A potter forms a piece of clay into a right circular cylinder. As she rolls it, the height h of the cylinder increases and the radius r decreases. Assume that no clay is lost in the process. Suppose the height of the cylinder is increasing by 0.5 centimeters per second. What is the rate at which the radius is changing when the radius is 3 centimeters and the height is 8 centimeters?
So the volume is a constant ----> d(volume)/dt = 0
V = πr^2 h
given: dh/dt = .5 cm/s
find: dr/dt when r = 3 and h = 8
dV/dt = πr^2 dh/dt + h(2πr) dr/dt
0 = π(9)(.5) + 8(6π)dr/dt
dr/dt = -4.5π/48π = -3/32 cm/s
the radius is decreasing at a rate of 3/32 cm/s
THANK U!!!!!!
Well, it seems like the potter is in quite the spin! Let's use some math to figure out the rate at which the radius is changing.
We know that the volume of a cylinder is given by the formula V = πr²h, and it remains constant in this case. So, we can differentiate both sides of the equation with respect to time t to find the rate of change.
dV/dt = d(πr²h)/dt
Since the height h is increasing by 0.5 centimeters per second, we can substitute dh/dt = 0.5 into the equation:
dV/dt = π(r²)(0.5)
Now, we need to figure out the rate at which the radius is changing when the radius is 3 centimeters and the height is 8 centimeters.
At this given moment, we can substitute r = 3 and h = 8 into the equation:
dV/dt = π(3²)(0.5)
Simplifying, we get:
dV/dt = 9π(0.5)
And finally, evaluating the expression, we find:
dV/dt = 4.5π
So, when the radius is 3 cm and the height is 8 cm, the rate at which the radius is changing is approximately 4.5π cm²/s.
Hope that brings a smile to your face, even if it's as twisted as the potter's clay!
To find the rate at which the radius is changing, we can use the relationship between the radius and the height of the cylinder. We know that the volume of a right circular cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.
Differentiating this equation with respect to time will give us the rate at which the volume is changing with respect to time. Since we are given the rate at which the height is changing, we can solve for the rate at which the radius is changing.
We have:
V = πr^2h
Differentiating both sides of the equation with respect to time (t):
dV/dt = d(πr^2h)/dt
dV/dt = π(2rh(dr/dt) + r^2(dh/dt))
Since we are looking for the rate at which the radius is changing (dr/dt) when the radius is 3 centimeters (r = 3 cm) and the height is 8 centimeters (h = 8 cm), we can substitute the known values into the equation.
Substituting the given values:
dV/dt = π(2(3)(8)(dr/dt) + (3^2)(0.5))
dV/dt = π(48(dr/dt) + 9(0.5))
We can simplify the equation by using the known values and simplifying the expression:
dV/dt = π(48(dr/dt) + 4.5)
Finally, we can solve for the rate at which the radius is changing (dr/dt) when the radius is 3 centimeters and the height is 8 centimeters.