The volume of a sphere with radius r cm decreases at a rate of 22 cm /s . Find the rate of change of r when r =3 cm?

Answer 1

If you cant's do your own DE, then drop the class and get a refund if you still can.  How did you manage to get by pre-calculus?

Answer 2

Write V as a function of r.

Calculate dV/dt as a function of r and dr/dt.

Fill in the known values for dV/dt and r. Solve for dr/dt.

Answer 3

Excuse me!  BUT ME DOES NOT CARE TO DO YOUR WORK FOR YOU! You will never learn anything that way!

Answer 4

Mistake: it should be "22 cm^3/s".

Solution:

4pi(r^3)/3=V

=>

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

=>

4pi(3^2)dr/dt=-22

=>

dr/dt=-11/(18pi)~ -0.19 cm/s.

Answer 5

The volume changes at 22cm³/s or is that (22cm/s)³ ? Meaningless.

Answer 6

dV/dt = dV/dr*dr/dt = 4πr^2 dr/dt

dr/dt = 22/(4π*3^2) = 11/(18π) ~ 0.1945 cm/s

Answer 7

Volume of the sphere as a function of the radius r:

V(r) = 4/3 * π * r³

Rate of change of the volume = First derivative of V as a function of time (t)

dV/dt = 4/3 * π * 3r² * dr/dt <=== Note: r = 3 cm; dV/dt = 22 cm³/s

22 = 4/3 * π * 3(3²) * dr/dt

22 = 36π * dr/dt

===> dr/dt = 22 / (36π) ≃ 0.1945 cm/s (to 4 d.p.)

The radius decreases by 0.1945 cm/s

Answer 8

V = (4/3)πr³

so, dV/dt = 4πr².dr/dt

Then, with dV/dt = -22 and r = 3 we have:

-22 = 4π(3)².dr/dt

i.e. dr/dt = -11/18π => -0.19

Hence, the radius decreases at a rate of 0.19 cm/s

Note: the volume is decreasing at 22 cm³/s

:)>

Answer 9

dV/dt = d/dt((4/3)πr³) = (4/3)π d/dt(r³) =

=  (4/3)π d/dt(r³) * (dr/dr)

=  (4/3)π d/dr(r³) * (dr/dt)

=  (4/3)π * (3r²) * (dr/dt)

=  4 r² π * (dr/dt)

(dr/dt) = (dV/dt) / (4 r² π)

where dV/dt=-22 cm³/s

The rate of change of r when r =3 cm is

(dr/dt) = -22 / (4*3² π) = -11/(18π) = -0.1945... cm/s