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A bicycle wheel has a diameter D = 60 cm and the tyre has a cross sectional radius of r = 2.5 cm. The volume V of the inner tube is given by

V = pi^2 D r^2 + (8/3*pi*r^3)

If air is being pumped into the tyre so that the volume of the inner tube is increasing at the rate of 12 cm3/sec, what is the rate at which the cross sectional radius r of the inner tube is increasing in cm/sec? (Assume the diameter D is not changing.)

Correct a decimal number to 3 significant figures.

  • calculus -

    given V as above,

    dV/dt = (2pi^2 D r + 8pi r^2) dr/dt

    now plug in the numbers, assuming that at the time of interest, r = 2.5cm :

    12 = (2pi^2*60*2.5 + 8pi*2.5^2) dr/dt
    dr/dt = 6/(25pi(6pi+1))= 0.0038 cm/s

  • calculus -

    Hi thanks for helping. But when differentiate pi^2 isn't that is 2pi? How come your answer is 2pi^2? and when different r^2 isn't that is 2r? How come your equation is dv/dt = 2pi^2 D r + 8/3*pi*r^2? Because my answer is 2pi D 2r + 8/3*pi*r*^2. anyone can explain? Thanks!

  • calculus -

    i get the answer already! thank you!

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