A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 6 centimeters.?

Question

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 6 centimeters. Assuming the balloon is filled with helium at a rate of 10 cubic centimeters per second, calculate how fast the radius is growing at the instant it pops. (The volume of a sphere of radius r is V = (4/3) πr3. Give the answer to two decimal places.)

Update:

I got .2 myself, but I want to make sure...

? · Accepted Answer

radius = 6
V = 4/3 pi r ^3
dV/dt = 10 cm^3/sec

dV/dt = 4/3 pi (3r^2) dr/dt   This is where Wile and I disagree.  I believe the d/dt of r^3 is in the expression
dV/dt = 4 pi r^2 dr/dt
10 = 4 (3.14) 6^2 dr/dt

dr/dt = 10 / ((4)( pi )(36) = 10/(452.38934) = .02 cm/s

Wile E. · Answer

Radus, r = 6 cm.
Volume = V
dV/dt = 10 cu. cm./sec.

Find dr/dt:

V = 4/3 π r³

Differentiating Implicitly Over Time,

dV/dt = 4/3 π r(dr/dt)
10 = 4/3 π (6)(dr/dt)
10 = 24/3 π (dr/dt)
10 = 8π (dr/dt)
dr/dt = 10 / 8π
dr/dt = 0.398

The volume is increasing at a rate of about 0.40 cu. cm./sec.

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A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 6 centimeters.?

2 Answers