Anonymous asked in Education & ReferenceHomework Help · 1 decade ago

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 ago
    Best 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

  • Anonymous
    1 decade ago

    o *** that

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