Let $r$ be the polar function $r( heta)=\sin (6 heta)+ heta$. Here is its graph for $0 \leq heta \leq 2 \pi$ :

What is the rate of change of the $y$-coordinate with respect to $ heta$ at the point $P$ ?

Choose 1 answer:
(A) -5
(B) $\frac{3 \pi}{2}$
(C) 5
(D) $-\frac{3 \pi}{2}$

Question

Let $r$ be the polar function $r(	heta)=\sin (6 	heta)+	heta$. Here is its graph for $0 \leq 	heta \leq 2 \pi$ :

What is the rate of change of the $y$-coordinate with respect to $	heta$ at the point $P$ ?

Choose 1 answer:
(A) -5
(B) $\frac{3 \pi}{2}$
(C) 5
(D) $-\frac{3 \pi}{2}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the polar function $r( heta) = \sin(6 heta) + heta$, we need to find the rate of change of the $y$-coordinate with respect to $ heta$ at the point $P(0, -5)$. ‖ ‖ step 2 ⋮ The $y$-coordinate in polar coordinates is given by $y = r( heta) \sin( heta)$. ‖ ‖ step 3 ⋮ To find the rate of change of $y$ with respect to $ heta$, we need to compute $\frac{dy}{d heta}$. ‖ ‖ step 4 ⋮ Using the product rule, $\frac{dy}{d heta} = \frac{d}{d heta} [r( heta) \sin( heta)] = r'( heta) \sin( heta) + r( heta) \cos( heta)$. ‖ ‖ step 5 ⋮ First, compute $r'( heta) = \frac{d}{d heta} [\sin(6 heta) + heta] = 6\cos(6 heta) + 1$. ‖ ‖ step 6 ⋮ At the point $P(0, -5)$, we need to find the corresponding $ heta$. Since $r( heta) = 0$ at $P$, we solve $\sin(6 heta) + heta = 0$. ‖ ‖ step 7 ⋮ For $ heta = -5$, we have $r(-5) = \sin(6(-5)) + (-5) = \sin(-30) - 5 = 0$. ‖ ‖ step 8 ⋮ Now, compute $\frac{dy}{d heta}$ at $ heta = -5$: ‖ ‖ step 9 ⋮ $\frac{dy}{d heta} = (6\cos(6(-5)) + 1) \sin(-5) + (\sin(6(-5)) + (-5)) \cos(-5)$. ‖ ‖ step 10 ⋮ Simplifying, $\frac{dy}{d heta} = (6\cos(-30) + 1) \sin(-5) + (0) \cos(-5) = 6\cos(-30) \sin(-5) + \sin(-5)$. ‖ ‖ step 11 ⋮ Since $\cos(-30) = \cos(30)$ and $\sin(-5) = -\sin(5)$, we get $\frac{dy}{d heta} = 6\cos(30)(-\sin(5)) + (-\sin(5)) = -6\cos(30)\sin(5) - \sin(5)$. ‖ ‖ step 12 ⋮ Factoring out $-\sin(5)$, we get $\frac{dy}{d heta} = -\sin(5)(6\cos(30) + 1)$. ‖ ‖ step 13 ⋮ Since $\cos(30) = \frac{\sqrt{3}}{2}$, we get $\frac{dy}{d heta} = -\sin(5)(6 \cdot \frac{\sqrt{3}}{2} + 1) = -\sin(5)(3\sqrt{3} + 1)$. ‖ ‖ step 14 ⋮ Therefore, the rate of change of the $y$-coordinate with respect to $ heta$ 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 $ heta$ in polar coordinates involves differentiating the product of the polar function $r( heta)$ and $\sin( heta)$ using the product rule.⚹◊What is the derivative of the polar function $r( heta) = \sin(6 heta) + heta$ with respect to $ heta$?⍭ Generate me a similar question◊

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