Teresa Manning

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 $\theta =\frac{-\pi}{3}$?

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.

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(3\theta \right)-12\mathrm{cos}\left(\frac{\theta}{2}\right)]$

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(3\theta \right)]-\frac{d}{d\theta}[-12\mathrm{cos}\left(\frac{\theta}{2}\right)]$

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:

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(u\right)]-\frac{d}{d\theta}\left[12\mathrm{cos}\left(v\right)\right]$

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(u\right)\cdot \frac{d}{d\theta}\left[u\right]-(-12\mathrm{sin}\left(v\right)\cdot \frac{d}{d\theta}\left[v\right])$

Let's now substitute u and v back to our original functions.

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot \frac{d}{d\theta}\left[3\theta \right]-(-12\mathrm{sin}\left(\frac{\theta}{2}\right)\cdot \frac{d}{d\theta}\left[\frac{\theta}{2}\right])$

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot 3+12\mathrm{sin}\left(\frac{\theta}{2}\right)\cdot \frac{1}{2}=-6\mathrm{cos}\left(3\theta \right)+6\mathrm{sin}\left(\frac{\theta}{2}\right)$

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 $\theta =\frac{-\pi}{3}$:

$\frac{d}{d\theta}\left[r\left(\frac{-\pi}{3}\right)\right]$

$=-6\mathrm{cos}(-\pi )+6\mathrm{sin}(-\frac{\pi}{6})$

$\mathrm{cos}(-\pi )=-1$ and $\mathrm{sin}\left(\frac{-\pi}{6}\right)=-\frac{1}{2}$

This gives us that the slope, for $\theta =(-\frac{\pi}{3})$, is:

$-6(-1)+6\cdot (-\frac{1}{2})=6-3=3$

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(3\theta \right)-12\mathrm{cos}\left(\frac{\theta}{2}\right)]$

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(3\theta \right)]-\frac{d}{d\theta}[-12\mathrm{cos}\left(\frac{\theta}{2}\right)]$

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:

$\frac{d}{d\theta}\left[r\right]=\frac{d}{d\theta}[-2\mathrm{sin}\left(u\right)]-\frac{d}{d\theta}\left[12\mathrm{cos}\left(v\right)\right]$

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(u\right)\cdot \frac{d}{d\theta}\left[u\right]-(-12\mathrm{sin}\left(v\right)\cdot \frac{d}{d\theta}\left[v\right])$

Let's now substitute u and v back to our original functions.

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot \frac{d}{d\theta}\left[3\theta \right]-(-12\mathrm{sin}\left(\frac{\theta}{2}\right)\cdot \frac{d}{d\theta}\left[\frac{\theta}{2}\right])$

$\frac{d}{d\theta}\left[r\right]=-2\mathrm{cos}\left(3\theta \right)\cdot 3+12\mathrm{sin}\left(\frac{\theta}{2}\right)\cdot \frac{1}{2}=-6\mathrm{cos}\left(3\theta \right)+6\mathrm{sin}\left(\frac{\theta}{2}\right)$

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 $\theta =\frac{-\pi}{3}$:

$\frac{d}{d\theta}\left[r\left(\frac{-\pi}{3}\right)\right]$

$=-6\mathrm{cos}(-\pi )+6\mathrm{sin}(-\frac{\pi}{6})$

$\mathrm{cos}(-\pi )=-1$ and $\mathrm{sin}\left(\frac{-\pi}{6}\right)=-\frac{1}{2}$

This gives us that the slope, for $\theta =(-\frac{\pi}{3})$, is:

$-6(-1)+6\cdot (-\frac{1}{2})=6-3=3$

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