The mechanics at lincoln automotive are reboring a 6-in. deep cylinder to fit a new piston. The machine they are using increases the cylinder's radius 0.02 inches each second. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.8 in.?
I am wondering at what rate the bore increasesdepth for each .02 in radius? Boring involving going two ways, out, and down.
To find out how the cylinder volume is changing, you need to find the rate at which its volume is increasing, which is given by dV/dt (the derivative of volume with respect to time).
Let's set up the problem:
Given:
- The cylinder has a constant depth: depth = 6 inches
- The radius is increasing over time: dr/dt = 0.02 inches/second
- You want to find dV/dt when the bore (diameter) of the cylinder is 3.8 inches.
To solve this problem, you'll need to use the formulas for the volume and radius of a cylinder.
1. Volume of a cylinder: V = πr²h, where r is the radius and h is the height (depth in this case).
2. From the given information, you have dr/dt, which gives you the rate at which the radius is changing. To relate this with dV/dt, you need to find a relationship between r, V, and t.
Now, let's differentiate the formula for the volume of a cylinder with respect to time t using the chain rule.
dV/dt = dV/dr * dr/dt
To find dV/dr (the derivative of volume with respect to radius), differentiate the volume formula with respect to r:
dV/dr = d/dt (πr²h)
= 2πrh
Now, substitute the given values and calculate dV/dr.
Recall:
- r (radius) = diameter/2 = 3.8/2 = 1.9 inches
- h (depth) = 6 inches
- dr/dt = 0.02 inches/second
Substituting these values into the formula:
dV/dr = 2π(1.9)(6)
≈ 22.68π cubic inches
Now, you can find dV/dt by substituting the values of dr/dt and dV/dr:
dV/dt = (dV/dr) * (dr/dt)
≈ 22.68π * (0.02)
≈ 0.4536π cubic inches/second
Hence, the cylinder volume is increasing at a rate of approximately 0.4536π cubic inches per second when the bore (diameter) is 3.8 inches.