If ` u = x^(2) + y^(2) " and " x = s + 3t, y = 2s - t , " then " (d^(2) u)/(ds^(2))` is equal to

Options

A. 12
B. 32
C. 36
D. 10

Explanation

To solve the problem, we need to find the second derivative of \( u \) with respect to \( s \), given that \( u = x^2 + y^2 \), \( x = s + 3t \), and \( y = 2s - t \).

### Step-by-Step Solution:

1. **Define the functions**:
\[ u = x^2 + y^2 \]
where
\[ x = s + 3t \quad \text{and} \quad y = 2s - t \]

2. **Differentiate \( u \) with respect to \( s \)**:
We will use the chain rule to differentiate \( u \):
\[ \frac{du}{ds} = \frac{du}{dx} \cdot \frac{dx}{ds} + \frac{du}{dy} \cdot \frac{dy}{ds} \]
First, calculate \( \frac{du}{dx} \) and \( \frac{du}{dy} \):
\[ \frac{du}{dx} = 2x \quad \text{and} \quad \frac{du}{dy} = 2y \]

3. **Calculate \( \frac{dx}{ds} \) and \( \frac{dy}{ds} \)**:
Differentiate \( x \) and \( y \) with respect to \( s \):
\[ \frac{dx}{ds} = 1 \quad \text{(since the derivative of \( s \) is 1 and \( 3t \) is constant with respect to \( s \))} \]
\[ \frac{dy}{ds} = 2 \quad \text{(since the derivative of \( 2s \) is 2 and \( -t \) is constant with respect to \( s \))} \]

4. **Substitute into the derivative of \( u \)**:
Now substitute \( \frac{du}{dx} \), \( \frac{du}{dy} \), \( \frac{dx}{ds} \), and \( \frac{dy}{ds} \) into the equation:
\[ \frac{du}{ds} = 2x \cdot \frac{dx}{ds} + 2y \cdot \frac{dy}{ds} = 2x \cdot 1 + 2y \cdot 2 = 2x + 4y \]

5. **Differentiate \( \frac{du}{ds} \) again to find \( \frac{d^2u}{ds^2} \)**:
We differentiate \( \frac{du}{ds} \) again:
\[ \frac{d^2u}{ds^2} = \frac{d}{ds}(2x + 4y) = 2\frac{dx}{ds} + 4\frac{dy}{ds} \]
Substitute \( \frac{dx}{ds} = 1 \) and \( \frac{dy}{ds} = 2 \):
\[ \frac{d^2u}{ds^2} = 2 \cdot 1 + 4 \cdot 2 = 2 + 8 = 10 \]

### Final Answer:
\[ \frac{d^2u}{ds^2} = 10 \]