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Question
A spherical balloon is being inflated at the rate of 99 cm3^3/s. The rate at which the surface area increases when the radius is 33 cm is:
(A)66 cm2^2/s
(B)1212 cm2^2/s
(C)3π3\pi cm2^2/s
(D)99 cm2^2/s
Solution Path
Chain rule: dV/dt→dr/dt=1/(4π)dV/dt \to dr/dt = 1/(4\pi), then dS/dt=8πr⋅dr/dt=6dS/dt = 8\pi r \cdot dr/dt = 6 cm2^2/s.
01Question Setup
1/4
Spherical balloon inflated at dVdt=9\frac{dV}{dt} = 9 cm3^3/s. Find dSdt\frac{dS}{dt} when r=3r = 3 cm.
dSdt=  ?\frac{dS}{dt} = \;?
02Find dr/dt
2/4
V=43πr3⇒dVdt=4πr2drdtV = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. At r=3r = 3: 9=36πdrdt9 = 36\pi \frac{dr}{dt}, so drdt=14π\frac{dr}{dt} = \frac{1}{4\pi}.
drdt=14π\frac{dr}{dt} = \frac{1}{4\pi}
03Find dS/dtKEY INSIGHT
3/4
S=4πr2⇒dSdt=8πrdrdt=8π(3)14π=6S = 4\pi r^2 \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 8\pi(3)\frac{1}{4\pi} = 6 cm2^2/s.
dSdt=6\frac{dS}{dt} = 6 cm2^2/s
04Final Answer
4/4
dSdt=6\frac{dS}{dt} = 6 cm2^2/s. Answer is (A).
6\boxed{6}
Concepts from this question2 concepts unlocked
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Tangent and Normal Geometry

EASY

The slope of the tangent to y = f(x) at a point (x_1, y_1) is f'(x_1), and the slope of the normal is -1/f'(x_1). The tangent line equation is y - y_1 = f'(x_1)(x - x_1).

y−y1=f′(x1)(x−x1)(tangent),y−y1=−1f′(x1)(x−x1)(normal)y - y_1 = f'(x_1)(x - x_1) \quad \text{(tangent)}, \quad y - y_1 = -\frac{1}{f'(x_1)}(x - x_1) \quad \text{(normal)}

Tangent and normal equations appear directly in JEE problems involving curves, and are also used to find subtangent, subnormal lengths and angle of intersection between curves

Tangent and normal equationsCurve intersection anglesLength of subtangent/subnormal
Practice (12 Qs) →

First Derivative Test

EASY

The sign change of f'(x) around a critical point determines whether it is a local maximum or minimum. If f'(x) changes from positive to negative, the point is a local max; if negative to positive, it is a local min.

f′(x) changes +→−  ⟹  local max,f′(x) changes −→+  ⟹  local minf'(x) \text{ changes } + \to - \implies \text{local max}, \quad f'(x) \text{ changes } - \to + \implies \text{local min}

JEE problems frequently ask to classify critical points of polynomial and trigonometric functions, and the first derivative test works even when the second derivative is zero or hard to compute

Local extrema classificationPolynomial optimizationCurve sketching
Practice (14 Qs) →
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