# Answer to Question #4613 in Calculus for Owais

Question #4613

Hello,

When the dr/dt for sphere is constant and r=at + b, how does dV/dt change over time?

This is my work so far:

dr/dt = a

V=(4/3)(pi)r^3

dV/dt = 4(pi)r^2 (dr/dt)

dV/dt = 4(pi)(at + b)^2 (a)

Well, if radius is steadily decreasing, is change in volume decreasing constantly.

I know in my work, since there is a variable t, volume would not change at a constant rate. But am I thinking about it correctly?

When the dr/dt for sphere is constant and r=at + b, how does dV/dt change over time?

This is my work so far:

dr/dt = a

V=(4/3)(pi)r^3

dV/dt = 4(pi)r^2 (dr/dt)

dV/dt = 4(pi)(at + b)^2 (a)

Well, if radius is steadily decreasing, is change in volume decreasing constantly.

I know in my work, since there is a variable t, volume would not change at a constant rate. But am I thinking about it correctly?

Expert's answer

Let's assume that a and b are greater than zero.

As you noted, dV/dt = 4a(pi)(at + b)^2. It's greater than zero anyway. So, V would increase while increasing t. Let's get d(dV/dt)/dt:

d(dV/dt)/dt = 8(a^2)(pi)(at+b). It's also greater then zero, so, dV/dt would increase while increasing t, similarly.

As you noted, dV/dt = 4a(pi)(at + b)^2. It's greater than zero anyway. So, V would increase while increasing t. Let's get d(dV/dt)/dt:

d(dV/dt)/dt = 8(a^2)(pi)(at+b). It's also greater then zero, so, dV/dt would increase while increasing t, similarly.

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