Find the slope of the tangent line to the given polar curve at the point specified by the value of $\theta$. $r = 4\sin(\theta), \quad \theta = \frac{\pi}{6}$

Neetesh Kumar | January 3, 2025

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This is the solution to Math 1c
Assignment: 10.3 Question Number 18
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Step-by-step solution:

Step 1: Slope formula for polar curves

The slope of the tangent line in polar coordinates is given by:

$\frac{dy}{dx} = \frac{r'\sin(\theta) + r\cos(\theta)}{r'\cos(\theta) - r\sin(\theta)}$

Here:

• $r = 4\sin(\theta)$
• $\theta = \frac{\pi}{6}$

Step 2: Differentiate $r$ with respect to $\theta$

Differentiate $r = 4\sin(\theta)$:

$r' = \frac{dr}{d\theta} = 4\cos(\theta)$

Step 3: Substitute values into the slope formula

Substitute $r = 4\sin(\theta)$, $r' = 4\cos(\theta)$, and $\theta = \frac{\pi}{6}$ into the formula:

1. Compute $r$ at $\theta = \frac{\pi}{6}$:

   $r = 4\sin\left(\frac{\pi}{6}\right) = 4\cdot\frac{1}{2} = 2$

2. Compute $r'$ at $\theta = \frac{\pi}{6}$:

   $r' = 4\cos\left(\frac{\pi}{6}\right) = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3}$

3. Substitute into the formula for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{r'\sin(\theta) + r\cos(\theta)}{r'\cos(\theta) - r\sin(\theta)}$

   Substituting the values:

   $\frac{dy}{dx} = \frac{(2\sqrt{3})\cdot\frac{1}{2} + 2\cdot\frac{\sqrt{3}}{2}}{(2\sqrt{3})\cdot\frac{\sqrt{3}}{2} - 2\cdot\frac{1}{2}}$

Step 4: Simplify the numerator and denominator

1. Simplify the numerator:

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

2. Simplify the denominator:

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

3. Final expression for the slope:

   $\frac{dy}{dx} = \frac{2\sqrt{3}}{2} = \sqrt{3}$

Final Answer:

The slope of the tangent line at $\theta = \frac{\pi}{6}$ is:

$\frac{dy}{dx} = \boxed{\sqrt{3}}$

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