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just wondering if i did this correctly

a spherical balloon is to be filled with water so that there is a constant increase in the rate of its surface area of 3cm2/sec .

a) Find the rate of increase in the radius when the radius is 3cm.

b) Find the volume when the volume is increasing at a rate of 10cm3=sec.

A=surface area, V=volume

a) i thought was fairly easy and got 1/(8Pi) cm/sec

b) was a bit more trickier here is what i did

(1) [tex]\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} [/tex]

(2) [tex]\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex]

from the question we have

[tex]\frac{dV}{dt}=10[/tex]

[tex]\frac{dA}{dt}=3[/tex]

[tex]\frac{dV}{dr}=4 \pi r^2[/tex]

[tex]\frac{dA}{dr}=8 \pi r[/tex]

from (2)

[tex]3=8 \pi r \frac{dr}{dt}[/tex]

so [tex]\frac{dr}{dt}=\frac{3}{8 \pi r}[/tex]

sub into (1) we get that r=20/3

so [tex]V=\frac{4}{3}\pi r^3=\frac{4 \pi}{3}(20/3)^3[/tex]

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# Homework Help: Related rates of increase

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