The problem from the image reads:

Problem (10): Given: $x = 5 + \sec 3\theta$ and $y = 1 - \tan 3\theta,$ find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$.

Solution:

We will follow these steps to solve the problem:

1. Find the first derivative, $\frac{dy}{dx}$:

Using the chain rule, we first find $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ and then divide them.

$\frac{dx}{d\theta}$:

$x = 5 + \sec 3\theta \Rightarrow \frac{dx}{d\theta} = 3\sec 3\theta \tan 3\theta.$

$\frac{dy}{d\theta}$:

$y = 1 - \tan 3\theta \Rightarrow \frac{dy}{d\theta} = -3\sec^2 3\theta.$

Thus, the first derivative $\frac{dy}{dx}$ is: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-3\sec^2 3\theta}{3\sec 3\theta \tan 3\theta} = \frac{-\sec 3\theta}{\tan 3\theta} = -\cot 3\theta.$

2. Find the second derivative, $\frac{d^2y}{dx^2}$:

To find $\frac{d^2y}{dx^2}$, we need to differentiate $\frac{dy}{dx}$ with respect to $\theta$, and then divide by $\frac{dx}{d\theta}$.

Differentiate $\frac{dy}{dx} = -\cot 3\theta$:

$\frac{d}{d\theta}(-\cot 3\theta) = 3\csc^2 3\theta.$

Now, divide by $\frac{dx}{d\theta} = 3\sec 3\theta \tan 3\theta$: $\frac{d^2y}{dx^2} = \frac{3\csc^2 3\theta}{3\sec 3\theta \tan 3\theta} = \frac{\csc^2 3\theta}{\sec 3\theta \tan 3\theta}.$

3. Evaluate at $\theta = \frac{\pi}{4}$:

At $\theta = \frac{\pi}{4}$, we have: $3\theta = \frac{3\pi}{4}, \quad \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}, \quad \tan\left(\frac{3\pi}{4}\right) = -1, \quad \csc^2\left(\frac{3\pi}{4}\right) = 2.$

Thus, at $\theta = \frac{\pi}{4}$: $\frac{d^2y}{dx^2} = \frac{2}{(-\sqrt{2})(-1)} = \frac{2}{\sqrt{2}} = \sqrt{2}.$

Final Answer:

$\frac{d^2y}{dx^2} \text{ at } \theta = \frac{\pi}{4} = \sqrt{2}.$

Would you like further details or have any other questions?

Here are some related questions to explore further:

1. What is the geometric meaning of the second derivative $\frac{d^2y}{dx^2}$?
2. How does the chain rule apply when functions are dependent on a third variable?
3. What are the trigonometric identities for $\sec$, $\csc$, and $\cot$ at different angles?
4. What is the significance of evaluating derivatives at specific points like $\theta = \frac{\pi}{4}$?
5. How does $\frac{dy}{dx}$ represent the rate of change in parametric equations?

Tip: When differentiating trigonometric functions, always double-check the identities and signs of the functions.