Need help with an implicit differentiation/related rate word problem?

Question

A machine is rolling a metal 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 128pi cubic inches. At what rate ios the length h changing when the radius r is 2.5 inches? [hint: V = pi(r^2)h]

please explain the problem. just an answer wont help

Answer 1

V= pi*r^2*h = 128
dV= pi(h* 2r *dr +r^2dh)=0
dh/dt= -2h/r dr/dt
you have to calculate h when r= 2.5
128= pi(2.5)^2*h so h= 6.52 inch
dh/dt= 2*6.52/2.5*0.05= 0.261 inch/s (positive) h increasing

Answer 2

Related Rate Word Problems

Answer 3

You're given that V is constant, at 128π cubic inches. Therefore

π r² h = 128π
r² h = 128

You're given that dr/dt = -0.05 and you are asked to find dh/dt when r = 2.5. So take the derivative with respect to t of the volume relation above:

d/dt (r² h) = d/dt(128)

2r dr/dt h + r² dh/dt = 0

dh/dt = (-2h dr/dt)/r

To evaluate this, you will need to find out what h is when r = 2.5 (by using r² h = 128)

Alternatively, you could solve r² h = 128 for h:

h = 128/r² = 128 * r^(-2) Then differentiate with respect to t:

dh/dt = 128*(-2)*r^(-3) dr/dt

and evaluate, using the given values for r and dr/dt

Answer 4

dD/dt = -0.3, even in spite of the indisputable fact that, by technique of creating use of actuality the question is searching how straight away it fairly is reducing (that be wide awake's significant), we are able to assert that dD/dt = 0.3 (it is, it fairly is increasing at -0.3 cm/min or in diverse words reducing at 0.3 cm/min) V = (4/3) * pi * r^3 V = (4/3) * pi * (D/2)^3 V = (4/3) * pi * (a million/8) * D^3 V = (4/8) * (a million/3) * pi * D^3 V = (pi/6) * D^3 dV/dt = 3 * (pi/6) * D^2 * dD/dt dV/dt = (pi/2) * D^2 * dD/dt D = 17 dD/dt = 0.3 dV/dt = (pi/2) * 17^2 * 0.3 dV/dt = (pi/2) * 289 * (3/10) dV/dt = (3pi/20) * 289 dV/dt = 867 * pi / 20