To find the slope of the tangent line to the polar curve $r = 1 - \sin(4\theta)$ at $\theta = \frac{\pi}{4}$, we need to calculate $\frac{dy}{dx}$ in polar coordinates.

For polar curves, the slope of the tangent line (i.e., $\frac{dy}{dx}$) 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 1: Compute $\frac{dr}{d\theta}$

Given: $r = 1 - \sin(4\theta)$

Differentiate $r$ with respect to $\theta$: $\frac{dr}{d\theta} = -4 \cos(4\theta)$

Step 2: Substitute $r$ and $\frac{dr}{d\theta}$ at $\theta = \frac{\pi}{4}$

1. Calculate $r$ at $\theta = \frac{\pi}{4}$: $r = 1 - \sin\left(4 \cdot \frac{\pi}{4}\right) = 1 - \sin(\pi) = 1 - 0 = 1$

2. Calculate $\frac{dr}{d\theta}$ at $\theta = \frac{\pi}{4}$: $\frac{dr}{d\theta} = -4 \cos\left(4 \cdot \frac{\pi}{4}\right) = -4 \cos(\pi) = -4 \cdot (-1) = 4$

Step 3: Substitute values into $\frac{dy}{dx}$ formula

Now, using: $\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}$

Substitute $\theta = \frac{\pi}{4}$, $r = 1$, and $\frac{dr}{d\theta} = 4$:

• $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Substitute these into the formula:

$\frac{dy}{dx} = \frac{4 \cdot \frac{\sqrt{2}}{2} + 1 \cdot \frac{\sqrt{2}}{2}}{4 \cdot \frac{\sqrt{2}}{2} - 1 \cdot \frac{\sqrt{2}}{2}}$

Simplify each term:

1. The numerator becomes: $4 \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2} + \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}$

2. The denominator becomes: $4 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 2\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$

Thus:

$\frac{dy}{dx} = \frac{\frac{5\sqrt{2}}{2}}{\frac{3\sqrt{2}}{2}} = \frac{5}{3}$

Final Answer

The slope of the tangent line to the polar curve at $\theta = \frac{\pi}{4}$ is:

$\frac{dy}{dx} = \frac{5}{3}$

Would you like further explanation on any part of this solution?

Follow-up Questions

1. How would the formula change if we wanted to find the slope for a different polar curve?
2. What is the significance of the slope of the tangent line in the context of polar coordinates?
3. How is $\frac{dy}{dx}$ in polar coordinates related to rectangular coordinates?
4. Can we use this method for finding slopes on curves defined parametrically?
5. How does symmetry affect the slope of the tangent line in polar curves?

Tip

When solving for slopes in polar coordinates, remember to convert $\sin\theta$ and $\cos\theta$ based on the specific angle given, as they often simplify the computation.