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Multiple Choice
What is the slope of the tangent line to the polar curve at the point where ?
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Verified step by step guidance
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Step 1: Recall that the slope of the tangent line to a polar curve is given by the formula \( \frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta} \), where \( r' \) is the derivative of \( r \) with respect to \( \theta \).
Step 2: Differentiate the given polar equation \( r = 4\theta^2 \) with respect to \( \theta \). This gives \( r' = \frac{d}{d\theta}(4\theta^2) = 8\theta \).
Step 3: Substitute \( r \) and \( r' \) into the slope formula. At \( \theta = \frac{\pi}{4} \), \( r = 4(\frac{\pi}{4})^2 \) and \( r' = 8(\frac{\pi}{4}) \). Compute these values symbolically without simplifying.
Step 4: Plug \( \theta = \frac{\pi}{4} \), \( r \), and \( r' \) into the slope formula \( \frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta} \). Use trigonometric values \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \).
Step 5: Simplify the numerator and denominator of the slope formula symbolically to express the slope \( \frac{dy}{dx} \) in terms of \( \pi \). This will yield the final slope value.