# Calculus

I have several problems that I don't know how to do. I will post one now and hope that someone can help me.

The problem is:
A cylinder made of playing dough with dimensions h=10cm, and r=1cm, is being stretched from both ends.
(a) If V is the volume of the dough and during this stretching process the dough remains cylinder, show that (dh/dt)=(-2V)/(pi r cubed)*(dr/dt)=(-2h)/(r)*(dr/dt)
(b) If the rate of stretching is (dh/dt)=0.1 mm/s, find the rate of change of its base diameter r when h=12.1 cm.

I have no clue where to start and what to do with the problem. All I have written is a drawn image of the cylinder and the Volume equation of a cylinder which is V=(pi)*r(squared)*h.

Assume the total volume of the playdough stays the same. As the cylinder lengthens, it must get narrower.

pi r^2 h = constant = V

h = (V/pi) r^-2
dh/dt = - 2(V/pi)r^-3 (dr/dt)
= -2 (pi r^2 h/pi) r^-3 (dr/dt
= -2 (h/r) (dr/dt)

How did you get h?

Thank you very much, although I would like to know how did you get h?

h is related to the volume V (which is constant) and the variable r. When h increases, r decreases. The formula which relates h to r was provided. A formula that relates dr/dt to dh/dt was derived.

To get the function h(r), I used the constant-volume equation
pi r^2 h = constant = V
and a bit of algebra.

Alright! Thank you! And for part B. Would I just plug in the numbers?

For Part b, yes, just plug in the numbers.
dh/dt = -2 (h/r) (dr/dt)
You are given dh/dt and h, and have to first solve for r to use that formula.
Use the fact that hr^2 = constant = 10 cm^3. So when h=12.1 cm,
r = sqrt (10/12.1) = 0.9091 cm
dr/dt = (-1/2)(r/h)(dh/dt)
= (-0.5)(0.9091/12.1)(0.01 cm/s)

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