# Calculus: need clarification to where the #'s go

Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 \mbox{cm}^3\mbox{/s}. How fast is the surface area of the balloon increasing when its radius is 14 \mbox{cm}? Recall that a ball of radius r has volume \displaystyle V=\frac{4}{3}\pi r^3 and surface area S=4\pi r^2.

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1. just use the formula

v = 4/3 pi r^3
a = 4pi r^2, so
dv/dt = 4pi r^2 dr/dt = a dr/dt
da/dt = 8pi r dr/dt

v(14) = 11494
a(14) = 2463

So, we have

80 = 2463 dr/dt
da/dt = 8pi(14)(80/2463) = 11.43

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2. what is an mbox?

V = (4/3) pi r^3
S = dV/dr = 4 pi r^2
dS/dr = 8 pi r
so
dS/dt = 8 pi r dr/dt

now find dr/dt
if dV/dt = 80 cm^3/s

dV = S * dr = 4 pi r^2 dr
so
dV/dt = 4 pi r^2 dr/dt
so
dr/dt = ( 1/4 pi r^2 )dV/dt = (20/pi 14^2)
so
dS/dt = 8 pi 14 (20/14^2 pi)
= 80/7
=

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2. 👎 0

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