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?
- 2 months ago
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?
- AmyLv 72 months ago
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.
- zipperLv 72 months ago
Excuse me! BUT ME DOES NOT CARE TO DO YOUR WORK FOR YOU! You will never learn anything that way!
- PinkgreenLv 72 months ago
Mistake: it should be "22 cm^3/s".
dr/dt=-11/(18pi)~ -0.19 cm/s.
- How do you think about the answers? You can sign in to vote the answer.
- oldschoolLv 72 months ago
The volume changes at 22cm³/s or is that (22cm/s)³ ? Meaningless.
- Ian HLv 72 months ago
dV/dt = dV/dr*dr/dt = 4πr^2 dr/dt
dr/dt = 22/(4π*3^2) = 11/(18π) ~ 0.1945 cm/s
- JeremyLv 62 months ago
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
- Wayne DeguManLv 72 months ago
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
- SlowfingerLv 62 months ago
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