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

NetherCraft 0

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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3 Answers

  • 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 🙂


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