Q.60 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 _________ cm2 s–1. (answer in integer)
Rate of Increase of Surface Area of Cube: 3000 cm²/s Solution
Problem Breakdown
The edge length x of the cube starts at 10 cm and increases at dx/dt = 25 cm/s.
Surface area: S = 6x²
Differentiate with respect to time: dS/dt = 12x × dx/dt
Step-by-Step Solution
- Substitute initial
x = 10cm anddx/dt = 25cm/s into the derivative equation. - Calculate:
dS/dt = 12 × 10 × 25 = 3000cm²/s. - The integer answer matches exactly, confirming the calculation.
Why Related Rates?
This applies chain rule in calculus for variables changing over time. Edge rate directly scales surface area change proportional to current edge length.
Introduction to Rate of Increase of Surface Area of Cube
Understanding the rate of increase of surface area of cube when edge length changes is key for CSIR NET Life Sciences math sections involving calculus. This problem tests related rates: edge grows at 25 cm/s from 10 cm initial length, find surface area increase rate in cm²/s (integer answer).
Cube Surface Area Formula and Differentiation
Cube surface area S = 6x², where x is edge length.
For variable edge, differentiate: dS/dt = 12x × dx/dt
Given dx/dt = 25 cm/s and initial x = 10 cm, compute 12 × 10 × 25 = 3000 cm²/s.
Detailed Calculation Steps
- Identify
S = 6x². - Implicit differentiation gives
dS/dt = 12x × dx/dt. - Plug values:
12 × 10 × 25 = 3000.
No options provided; direct integer result is 3000. Common pitfalls include forgetting the 12 factor or using volume formula V = x³.
Related Rates Applications in Exams
Similar problems appear in CSIR NET, JEE: volume rates, sphere balloons. Practice verifies edge rate scales surface quadratically via chain rule.
1 Comment
Vanshika Sharma
December 30, 2025by differentiating we get 3000cm^2/s