Math Problem Statement

Find the slope of the tangent line to the polar curve r = 1 - sin(4θ) at θ = π/4.

Solution

To find the slope of the tangent line to the polar curve r=1sin(4θ)r = 1 - \sin(4\theta) at θ=π4\theta = \frac{\pi}{4}, we need to calculate dydx\frac{dy}{dx} in polar coordinates.

For polar curves, the slope of the tangent line (i.e., dydx\frac{dy}{dx}) is given by:

dydx=drdθsinθ+rcosθdrdθcosθrsinθ\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 drdθ\frac{dr}{d\theta}

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

Differentiate rr with respect to θ\theta: drdθ=4cos(4θ)\frac{dr}{d\theta} = -4 \cos(4\theta)

Step 2: Substitute rr and drdθ\frac{dr}{d\theta} at θ=π4\theta = \frac{\pi}{4}

  1. Calculate rr at θ=π4\theta = \frac{\pi}{4}: r=1sin(4π4)=1sin(π)=10=1r = 1 - \sin\left(4 \cdot \frac{\pi}{4}\right) = 1 - \sin(\pi) = 1 - 0 = 1

  2. Calculate drdθ\frac{dr}{d\theta} at θ=π4\theta = \frac{\pi}{4}: drdθ=4cos(4π4)=4cos(π)=4(1)=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 dydx\frac{dy}{dx} formula

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

Substitute θ=π4\theta = \frac{\pi}{4}, r=1r = 1, and drdθ=4\frac{dr}{d\theta} = 4:

  • sin(π4)=22\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
  • cos(π4)=22\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

Substitute these into the formula:

dydx=422+122422122\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: 422+22=22+22=5224 \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: 42222=2222=3224 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 2\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}

Thus:

dydx=522322=53\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 θ=π4\theta = \frac{\pi}{4} is:

dydx=53\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 dydx\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 sinθ\sin\theta and cosθ\cos\theta based on the specific angle given, as they often simplify the computation.

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Differentiation
Slope of Tangent Line

Formulas

Slope of tangent line in polar coordinates: dy/dx = (dr/dθ * sin θ + r * cos θ) / (dr/dθ * cos θ - r * sin θ)

Theorems

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Suitable Grade Level

College Calculus