Question

Volume of a sphere , V is increasing at a constant rage dV/dt=k. At the moment when volume of is V0, the rate of change of radius is

Answer 1

The volume of sphere V is increasing at the constant rate , dV/dt = k......(1)

Let radius of sphere is r.

from formula of sphere , V = 4/3 πr³

differentiating both sides with respect to time,

dV/dt = d(4/3 πr³)/dt

dV/dt = d(4/3 πr³)/dt= 4/3π × d(r³)/dt

dV/dt = d(4/3 πr³)/dt= 4/3π × d(r³)/dt = 4/3 π × 3r² dr/dt [ from application of differentiation, if y = f(x)ⁿ, then dy/dx = nf(x)ⁿ-¹ ]

= 4πr² × dr/dt

so we get , dV/dt = 4πr² dr/dt

from equation (1),

dV/dt = k = 4πr² dr/dt

dV/dt = k = 4πr² dr/dtor, dr/dt = k/4πr² ....(2)

a/c to question, at the moment volume of sphere is v0 .

then radius of sphere at that moment, r = $\sqrt[3]{\frac{3v_0}{4\pi}}$ [ using formula v0 = 4/3 πr³ ]

now putting value of r in equation (2),

we get, $\bf{\frac{dr}{dt}=\frac{k}{4\pi\left(\sqrt[3]{\frac{3v_0}{4\pi}}\right)^2}}$

$=\frac{k}{\sqrt[3]{36\pi v_0^2}}$

Answer 2

Dear Student,

◆ Answer -

dr/dt = k / ∛(36π.V²)

● Explanation -

Volume of the sphere is given by -

V = (4π/3) r^3

r^3 = 3/4π × V

r = ∛(3/4π) ∛V

Differentiating w.r.t. t,

dr/dt = ∛(3/4π) × dV/dt × 1/3 × V^(-2/3)

dr/dt = k × 1/3 × ∛(3/4π) / ∛V^2

dr/dt = k ∛(3/4π) / (∛27 × ∛V²)

dr/dt = k / ∛(36π.V²)

Hope this helps you. Thanks for asking..