A machine lengthens a metal cylinder by rolling it under pressure. The radius of the cylinder decreases at a constant rate of 0.05 inches per second while the volume stays constant at 128π cubic inches. At what rate is the length L changing when the radius r is 1.8 inches? [Note: V = πr 2L.]
V = pi r^2 L
ln V = ln pi + 2 lnr + lnL
d/dt (lnV) = 0 + (2/r) dr/dt + (1/L) dL/dt
dL/dt = - (2L/r) dr/dt
When r = 1.8,
128 pi = pi*(1.8)^2 L
L = 39.51 inch
You know dr/dt. Use the dL/dt formula
thanks that helps alot
I forgot to add this step:
0 = (2/r) dr/dt + (1/L) dL/dt
which is true because V does not change.
To find the rate at which the length L is changing, we can use implicit differentiation. Since the volume V is constant, we can differentiate both sides of the volume equation with respect to time (t) using the chain rule.
Given that V = πr^2L, we have:
dV/dt = 0
Differentiating both sides with respect to t:
d/dt (πr^2L) = d/dt (0)
Using the chain rule, we get:
dV/dt = π(2r * dr/dt * L + r^2 * dL/dt)
Now, we substitute the given values into the equation. Since the radius r is decreasing at a constant rate of 0.05 inches per second, we have:
dr/dt = -0.05 inches per second
The volume V is given as 128π cubic inches, so:
π(2r * dr/dt * L + r^2 * dL/dt) = 0
Substituting r = 1.8 inches and solving for dL/dt, we can compute the rate at which the length L is changing:
π(2(1.8) * (-0.05) * L + (1.8)^2 * dL/dt) = 0
Simplifying the equation with the given values:
π(-0.18 * L + 3.24 * dL/dt) = 0
Dividing both sides by π:
-0.18L + 3.24dL/dt = 0
Rearranging the equation:
3.24dL/dt = 0.18L
Finally, solving for dL/dt:
dL/dt = 0.18L / 3.24
When the radius r is 1.8 inches, we can substitute this value into the equation to calculate the rate at which the length L is changing.