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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

Postby mabasamunashe on Sat Oct 19, 2013 5:20 am

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
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Re: rate of change  TOPIC_SOLVED

Postby buddy on Sat Oct 19, 2013 11:19 am

mabasamunashe wrote:[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.

mabasamunashe wrote: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.

mabasamunashe wrote: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.
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