Question

The volume of a sphere v is increasing at a constant rate dv/dt =k.at the moment when volume is v0 the rate of change of radius?

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

= 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

or, 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

Answer:

dv/dt=k/(4πr₀²)

Explanation:

Let the radius be r,

At v₀ let radius be r₀,

dv/dt=k (given),

as we know ,

volume of sphere, v=4/3(πr³)

Therefore,

dv/dt=4/3(3πr²dr/dt)

k=4πr²dr/dt

At v₀=r₀ putting the value we get,

k=4πr₀²dr/dt

dr/dt=k/(4πr₀²)