# The volume of a cone of radius r and height of h is given by V=(1/3)πr²h….?

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The volume of a cone of radius r and height of h is given by V=(1/3)πr²h. If the radius and the height both increase at a constant rate of 1/2 centimeter per second, at what rate, in cubic centimeters per second, is the volume increasing when the height is 9cm and radius is 6cm?

Here are the answer choices…

a) π/2

b) 10π

c) 24π

d) 54π

e) 108π

If you could show the work you did to get the answer that would be VERY helpful! 🙂

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• V=(1/3)pi r^2h

dv/dt=?

dr/dt=.5

dh/dt=.5

h=9

r=6

dv/dt=(2/3)(pi)(r)(dr/dt)(h)+(1/3)(pi)(r^2)(dh/dt)

dv/dt=(2/3)(pi)(6)(.5)(9)+(1/3)(pi)(6^2)(9)

dv/dt=24 pi

• Let:

r = base radius

h = height

V = volume

Given:

dr/dt = 1/2 cm/sec

dh/dt = 1/2 cm/sec

Find:

dV/dt when h = 9cm and r = 6 cm

V = πr²h/3

Take the derivative with respect to t.

dV/dt = (dV/dr)(dr/dt) + (dV/dh)(dh/dt)

dV/dt = (2πrh/3)(1/2) + (πr²/3)(1/2) = πrh/3 + πr²/6

Plug in the values for r and h and solve.

dV/dt = π69/3 + π*6²/6 = 18π + 6π = 24π cc/sec

The answer is c).

• use pythagoras to artwork out the vertical top, then the equation. do u understand the radius? or the perspective of the slope? is there the different formula u can use to artwork out the two the perspective of the slope or the radius/top? (sorry, im no longer lots help. im in basic terms in year 10 in college 🙂