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Find the points of horizontal tangency (if any) to the polar curve. $$ r=a \sin \theta \cos ^{2} \theta $$

Short Answer

Expert verified
The exact solution depends highly on the value of 'a' and may involve complex trigonometric solutions. The solution would involve handling various trigonometric identities and could involve multiple values of \(\theta\).

Step by step solution

01

Compute the derivative of r with respect to \(\theta\)

First, we compute the derivative of \( r \) with respect to \(\theta\). The given equation is \( r = a \sin \theta \cos^2 \theta \), so applying the product and chain rules for differentiation gives \(\frac{dr}{d\theta} = a \cos \theta \cos^2 \theta - 2a \sin^2 \theta \cos \theta \).
02

Set the slope to zero

The slope of the tangent line in polar coordinates is given by \( \frac{dr}{d\theta}/ (1 - r \sin \theta) \). For horizontal tangents, this has to be zero. We solve \(a \cos \theta \cos^2 \theta - 2a \sin^2 \theta \cos \theta / (1 - a \sin \theta \cos^2 \theta \sin \theta) = 0\) for \(\theta\).
03

Solve the equation

Solving the equation will give us the values of theta at which the slope of the tangent line to the polar curve is zero, thus identifying the points of horizontal tangency. Factor out common terms where possible, and find values of \(\theta\) that satisfy the equation. Note that some values may not yield real points on the curve, thus should be disregarded.

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