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[SOLVED] Spherical Balloon - Related Rates Problem
A spherical balloon is inflated so that its radius increases at a rate of 1 cm/min. How fast is the volume increasing when:
a) the diameter is 2000 cm
b) the surface area is 324 pi cm^2 ---> I have solved this already
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)(dr/dt)
For part (a) the diameter is 2000 cm.
So I make it 1000 cm for Radius
Now, I put it in the equation:
dV/dt = 4pi(r^2)(dr/dt)
where dr/dt = 1cm/min
dV/dt = 4pi(1000^2)(1 cm/min)
= 4 000 000 pi cm^3 / min
But, the answer in the book is 40 000 pi cm^3 / min.
Please help me out. Thanks.
Homework Statement
A spherical balloon is inflated so that its radius increases at a rate of 1 cm/min. How fast is the volume increasing when:
a) the diameter is 2000 cm
b) the surface area is 324 pi cm^2 ---> I have solved this already
Homework Equations
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)(dr/dt)
The Attempt at a Solution
For part (a) the diameter is 2000 cm.
So I make it 1000 cm for Radius
Now, I put it in the equation:
dV/dt = 4pi(r^2)(dr/dt)
where dr/dt = 1cm/min
dV/dt = 4pi(1000^2)(1 cm/min)
= 4 000 000 pi cm^3 / min
But, the answer in the book is 40 000 pi cm^3 / min.
Please help me out. Thanks.