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! 🙂
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 🙂