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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 constant 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?

Respuesta :

LammettHash
In general, the volume

[tex]V=\pi r^2h[/tex]

has total derivative

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

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

[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}[/tex]

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

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

[tex]\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}[/tex]

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

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

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