The weight W of a certain stock of fish is given by W = nw, where n is the size of stock and w is the average weight of a fish. If n and w change with time t as n = `2t^(2)+3` and `w=t^(2)-t+2`, then the rate of change of W with respect to t at t = 1, is

Options

A. 1
B. 13
C. 5
D. 8

Explanation

To solve the problem step by step, we need to find the rate of change of the weight \( W \) of a certain stock of fish with respect to time \( t \). The weight \( W \) is given by the product of the size of the stock \( n \) and the average weight of a fish \( w \).

### Step 1: Define the Variables
Given:
- \( W = n \cdot w \)
- \( n = 2t^2 + 3 \)
- \( w = t^2 - t + 2 \)

### Step 2: Differentiate \( W \) with Respect to \( t \)
Using the product rule of differentiation, we have:
\[ \frac{dW}{dt} = n \cdot \frac{dw}{dt} + w \cdot \frac{dn}{dt} \]

### Step 3: Calculate \( \frac{dw}{dt} \) and \( \frac{dn}{dt} \)
1. Differentiate \( w \):
\[ w = t^2 - t + 2 \implies \frac{dw}{dt} = 2t - 1 \]

2. Differentiate \( n \):
\[ n = 2t^2 + 3 \implies \frac{dn}{dt} = 4t \]

### Step 4: Substitute \( n \), \( w \), \( \frac{dw}{dt} \), and \( \frac{dn}{dt} \) into the equation
Now substituting these values into the equation for \( \frac{dW}{dt} \):
\[ \frac{dW}{dt} = (2t^2 + 3)(2t - 1) + (t^2 - t + 2)(4t) \]

### Step 5: Simplify the Expression
1. Expand \( (2t^2 + 3)(2t - 1) \):
\[ = 4t^3 - 2t^2 + 6t - 3 \]

2. Expand \( (t^2 - t + 2)(4t) \):
\[ = 4t^3 - 4t^2 + 8t \]

3. Combine the two expansions:
\[ \frac{dW}{dt} = (4t^3 - 2t^2 + 6t - 3) + (4t^3 - 4t^2 + 8t) \]
\[ = 8t^3 - 6t^2 + 14t - 3 \]

### Step 6: Evaluate at \( t = 1 \)
Now, substitute \( t = 1 \) into \( \frac{dW}{dt} \):
\[ \frac{dW}{dt} \bigg|_{t=1} = 8(1)^3 - 6(1)^2 + 14(1) - 3 \]
\[ = 8 - 6 + 14 - 3 = 13 \]

### Final Answer
The rate of change of \( W \) with respect to \( t \) at \( t = 1 \) is \( 13 \).