# Hi, I am working on my Calculus hw. and I can't seem to solve this problem?

The mechanics at Lincol Automotive are reboring a 6 in. deep cylinder to fit a new piston. The machine they are using increases the cylinder's radius one-thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter0 is 3.800 in?

Can anyone please explain how to solve this problem? I would truly appreciate it. Thanks :]

### 3 Answers

- 1 decade agoBest Answer
The depth of the cylinder is constant (d = 6"). The radius is increasing at a constant rate (dR/dt = .000333 inches per minute)

The volume of a cylinder is given by

V = PI*R^2*h. Since everything in the equation is constant with respect to time except radius (R)

Let A = dR/dt. Now integrate A to find that R(t) = Ro + A*t

Now input R(t) into the volume equation.

V = PI*(Ro + At)^2*h

now differentiate V with respect to t.

dV/dt = PI*h*2*(Ro + At)*A

Now all you need to do is find the time (t) at which the diameter is equal to 3.8"m, which is equivalent to the time at which R = 1.9, or when

Ro + (A*t) = 1.9

Solving the above for t gives

t = (1.9 - Ro)/A

Thus, by substitution, dV/dt when d = 3.8 is given by:

dV/dt = A*PI*h*2*(Ro + A((1.9 - Ro)/A)) the A's and Ro inside the parenthesis cancel down to:

dV/dt = A*PI*h*2*1.9, or plugging back in the numbers,

dV/dt = ( 0.000333 in/s ) * PI * ( 6 in ) * 2 * ( 1.9 in )

dV/dt = 0.0238 in^3/s

- 1 decade ago
suppose r =3.8/2 in, h=6in

V1 = 3.14* r^2 *h

V2 = 3.14* (r+.001)^2 *h

rate of increase of volume when bore diam is 3.8 inch

= (V2 - V1)/3 cubic inch per min

- Anonymous1 decade ago
o *** that