A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 in per second and the volume V is 128pi cubic in. At what rate is the length h changing when the radius r is 1.8 in?

Question

earthcitizen · Answer

$$V=A*h$$

earthcitizen · Answer

can you state options to the question ?

mathmate · Answer

$$V=\pi r^2h$$
Assuming volume is constant (i.e. material incompressible,
differentiate both sides with respect to t (time)
$$\frac{dV}{dt} = \pi r(2\frac{dr}{dt}+r \frac{dh}{dt})$$
Since V is constant,  dV/dt =0, so we get
$$2\frac{dr}{dt}+r \frac{dh}{dt} = 0$$
Now r and dr/dt are both known, can you solve for dh/dt?

mathmate · Answer

Sorry, the first equation should read:
$$\frac{dV}{dt} = \pi r(2h \frac{dr}{dt}+r \frac{dh}{dt})$$
and the second:
$$2h \frac{dr}{dt} + r \frac{dh}{dt} = 0$$
h can be calculated from
V=pi r^2h
h=V/(pi r^2)

earthcitizen · Answer

@mathmate, what methods was used here ?

mathmate · Answer

It is basically partial differentiation, since both h and r are functions of time.
The differentiation is done similar to the differentiation of products, d(uv)=vdu+udv while holding the other variable constant.

earthcitizen · Answer

the rhs was derived by implicit differentiation ?

mathmate · Answer

It is slightly different.
Implicit differentiation usually involves only two variables, while here we have a third variable which is the independent variable, both x and y are dependent on t.

The tree diagram looks like:
V - r  - t
    \ h  - t
That is why we need to differentiate both r and h with respect to t, keeping the other variable constant.

mathmate · Answer

Perhaps a Wiki article can explain better than I can: http://en.wikipedia.org/wiki/Partial_derivative I apologize for not having used the partial derivative notaion as in: $$2h \frac{\partial r}{\partial t} + r \frac{\partial h}{\partial t} = 0$$

earthcitizen · Answer

smashing!