On heating, the volume of a metal cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters?

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Accepted Answer

> [!TLDR]
> This is a related-rates problem: differentiate volume with respect to time to find $dx/dt$, then differentiate surface area with respect to time using the chain rule.

> [!FORMULA]
> For a cube of edge $x$: volume $V=x^3$, surface area $S=6x^2$. Chain rule for related rates:
> $$\frac{dV}{dt} = \frac{dV}{dx}\cdot\frac{dx}{dt}, \qquad \frac{dS}{dt} = \frac{dS}{dx}\cdot\frac{dx}{dt}$$

1. **Given data.**
   $$\frac{dV}{dt} = 9 \ 	ext{cm}^3/	ext{s}, \qquad x = 10 \ 	ext{cm (at the instant in question)}$$

2. **Relate $V$ to $x$ and differentiate w.r.t. time $t$ (chain rule, since $x$ depends on $t$).**
   $$V = x^3 \ \implies\ \frac{dV}{dt} = 3x^2\,\frac{dx}{dt}$$

3. **Solve for $\dfrac{dx}{dt}$ at $x=10$.**
   $$9 = 3(10)^2 \cdot \frac{dx}{dt} = 300\,\frac{dx}{dt} \ \implies\ \frac{dx}{dt} = \frac{9}{300} = \frac{3}{100} = 0.03 \ 	ext{cm/s}$$

4. **Relate $S$ to $x$ and differentiate w.r.t. time.**
   $$S = 6x^2 \ \implies\ \frac{dS}{dt} = 12x\,\frac{dx}{dt}$$

5. **Substitute $x=10$ and $\dfrac{dx}{dt}=0.03$.** …

Q.On heating, the volume of a metal cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters?

Concept understanding — Instantaneous Rate of Change and Derivative

Instantaneous Rate of Change and Derivative

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Q.On heating, the volume of a metal cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters?

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