Questions: Find the slope of the line tangent to the polar curve r=4 sec^2 theta at the point theta=-3π/4. Write the exact answer. Do not round.

Solution Steps

To find the slope of the tangent line to a polar curve at a given point, we first need to convert the polar equation to Cartesian coordinates. Then, we use the derivatives of these coordinates with respect to \(\theta\) to find the slope of the tangent line. Specifically, the slope \(m\) is given by \(\frac{dy/d\theta}{dx/d\theta}\).

The given polar curve is defined as: \[ r = 4 \sec^2(\theta) \]

Using the relationships \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), we can express \(x\) and \(y\) in terms of \(\theta\): \[ x = 4 \cos(\theta) \sec^2(\theta) = 4 \cos(\theta) \frac{1}{\cos^2(\theta)} = 4 \frac{1}{\cos(\theta)} = 4 \sec(\theta) \] \[ y = 4 \sin(\theta) \sec^2(\theta) = 4 \sin(\theta) \frac{1}{\cos^2(\theta)} = 4 \tan(\theta) \sec(\theta) \]

Next, we compute the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \): \[ \frac{dx}{d\theta} = -4 \sin(\theta) \sec^2(\theta) + 8 \cos(\theta) \tan(\theta) \sec^2(\theta) \] \[ \frac{dy}{d\theta} = 8 \sin(\theta) \tan(\theta) \sec^2(\theta) + 4 \cos(\theta) \sec^2(\theta) \]

The slope \(m\) of the tangent line is given by: \[ m = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]

Substituting \(\theta = -\frac{3\pi}{4}\) into the slope expression yields: \[ m = 3 \]

Final Answer