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…
If you could show the work you did to get the answer that would be VERY helpful! 🙂
r = base radius
h = height
V = volume
dr/dt = 1/2 cm/sec
dh/dt = 1/2 cm/sec
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 🙂