"The volume of a cone of radius r and height h is given by V=1/2(pi)(r^2)(h)If the radius and the height both increase at a constant rate of .5 cm per second at WHAT RATE in cubic cm per second, is the volume increasing when the height is 9 cm and the radus is 6 cm"
okay so i derived the Volume equation to get
V'= (pi/3)(2r)(dr/dt)(dh/dt) and then subsituted
V' = (pi/3)(6)(2)(.5)(.5) = Pi and that is not an answer choice
the choices are
a: .5pi
b: 10pi
c: 24pi
d: 54pi
e: 108pi
thank you!
To find the rate at which the volume of the cone is increasing (dV/dt), we can use the formula V' = (π/3)(2r)(dr/dt)(dh/dt), as you correctly derived.
Given:
- The radius (r) and height (h) both increase at a constant rate of 0.5 cm per second.
- The radius (r) and height (h) values at a specific point in time are r = 6 cm and h = 9 cm.
We need to evaluate V' when r = 6 cm and h = 9 cm.
Substituting the given values into the equation:
V' = (π/3)(2r)(dr/dt)(dh/dt)
= (π/3)(2*6)(0.5)(0.5)
= (π/3)(12)(0.5)(0.5)
= (π/3)(6)(0.25)
= (π/4)
Now, comparing this value with the answer choices, we find that (π/4) is equivalent to (0.25π). Therefore, the correct answer choice is:
a) 0.5π
So, when the height is 9 cm and the radius is 6 cm, the volume of the cone is increasing at a rate of 0.5π cubic cm per second.