Question

The length of the edge of a variable cube is increasing at the rate of 25 cm s\(^{-1}\). If the initial length of the edge of the cube is 10 cm, the rate of increase of the surface area of the cube is .............. cm\(^2\) s\(^{-1}\). (answer in integer)

Hint:

To find the rate of change of the surface area of a cube, differentiate the surface area formula \( A = 6a^2 \) with respect to time and substitute the given values for \( a \) and \( \frac{da}{dt} \).

Answer 1

Step 1: Understanding the problem.
The surface area \( A \) of a cube is given by the formula: \[ A = 6a^2 \] Where \( a \) is the length of the edge of the cube. We are given that \( \frac{da}{dt} = 25 \, \text{cm/s} \), and we need to find \( \frac{dA}{dt} \), the rate of change of the surface area.
Step 2: Differentiating the surface area formula.
To find \( \frac{dA}{dt} \), we differentiate the surface area equation with respect to time: \[ \frac{dA}{dt} = 12a \frac{da}{dt} \] Step 3: Substituting the known values.
At the initial length of the edge of the cube \( a = 10 \, \text{cm} \) and \( \frac{da}{dt} = 25 \, \text{cm/s} \): \[ \frac{dA}{dt} = 12 \times 10 \times 25 = 3000 \, \text{cm}^2 \, \text{s}^{-1} \] Final Answer: \[ \boxed{3000} \]