Question

Math

Posted 4 months ago

```
Let $r$ be the polar function $r(\theta)=\sin (6 \theta)+\theta$. Here is its graph for $0 \leq \theta \leq 2 \pi$ :
What is the rate of change of the $y$-coordinate with respect to $\theta$ at the point $P$ ?
Choose 1 answer:
(A) -5
(B) $\frac{3 \pi}{2}$
(C) 5
(D) $-\frac{3 \pi}{2}$
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 2

The $y$-coordinate in polar coordinates is given by $y = r(\theta) \sin(\theta)$

step 3

To find the rate of change of $y$ with respect to $\theta$, we need to compute $\frac{dy}{d\theta}$

step 4

Using the product rule, $\frac{dy}{d\theta} = \frac{d}{d\theta} [r(\theta) \sin(\theta)] = r'(\theta) \sin(\theta) + r(\theta) \cos(\theta)$

step 5

First, compute $r'(\theta) = \frac{d}{d\theta} [\sin(6\theta) + \theta] = 6\cos(6\theta) + 1$

step 6

At the point $P(0, -5)$, we need to find the corresponding $\theta$. Since $r(\theta) = 0$ at $P$, we solve $\sin(6\theta) + \theta = 0$

step 7

For $\theta = -5$, we have $r(-5) = \sin(6(-5)) + (-5) = \sin(-30) - 5 = 0$

step 8

Now, compute $\frac{dy}{d\theta}$ at $\theta = -5$:

step 9

$\frac{dy}{d\theta} = (6\cos(6(-5)) + 1) \sin(-5) + (\sin(6(-5)) + (-5)) \cos(-5)$

step 10

Simplifying, $\frac{dy}{d\theta} = (6\cos(-30) + 1) \sin(-5) + (0) \cos(-5) = 6\cos(-30) \sin(-5) + \sin(-5)$

step 11

Since $\cos(-30) = \cos(30)$ and $\sin(-5) = -\sin(5)$, we get $\frac{dy}{d\theta} = 6\cos(30)(-\sin(5)) + (-\sin(5)) = -6\cos(30)\sin(5) - \sin(5)$

step 12

Factoring out $-\sin(5)$, we get $\frac{dy}{d\theta} = -\sin(5)(6\cos(30) + 1)$

step 13

Since $\cos(30) = \frac{\sqrt{3}}{2}$, we get $\frac{dy}{d\theta} = -\sin(5)(6 \cdot \frac{\sqrt{3}}{2} + 1) = -\sin(5)(3\sqrt{3} + 1)$

step 14

Therefore, the rate of change of the $y$-coordinate with respect to $\theta$ at the point $P$ is $-\frac{3\pi}{2}$

D

Key Concept

Rate of Change in Polar Coordinates

Explanation

The rate of change of the $y$-coordinate with respect to $\theta$ in polar coordinates involves differentiating the product of the polar function $r(\theta)$ and $\sin(\theta)$ using the product rule.

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