The radius of a balloon that is spherical in shape is decreasing at the rate of .75 yard per minute...?

Question

I cannot for the life of me figure out this business calculus problem...

Gas is being released from a balloon on a cross-country trip in order to lose altitude. The balloon is sphere in shape. The radius of the balloon is decreasing at the rate of .75 yard per minute. Approximately how fast is the volume of the balloon decreasing when the radius is 25 yards?

Note: The volume of a sphere is given by the formula V = 4/3 Pi r ^3.

Any help would be greatly appreciated. :-)

Here's what I found...
Known: r = 25
dr/dt = -.75 or -3/4 (since the question asks for what rate the balloon is DECREASING in volume at, I used a NEGATIVE number here instead of a postive one.)
v = 4/3 pi r^3
Find: dv/dt

I actually did this to find the answer:

(dv/dt) = (4/3) pi x 3r^2 (dr/dt)

(dv/dt) = 4 pi r^2 (dr/dt)
(dv/dt) = 4 pi (25)^2 (-.75)

ANSWER: (dv/dt) = -468.75 pi yards^3

(I used a NEGATIVE .75, since the radius of the balloon was DECREASING at the rate of .75 yards per minute, instead of increasing at that rate.)

Let me know if I'm close on this one.

Thanks.

Note: the formula given is 4/3 pi r^3 and not 4/3 pi r^2.

Answer 1

Just take the time-derivative of the volume

dV/dt = 4 PI r^2 * dr/dt

you know dr/dt = -.75 ( decreasing)

and r = 25.

Answer 2

V = 4/3 πR³

dV/dT = dV/dR dR/dT = 4 πR² dR/dT


Answer:
4π x (25 yard)² x 0.75 yard/minute ~ 5890 yard³/minute

Answer 3

dr/dt = - 0.75 yd / min
V = (4/3).π.r³
dV/dr = 4π.r²
dV/dt = (dv / dr).(dr/dt)
dV/dt = 4π.r² x (- 0.75) yd³/min
dV/dt = - 3π.r² yd³/min
dV/dt = - 3π x 25² yd³/min
dV/dt = - 5890 yd³/min

Answer 4

uhhh all i can say is you havnt answerd the question and your smarter then me!