## rate of change of volume if r increases at 2cm/m

Limits, differentiation, related rates, integration, trig integrals, etc.
mabasamunashe
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### rate of change of volume if r increases at 2cm/m

A solid consist of a hemisphere of radius r joined joined to a cone of constant height 60cm.The base of the cone is a plane face of a hemisphere.If r increases at a rate of 2cm/m.Calculate the rate of increase of the volume of the solid if r=60.leave answer in terms of pi.
[volume of sphere=(4pir^3/3 ;volume of cone=pi^h]
i've been confused by the formular for volume of a cone due to elimination of r

buddy
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### Re: rate of change

[volume of sphere=(4pir^3/3 ;volume of cone=pi^h]
The volume of a cone with radius r (for the base) and height h is really (1/3)(pi)(r^2)(h), not (pi)^h.
i've been confused by the formular for volume of a cone due to elimination of r
Use the right formula. that'll work better.
A solid consist of a hemisphere of radius r joined joined to a cone of constant height 60cm.The base of the cone is a plane face of a hemisphere.If r increases at a rate of 2cm/m.Calculate the rate of increase of the volume of the solid if r=60.leave answer in terms of pi.
Do half of a sphere: (1/2)(4/3)(pi)(r^3) = (2/3)(pi)(r^3)
plus a cone (so this is like a waffle cone w/ a scoop of ice cream on top): (1/3)(pi)(r^2)(h)
They give you that dh/dt = 0 & h = 60 for the cone & dr/dt = 2. You have to find dV/dt when r = 60.
So differentiate V, plug in 60 for h, 0 for dh/dt and 2 for dr/dt. Plug in 60 for r. Solve for dV/dt.