\left(\frac{\theta }{2}\right)\cdot \frac{1}{2}=-6\mathrm{cos}\left(3\theta \right)+6\mathrm{sin}\left(\frac{\theta }{2}\right)$
Now we have an expression for $\frac{d}{d\theta }r\left(\theta \right)$, where we can put in any values we want, and get the slope for whatever $\theta$-value we want. Thus, let's put in $\theta =\frac{-\pi }{3}$:
$\frac{d}{d\theta }\left[r\left(\frac{-\pi }{3}\right)\right]$
$=-6\mathrm{cos}\left(-\pi \right)+6\mathrm{sin}\left(-\frac{\pi }{6}\right)$
$\mathrm{cos}\left(-\pi \right)=-1$ and $\mathrm{sin}\left(\frac{-\pi }{6}\right)=-\frac{1}{2}$
This gives us that the slope, for $\theta =\left(-\frac{\pi }{3}\right)$, is:
$-6\left(-1\right)+6\cdot \left(-\frac{1}{2}\right)=6-3=3$

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