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Multiple Choice
What is the slope of the tangent line to the polar curve when ?
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Step 1: Recall that the slope of the tangent line to a polar curve is given by \( \frac{dy}{dx} \), where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Start by expressing \( r \) in terms of \( \theta \): \( r = 2\theta^2 \).
Step 2: Compute \( x \) and \( y \) in terms of \( \theta \): \( x = r \cos(\theta) = 2\theta^2 \cos(\theta) \) and \( y = r \sin(\theta) = 2\theta^2 \sin(\theta) \).
Step 3: Differentiate \( x \) and \( y \) with respect to \( \theta \) using the product rule. For \( x \), \( \frac{dx}{d\theta} = \frac{d}{d\theta}(2\theta^2 \cos(\theta)) = 4\theta \cos(\theta) - 2\theta^2 \sin(\theta) \). For \( y \), \( \frac{dy}{d\theta} = \frac{d}{d\theta}(2\theta^2 \sin(\theta)) = 4\theta \sin(\theta) + 2\theta^2 \cos(\theta) \).
Step 4: Use the chain rule to find \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \). Substitute \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) into this formula.
Step 5: Evaluate \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{2} \). Substitute \( \theta = \frac{\pi}{2} \) into \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \), and simplify the expression for \( \frac{dy}{dx} \). This will give the slope of the tangent line at the specified point.