A metallic disc is being heated. Its area (in `m^(2)`) at any time t (in sec) is given by `A=5t^(2)+4t`. Calculate the rate of increase in area at `t=3sec`.

Answer

To find the rate of increase in the area of the metallic disc at \( t = 3 \) seconds, we need to differentiate the area function \( A(t) = 5t^2 + 4t \) with respect to time \( t \).

### Step-by-Step Solution:

1. **Write down the area function**:
\[ A(t) = 5t^2 + 4t \]

2. **Differentiate the area function with respect to time \( t \)**:
To find the rate of increase of area, we need to compute \( \frac{dA}{dt} \).
\[ \frac{dA}{dt} = \frac{d}{dt}(5t^2) + \frac{d}{dt}(4t) \]

3. **Apply the power rule for differentiation**:
- The derivative of \( 5t^2 \) is \( 10t \) (using the power rule \( \frac{d}{dt}(t^n) = nt^{n-1} \)).
- The derivative of \( 4t \) is \( 4 \).
\[ \frac{dA}{dt} = 10t + 4 \]

4. **Substitute \( t = 3 \) seconds into the derivative**:
Now, we will find the rate of increase in area at \( t = 3 \) seconds.
\[ \frac{dA}{dt} \bigg|_{t=3} = 10(3) + 4 \]

5. **Calculate the value**:
\[ \frac{dA}{dt} \bigg|_{t=3} = 30 + 4 = 34 \]

6. **State the final answer**:
The rate of increase in area at \( t = 3 \) seconds is:
\[ \frac{dA}{dt} = 34 \, \text{m}^2/\text{s} \]