Question

Math

Posted 5 months ago

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

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To find the slope of the tangent line to the polar curve $r=4\sin(3\theta)$ at $\theta=\frac{\pi}{3}$, we first differentiate $r$ with respect to $\theta$

step 2

The derivative of $r$ with respect to $\theta$ is given by $\frac{dr}{d\theta} = \frac{d}{d\theta}(4\sin(3\theta))$

step 3

Applying the chain rule, $\frac{dr}{d\theta} = 4 \cdot 3\cos(3\theta) = 12\cos(3\theta)$

step 4

Evaluating this derivative at $\theta=\frac{\pi}{3}$, we get $\frac{dr}{d\theta}\bigg|_{\theta=\frac{\pi}{3}} = 12\cos(\pi) = -12$

step 5

The slope of the tangent line in polar coordinates is given by $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}$

step 6

Substituting $r=4\sin(3\theta)$ and $\frac{dr}{d\theta}=-12$ 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 $\theta$, evaluating at the given angle, and then using the formula for the slope in polar coordinates.

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