Applications of Derivatives
Application of Derivatives
nta_pyq_2025_apr
Grade 12
Question:
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is $1$ cm, the ice-cream melts at the rate of $81$ cm$^3$/min and the thickness of the ice-cream layer decreases at the rate of $\dfrac{1}{4\pi}$ cm/min. The surface area (in cm$^2$) of the chocolate ball (without the ice-cream layer) is:
$196\pi$
$256\pi$
$225\pi$
$128\pi$
Step-by-Step Solution
Key Concept: Use $V = \tfrac{4}{3}\pi r^3$ for the combined sphere (chocolate + ice-cream), differentiate to get $\tfrac{dV}{dt} = 4\pi r^2 \tfrac{dr}{dt}$, substitute the given rates to find $r$ (the outer radius), then subtract the ice-cream thickness to get the chocolate ball radius.
Let $r$ be the outer radius (chocolate + ice-cream). Volume $V = \dfrac{4}{3}\pi r^3$.
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
Given $\dfrac{dV}{dt} = 81$ cm$^3$/min and $\dfrac{dr}{dt} = \dfrac{1}{4\pi}$ cm/min:
$$81 = 4\pi r^2 \times \frac{1}{4\pi} = r^2$$
$$r^2 = 81 \Rightarrow r = 9$$
Radius of chocolate ball $= r - 1 = 8$ cm.
Surface area $= 4\pi(8)^2 = 256\pi$ cm$^2$.
Correct Answer: 2