Calculus/Rate of increase in volume and radius
Expert: Ahmed Salami - 9/7/2010
QuestionHi, I have this question in my physics book that says (I
promise I didn't make this scenario up, i'm not that
creative) "A child loves to watch as you fill a transparent
plastic bottle with shampoo. Every horizontal cross section
of this bottle is circular, but the diameters of the circles
have different values. You pour the brightly colored shampoo
into the bottle at a constant rate of 16.5 cm^3/s. At what
rate is its level in the bottle rising at a point where the
diameter of the bottle is 6.3 cm, what about 1.35 cm?"
I know the answer is found using derivatives, but I think
I've hit a wall as I am pretty stuck on this one. Any help
would be GREATLY appreciated.
AnswerHi David,
The rate at which the level h rises would depend on the relationship between the volume V at any time and the level and diameter d. The diameter might even depend on the level of liquid and so we could write the volume as a function of level only (to avoid resulting to partial derivatives).
V = f(h)
Finding the time derivatives,
dV/dt = f'(h) . dh/dt
where f'(h) is the derivative of V with respect to h.
We have the value of dV/dt but its clear that we cannot proceed since we do not have f'(h) by not knowing the relationship between V and h.
If we had, say, a cylindrical or cubic or even conical bottle then its quite straightforward to figure out.
You can always get back to me.
Regards