Question

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1. The volume of a sphere is increasing at the rate of 6 cm^3/hr. at what rate is the surface area increasing when the radius is 50cm?

2. a plane 3000ft from the earth's surface is flying east at the rate of 120mph. it passes directly over a car also going east at 60mph. how fast are they separating when the distance between them is 5000ft

Answer 1

1.

rate of volume increasing:

$\frac{dV}{dt}=\frac{d}{dt}(4\pi r^3/3)=4\pi r^2\frac{dr}{dt}=6$ cm/h

for $r=50$ cm:

$\frac{dr}{dt}=\frac{6}{4\pi \cdot50^2}=1.91\cdot10^{-4}$ cm/h

rate of change of surface are:

$\frac{dS}{dt}=\frac{d}{dt}(4\pi r^2)=8\pi r\frac{dr}{dt}=8\pi\cdot50\cdot1.91\cdot10^{-4}=0.24$ cm/h

2.

distance between plane and car:

$s=\sqrt{h^2+x^2}$

where h = 3000 ft is height of plane from earth's surface,

x is distance between them in east direction

$x=120t-60t=60t$

1 mile = 5280 feet

then:

$s=\sqrt{3000^2+(60\cdot5280t)^2}=5000$ ft

time for s = 5000 ft:

$t=\frac{\sqrt{5000^2-3000^2}}{60\cdot5280}=0.0126$ hours

rate of separating:

$\frac{ds}{dt}=\frac{d}{dt}(\sqrt{3000^2+3600\cdot5280^2t^2})=\frac{3600\cdot5280^2}{\sqrt{3000^2+3600\cdot5280^2t^2}}$

rate of separating when the distance between them is 5000ft:

$\frac{ds}{dt}(0.0126)=\frac{3600\cdot5280^2}{\sqrt{3000^2+3600\cdot5280^2\cdot0.0126^2}}=3801.6$ mph