annnnd....
a)1/2pi
b) 10pi
c) 24pi
d) 54pi
e) 108pi
explain please
a)1/2pi
b) 10pi
c) 24pi
d) 54pi
e) 108pi
Given:
- Volume of the cone, V = (1/3)Ï€r^2h
- Rate of increase of radius = dr/dt = 1/2 cm/s
- Rate of increase of height = dh/dt = 1/2 cm/s
- Height, h = 9 cm
- Radius, r = 6 cm
We want to find:
- The rate at which the volume is increasing, dV/dt (in cubic centimeters per second) when h = 9 cm and r = 6 cm.
Step 1: Find the expressions for dr/dt and dh/dt in terms of r and h.
Since both the radius and height increase at a constant rate of 1/2 cm/s, we know:
dr/dt = 1/2
dh/dt = 1/2
Step 2: Use the volume equation, V = (1/3)Ï€r^2h, to express V in terms of r and h.
V = (1/3)Ï€r^2h
Step 3: Differentiate V with respect to time (t) using the product rule.
dV/dt = (1/3)Ï€ * (2rh * dr/dt + r^2 * dh/dt)
Step 4: Substitute the given values into the expression for dV/dt.
dV/dt = (1/3)Ï€ * (2(6)(9) * (1/2) + (6^2) * (1/2))
= (1/3)Ï€ * (108 + 18)
= (1/3)Ï€ * 126
= 42Ï€
Therefore, when the height is 9 centimeters and the radius is 6 centimeters, the volume of the cone is increasing at a rate of 42Ï€ cubic centimeters per second.