1.

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»

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}}


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