Math 19: Calculus
Winter 2008
Instructor: Jennifer Kloke
Supplemental Notes on Related Rates
Wednesday, March 5
Remember three different types of quantities in any related rates problem. Namely:
(A) Quantities that do not change at all (ones that are the same no matter what the
time.) (B) Quantities that vary depending on time. (C) Quantities that are only true at a
specific moment in time.
Another example
Gravel is dumped from a conveyor belt at a rate of 30 f t3 /min, and it forms a pile in the
shape of a cone whose base diameter and height are always equal. How fast is the height of
the pile increasing when the pile is 10 ft. hight?
Solution:
Again, let’s label all the quantities of type A, B, and C first.
First we’re told that gravel is dumped at a rate of 30 ft3 /min. This is a quantity that
does not change when time changes so this is a quantity of type A.
Second, we’re told that the height and the base diameter are quantities in our problem.
These change with time (because more gravel is being dropped with time) so these are quantities of type B.
Third, we’re given the quantity 10 ft high at a given time - this is a quantity of type C.
Recall from class that we should assign a variable for every quantity of type B. Let’s let
h be the height of the pile and d be the base diameter of the pile.
Next we draw and label our picture with all the quantities of type A and B. I’d draw
something like:
1
Now, if any quantity of type A in our problem is a rate of change, it should be the
derivative of a variable (a quantity of type B). Our only quantity of type A is that we’re
adding gravel at 30 ft3 /min. This is a rate of change of the increase of volume of the pile
over time. We don’t have a variable for the volume of the pile so we should make one, let’s
= 30.
call it V and label it in our diagram. Now we know that dV
dt
Finally, before setting up any equation, let’s notice that what we are trying to find is the
rate of change in height when the height is 10 ft high. In other words, we want to find dh
dt
2
when the height is 10 ft. (the type C quantities come in here.)
So we’ve read the problem, introduced notation, and drawn and labeled a picture. Next
we need to create some equations that relate the variables we have (namely, h, d, and V ).
The problem tells us that the base diameter is always equal to the height so d = h. Next
we need to relate V to h or d. Recall that V is the volume of the pile (which is in the shape
of a cone. The front of the textbook tells us that V = 31 πr2 h. We don’t have r, the radius,
in our problem but we have the diameter d which is twice the radius. So d = 2r which we
can plug in to the above equation to get V = 31 π( d2 )2 h.
We’d like a single equation that relates the specific derivatives that we want to those that
and we want dh
. Let’s use the fact that d = h to rewrite the above
we know. We know dV
dt
dt
equation in terms of just V and h so that we can implicitly differentiate to get the equation
we want.
V
dV
dt
1 d 2
π( ) h
3 2
1 h 2
=
π( ) h
3 2
1
=
π(h2 /4)h
3
1 3
=
πh
12
=
=
1
dh
π3h2
12
dt
Great! We now know that at any time, the rate of change of the volume is always equal
quantity we have, we know that dV
= 30 always, so
dt
always.
1
π3h2 dh
. Moreover, from the type A
to 12
dt
1
120
then 30 = 12
π3h2 dh
or dh
= 30∗12
= πh
2
dt
dt
3πh2
in general, this would be our answer. But we want to know
If we just wanted to know dh
dt
dh
what dt is at a specific moment in time, namely when then height is exactly 10 ft. This is
where the quantities of type C come into play. At that specific moment in time, we know
120
6
that h = 10. So from the above equation dh
= π(10)
2 = 5π which is approximately .38 ft/min.
dt
3