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Application of Derivatives: 3 marks Questions (CBSE Class 12)

3 different 3 marks questions on Application of Derivatives from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (9)2 marks (12)3 marks (3)4 marks (4)5 marks (6)

Q263 marksShort AnswerApplication of Derivatives CBSE 2026 · 65/5/1

A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.

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Answer: Proved.

$V = \frac{4}{3} π r^{3}$ , $S = 4 π r^{2}$ .
Given $\frac{d V}{d t} = - k S = - 4 k π r^{2}$ , where $k > 0$ is constant.
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ , so $4 π r^{2} \frac{d r}{d t} = - 4 k π r^{2}$ .
$\frac{d r}{d t} = - k$ , a constant; so the radius decreases at a constant rate.

Q263 marksShort AnswerApplication of Derivatives CBSE 2026 · 65/5/2

A thin metallic wire in the shape of a circular ring has its enclosed area increasing at a uniform rate when heated. Show that the rate of change of circumference varies inversely as the radius.

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Answer: Proved.

Let r be the radius, $A = π r^{2}$ the enclosed area and $C = 2 π r$ the circumference.
Given $\frac{d A}{d t} = k$ , a constant: $2 π r \frac{d r}{d t} = k \Rightarrow \frac{d r}{d t} = \frac{k}{2 π r}$ .
$\frac{d C}{d t} = 2 π \frac{d r}{d t} = 2 π \cdot \frac{k}{2 π r} = \frac{k}{r}$ .
So the rate of change of circumference varies inversely as the radius.

Q263 marksShort AnswerApplication of Derivatives CBSE 2026 · 65/5/3

The volume of a wooden block in the shape of a cube increases at a constant rate as the air becomes moist during the rainy season. Show that the rate of change of its surface area varies inversely as the length of edge of the cube.

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Answer: Proved.

Let the edge be x, volume $V = x^{3}$ and surface area $S = 6 x^{2}$ .
Given $\frac{d V}{d t} = k$ , a constant: $3 x^{2} \frac{d x}{d t} = k \Rightarrow \frac{d x}{d t} = \frac{k}{3 x ^{2}}$ .
$\frac{d S}{d t} = 12 x \frac{d x}{d t} = 12 x \cdot \frac{k}{3 x ^{2}} = \frac{4 k}{x}$ .
So the rate of change of surface area varies inversely as the length of the edge.

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