Asked by kgothy
A balloon is in the shape of a cylinder with a hemisphere at both ends.The height of the cylinder is 4 times that of the radius.Air is pumped in the balloon at a rate of 60cm cubed per second .At what rate does the surface area change if the radius is 8cm?
Answers
Answered by
Steve
the figure is really just a sphere of radius r and a cylinder of radius r and height 4r. So, the volume is
V = 4/3 π r^3 + πr^2 (4r)
= 16/3 π r^3
So,
dV/dt = 16πr^2 dr/dt
That means that
dr dt = dV/dt / (16πr^2)
The surface area is
A = 4πr^2 + 2πr*4r
= 4πr^2 + 8πr^2
= 12πr^2
Thus
dA/dt = 24πr dr/dt
= 24πr * dV/dt / (16πr^2)
= 3/(2r) dV/dt
Now, just plug in r=8 and dV/dt = 60, giving
dA/dt = 3/16 * 60 = 11.25 cm^2/s
V = 4/3 π r^3 + πr^2 (4r)
= 16/3 π r^3
So,
dV/dt = 16πr^2 dr/dt
That means that
dr dt = dV/dt / (16πr^2)
The surface area is
A = 4πr^2 + 2πr*4r
= 4πr^2 + 8πr^2
= 12πr^2
Thus
dA/dt = 24πr dr/dt
= 24πr * dV/dt / (16πr^2)
= 3/(2r) dV/dt
Now, just plug in r=8 and dV/dt = 60, giving
dA/dt = 3/16 * 60 = 11.25 cm^2/s
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