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Exercise 6.1 · Q4

Q.An edge of a variable cube is increasing at the rate of 3 cm/s3 \text{ cm/s}. How fast is the volume of the cube increasing when the edge is 10 cm10 \text{ cm} long?

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We use related rates: the volume V=s3V = s^3 changes at dVdt=3s2dsdt\frac{dV}{dt} = 3s^2 \frac{ds}{dt}. With dsdt=3\frac{ds}{dt} = 3 cm/s and s=10s = 10 cm, the volume increases at 900900 cm³/s.

This is a classic related rates problem — a staple in calculus. The idea is simple: when one quantity (the edge length ss) changes with time, another quantity that depends on it (the volume VV) also changes. We connect their rates of change using differentiation with respect to time tt.

The key step is always: write the relationship between the quantities, then differentiate both sides with respect to time. Do not plug in the given numbers until after you differentiate — that’s a common trap.


  1. Write the relationship. For a cube of edge length ss, the volume is

V=s3.V = s^3.

  1. Differentiate with respect to time tt. Since both VV and ss are functions of tt, we use the chain rule:

dVdt=3s2⋅dsdt.\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}.

This equation tells us: the rate at which volume grows depends on the current edge length and the rate at which the edge itself grows.

  1. Identify the given rates and the instant.

    We are told:

    • dsdt=3\frac{ds}{dt} = 3 cm/s (constant rate of increase of the edge).
    • We want dVdt\frac{dV}{dt} when s=10s = 10 cm.
  2. Substitute the values.

    dVdt=3⋅(10)2⋅3=3⋅100⋅3=900.\frac{dV}{dt} = 3 \cdot (10)^2 \cdot 3 = 3 \cdot 100 \cdot 3 = 900. …

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