Consider the polar curve $r=4 \sin (3 heta)$.
What is the slope of the tangent line to the curve $r$ when $ heta=\frac{\pi}{3}$ ? Give an exact expression.
slope $=$ $\square$

Question

Consider the polar curve $r=4 \sin (3 	heta)$.
What is the slope of the tangent line to the curve $r$ when $	heta=\frac{\pi}{3}$ ? Give an exact expression.
slope $=$ $\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the slope of the tangent line to the polar curve $r=4\sin(3 heta)$ at $ heta=\frac{\pi}{3}$, we first differentiate $r$ with respect to $ heta$. ‖ ‖ step 2 ⋮ The derivative of $r$ with respect to $ heta$ is given by $\frac{dr}{d heta} = \frac{d}{d heta}(4\sin(3 heta))$. ‖ ‖ step 3 ⋮ Applying the chain rule, $\frac{dr}{d heta} = 4 \cdot 3\cos(3 heta) = 12\cos(3 heta)$. ‖ ‖ step 4 ⋮ Evaluating this derivative at $ heta=\frac{\pi}{3}$, we get $\frac{dr}{d heta}\bigg|_{ heta=\frac{\pi}{3}} = 12\cos(\pi) = -12$. ‖ ‖ step 5 ⋮ The slope of the tangent line in polar coordinates is given by $\frac{dy}{dx} = \frac{\frac{dr}{d heta}\sin( heta) + r\cos( heta)}{\frac{dr}{d heta}\cos( heta) - r\sin( heta)}$. ‖ ‖ step 6 ⋮ Substituting $r=4\sin(3 heta)$ and $\frac{dr}{d heta}=-12$ into the slope formula, we get $\frac{dy}{dx} = \frac{-12\sin(\frac{\pi}{3}) + 4\sin(\pi)}{-12\cos(\frac{\pi}{3}) - 4\sin(\pi)}$. ‖ ‖ step 7 ⋮ Simplifying the expression, we find $\frac{dy}{dx} = \frac{-12\cdot\frac{\sqrt{3}}{2}}{-12\cdot\frac{1}{2}} = \frac{-6\sqrt{3}}{-6} = \sqrt{3}$. ‖ ∻Answer∻ ⚹ $\sqrt{3}$ ⚹ ∻Key Concept∻ ⚹Finding the slope of a tangent line to a polar curve at a given angle⚹ ∻Explanation∻ ⚹The slope of the tangent line to a polar curve at a given angle is found by differentiating the polar equation with respect to $ heta$, evaluating at the given angle, and then using the formula for the slope in polar coordinates.⚹◊What is the formula to determine the slope of a polar curve at a given point?⍭ Generate me a similar question◊

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