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[ihatecalc]

i hate these problems, i never know how to do them
the problem states,

Mechanics 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 minutes. how rapidly is the cylinder volume increasing when the bore(diameter) is 3.800 inches?

this is what i have so far:

V = pi r^2 h
V' = 2(pi)(r)(r')(h')

h=6
r'=.001
r=3.800

where do i go from here? thank you!

 

[tkhunny]

V' = 2(pi)(r)(r')(h') Where do i go from here? thank you!

You back up and get the derivative right.

\(\displaystyle V\;=\;\pi*r^{2}*h\)

h = 6

\(\displaystyle V\;=\;6\pi*r^{2}\)

\(\displaystyle dV\;=\;12\pi*r*dr\)

Does that look a tittle easier?

Note: That r'h' thing you wrote is very worrysome. Please review the product rule for derivatives. You really were not very close. You'll need that later - but not for this problem.

 

[skeeter]

i hate these problems, i never know how to do them
the problem states,

Mechanics 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 minutes. how rapidly is the cylinder volume increasing when the bore(diameter) is 3.800 inches?

this is what i have so far:

V = pi r^2 h
V' = 2(pi)(r)(r')(h') h is a constant ... dV/dt = 2pi*r*h*(dr/dt)

h=6
r'=.001 correction ... dr/dt = .001"/3 min
r=3.800 correction ... d = 3.800", r = 1.900"

where do i go from here? substitute your known values into the equation given for dV/dt

 

[ihatecalc]

that doesnt give me the right answer.

when i have V' = 2(pi)rh dr/dt, i get .0716, when the real answer is .02something.

i think i need to find an equation where i can get dh/dt in terms of dr/dt so i can substitute. i think i do need to use the product rule, and have the derivative
V' = (pi)(r^2)(dh/dt) + (h)(2pi(r))(dr/dt)

help me please?

 

[skeeter]

that doesnt give me the right answer.

when i have V' = 2(pi)rh dr/dt, i get .0716, when the real answer is .02something.

i think i need to find an equation where i can get dh/dt in terms of dr/dt so i can substitute. i think i do need to use the product rule, and have the derivative
V' = (pi)(r^2)(dh/dt) + (h)(2pi(r))(dr/dt)

did you even bother to read my previous post?
dh/dt = 0 because the height of the cylinder is constant.

dV/dt = 2pi*r*h*(dr/dt)
dV/dt = 2pi*(1.9)*(6)*(.001/3) = approx 0.024 in³/min

 

[ihatecalc]

ohhh im sorry, i didnt see that .001 was .001/3.
thank you!!