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Taya2010 [7]
2 years ago
6

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

ant and r varies with time? 2. how is dv/dt related to dh/dt if r is constant and h varies with time? 3. how is dV/dt related to dh/dt and dr/dt if both h and r vary with time?
Mathematics
1 answer:
laiz [17]2 years ago
8 0
In general, the volume

V=\pi r^2h

has total derivative

\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)

If the cylinder's height is kept constant, then \dfrac{\mathrm dh}{\mathrm dt}=0 and we have

\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}

which is to say, \dfrac{\mathrm dV}{\mathrm dt} and \dfrac{\mathrm dr}{\mathrm dt} are directly proportional by a factor equivalent to the lateral surface area of the cylinder (2\pi r h).

Meanwhile, if the cylinder's radius is kept fixed, then

\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}

since \dfrac{\mathrm dr}{\mathrm dt}=0. In other words, \dfrac{\mathrm dV}{\mathrm dt} and \dfrac{\mathrm dh}{\mathrm dt} are directly proportional by a factor of the surface area of the cylinder's circular face (\pi r^2).

Finally, the general case (r and h not constant), you can see from the total derivative that \dfrac{\mathrm dV}{\mathrm dt} is affected by both \dfrac{\mathrm dh}{\mathrm dt} and \dfrac{\mathrm dr}{\mathrm dt} in combination.
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