To find the slope of the polar curve $r = -5 - 4\cos\theta$ at the point where $\theta = \frac{\pi}{2}$, we will need to follow these steps:

1. Convert the Polar Equation to Cartesian Coordinates:

The relationship between polar and Cartesian coordinates is given by: $x = r\cos\theta$ $y = r\sin\theta$

2. Find $\frac{dy}{dx}$ in Polar Coordinates:

The slope of the curve at a point is given by $\frac{dy}{dx}$. In polar coordinates, this can be expressed as: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

First, calculate $\frac{dr}{d\theta}$: $\frac{dr}{d\theta} = \frac{d}{d\theta} \left(-5 - 4\cos\theta\right) = 4\sin\theta$

Now, calculate $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$: $\frac{dx}{d\theta} = \frac{d}{d\theta} \left(r\cos\theta\right) = \frac{dr}{d\theta}\cos\theta - r\sin\theta$ $\frac{dy}{d\theta} = \frac{d}{d\theta} \left(r\sin\theta\right) = \frac{dr}{d\theta}\sin\theta + r\cos\theta$

Substitute the values: $\frac{dx}{d\theta} = 4\sin\theta\cos\theta - (-5 - 4\cos\theta)\sin\theta$ $\frac{dy}{d\theta} = 4\sin^2\theta + (-5 - 4\cos\theta)\cos\theta$

3. Evaluate at $\theta = \frac{\pi}{2}$:

At $\theta = \frac{\pi}{2}$, we have: $\cos\left(\frac{\pi}{2}\right) = 0, \quad \sin\left(\frac{\pi}{2}\right) = 1$ Substitute these into the expressions for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$: $\frac{dx}{d\theta} = 4(1)(0) - (-5 - 4(0))(1) = 5$ $\frac{dy}{d\theta} = 4(1)^2 + (-5 - 4(0))(0) = 4$

4. Compute the Slope $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{4}{5}$

5. Choose the Correct Answer:

The slope of the polar curve at the point where $\theta = \frac{\pi}{2}$ is $\frac{4}{5}$.

Therefore, the correct answer is C. four fifths ( $\frac{4}{5}$ ).

Would you like further details or have any questions?

Here are some additional related questions:

1. How do you convert a polar equation into Cartesian form?
2. What is the significance of the derivative $\frac{dr}{d\theta}$ in polar coordinates?
3. How does the slope of a polar curve differ from that of a Cartesian curve?
4. What is the general formula for $\frac{dy}{dx}$ in polar coordinates?
5. How do trigonometric identities simplify calculations in polar coordinates?

Tip: Always check the trigonometric values at the specific angles to avoid common mistakes in sign or magnitude.