Question

volume of a sphere v is increasing at constant rate differentiation of y with respect to t is equal to k at the moment when volume of is Vo(zero) the rate of change of radius is​

Answer 1

$\bf \huge \underline {\underline{ANSWER}}$

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 .

Answer 2

Explanation:

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}}

34π3v0

[ 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}}

dt/dr= 4π(34π3v0)2k

= \frac{k}{\sqrt[3]{36\pi v_0^2}} 336πv 02k