A machine is a rolling a metel cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume V is 128π cubic inches. At what rate is the length h changing when the radius r is 2.5 inches? [Hint: V = (π)(h)(r^2)]
2007-11-22 12:29:58 · 2 answers · asked by Oscar A 2 in Science & Mathematics ➔ Mathematics
V is a constant, so we have
π h r^2 = 128π
so h = 128 / r^2
so dh/dt = (128 (-2) / r^3) dr/dt
= (-256 / (2.5)^3) (-0.05)
= 0.82 inches per second (2 s.f.)
2007-11-22 12:38:25 · answer #1 · answered by Scarlet Manuka 7 · 1⤊ 0⤋
first, we have the equation V = PI * h * r^2. let's rearrange it,
h = V/( PI * r^2) right? then differentiate with respect to t (time) you'll get dh/dt = -2 V/ PI * r ^-3 * dr/dt , we know that dr/dt is -0.05inch/sec, r is 2.5 inches, so put them in the equation and solve for dh/dt, you'll get dh/dt = 0.82 inches/sec
2007-11-22 12:44:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋