Question

help please!
Picture below

Answer 1

Assume the radius to be r
Volume = (4/3)(pi)r^3
Find the derivative of the left and right hand side with respect to time t.
Can you tell me what you get?

Answer 2

i dont understand

Answer 3

"The radius of a sphere is increasing at a rate of 2 inches per minute. At what rate (in cubic inches per minute) is the volume increasing when the surface area of the sphere is 9π square inches?"

This is a related rates problem, meaning that you're required to play around with derivative rates to manipulate, isolate, and find a value for a particular rate given some initial conditions. In this case, you know some critical information already.

dr/dt = 2; Find dV/dt when S = 9π

So what we need to do is find the rate at which volume is changing. We would need the equation for finding the volume of a sphere then!

V = (4/3)πr^3 = (4π/3)(r^3)

Now let's take the derivative.

dV/dt = (4π/3)(3r^2)(dr/dt)

Now we have a slight problem here. We want to be able to solve for dV/dt, our unknown variable, with the given condition of a value for dr/dt. However, there's another unknown variable here: r. While we aren't given any value for r, we are given one for S. Therefore, it becomes necessary to convert r into terms of S in order to further solve our differential equation. The equation for surface area of a sphere would help!

S = 4π(r^2)

Now we should isolate r for this to work.

S = 4π(r^2)
r^2 = S/4π
r = ±√(S/4π)
r = √(S/4π) [Rate is increasing; value cannot be negative]

Now that we've found r in terms of S, we can plug that back into our derived differential equation to solve for dV/dt.

dV/dt = (4π/3)(3r^2)(dr/dt) [r = √(S/4π)]
dV/dt = (4π/3)(3(√(S/4π))^2)(dr/dt)
dV/dt = (4π/3)(3S/4π)(dr/dt)
dV/dt = S(dr/dt)

Notice what happened there! We had two fractions of (4π/3) and (3/4π) which canceled each other out, leaving the variable S of the second fraction isolated. Now we can plug in values and solve for dV/dt.

dV/dt = S(dr/dt)
dV/dt = (9π sq.in)(2 in/min)
dV/dt = 18π cub.in/min

Finally, we have the answer. The rate at which the volume is changing over time, depicted by dV/dt, equals 18π cubic inches per minute.

Answer 4

thank you!:)