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Exercise 6.1 · Q2

Q.The volume of a cube is increasing at the rate of 8 cm3/s8 \text{ cm}^3/\text{s}. How fast is the surface area increasing when the length of an edge is 12 cm12 \text{ cm}?

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We relate the rates of change of volume and surface area through the edge length. Using V=s3V = s^3 and S=6s2S = 6s^2, we find dSdt=83 cm2/s\frac{dS}{dt} = \frac{8}{3} \text{ cm}^2/\text{s} when s=12s = 12 cm.

This is a classic related rates problem. The key idea: when one quantity changes with time, other quantities linked to it also change. Here, volume and surface area both depend on the edge length ss, and we know how fast volume is increasing. We want how fast surface area is increasing at a specific moment.

The chain rule is our tool. If two quantities AA and BB are both functions of ss, and ss itself changes with time, then dAdt=dAds⋅dsdt\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt} and similarly for BB. So we can connect dVdt\frac{dV}{dt} to dSdt\frac{dS}{dt} through dsdt\frac{ds}{dt}.

Let’s work through it step by step.

  1. Write the formulas.

    Let the edge length be ss cm.

    Volume: V=s3V = s^3

    Surface area: S=6s2S = 6s^2 (a cube has 6 faces, each of area s2s^2).

  2. What we know.

    dVdt=8 cm3/s\frac{dV}{dt} = 8 \text{ cm}^3/\text{s} (given, constant rate).

    We want dSdt\frac{dS}{dt} when s=12s = 12 cm.

  3. Differentiate volume with respect to time.

    Using the chain rule:

dVdt=ddt(s3)=3s2⋅dsdt\frac{dV}{dt} = \frac{d}{dt}(s^3) = 3s^2 \cdot \frac{ds}{dt}

So 8=3s2⋅dsdt8 = 3s^2 \cdot \frac{ds}{dt}.

  1. Solve for dsdt\frac{ds}{dt} at the given edge length. When s=12s = 12:

8=3(12)2⋅dsdt=3⋅144⋅dsdt=432⋅dsdt8 = 3(12)^2 \cdot \frac{ds}{dt} = 3 \cdot 144 \cdot \frac{ds}{dt} = 432 \cdot \frac{ds}{dt}

dsdt=8432=154 cm/s\frac{ds}{dt} = \frac{8}{432} = \frac{1}{54} \text{ cm/s}

So the edge is growing slowly — about 0.0185 cm each second.

  1. Differentiate surface area with respect to time.

dSdt=ddt(6s2)=12s⋅dsdt\frac{dS}{dt} = \frac{d}{dt}(6s^2) = 12s \cdot \frac{ds}{dt}

  1. Plug in s=12s = 12 and dsdt=154\frac{ds}{dt} = \frac{1}{54}.

dSdt=12⋅12⋅154=144⋅154=14454=83 cm2/s\frac{dS}{dt} = 12 \cdot 12 \cdot \frac{1}{54} = 144 \cdot \frac{1}{54} = \frac{144}{54} = \frac{8}{3} \text{ cm}^2/\text{s}

Watch out

A common mistake is to forget that dsdt\frac{ds}{dt} is not constant — it changes as ss changes. You must compute it at the specific instant given. Also, don’t confuse the rate of change of volume with the rate of change of surface area; they have different units.

Tip

You can also solve this in one shot by eliminating dsdt\frac{ds}{dt}:

From dVdt=3s2dsdt\frac{dV}{dt} = 3s^2 \frac{ds}{dt} and dSdt=12sdsdt\frac{dS}{dt} = 12s \frac{ds}{dt}, divide the second by the first:

dS/dtdV/dt=12s3s2=4s\frac{dS/dt}{dV/dt} = \frac{12s}{3s^2} = \frac{4}{s}

So dSdt=4s⋅dVdt\frac{dS}{dt} = \frac{4}{s} \cdot \frac{dV}{dt}.

With s=12s=12 and dVdt=8\frac{dV}{dt}=8, we get dSdt=412⋅8=3212=83\frac{dS}{dt} = \frac{4}{12} \cdot 8 = \frac{32}{12} = \frac{8}{3}. This shortcut works because both derivatives share the same dsdt\frac{ds}{dt}.

✓Final answer

The surface area is increasing at 83 cm2/s\boxed{\frac{8}{3} \text{ cm}^2/\text{s}} when the edge is 12 cm.

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