We’ll start today with a warm-up problem in 4 parts. I will give you five
minutes to attempt it the first two parts of the problem. We will then do part 3
collectively. In part 4, we will compare your answer to part 3 with your answers
to part 1 and 2.
Example 0.1. You’re filling up a pyramidal container with water. The container is 10cm tall, and the base has an area of 100cm.
• Without calculating, will the water level rise faster when the tank is more
full, or when the tank is less full?
• Intuitively, how do you expect the rate of change of water level to depend
on the height h? Do you think the dependence should be constant, linear,
quadratic, or something else? Why?
• You’re pouring water in at a rate of 5cm3 /s. The current water level is
3cm. How quickly is the water level increasing? The volume of a square
pyramid is 13 bh, where b is the area of the base. Hint: The water level
in the pyramid can be obtained by subtracting the “empty space” from the
full pyramid.
• Was your guess of the dependence of dh
dt correct? Explain in your own
depends
on
h
the
way it does.
words why you think dh
dt
Now, let’s start solving the problem. We pour water into the pyramidal tank
3
at a rate of 5cm3 /s. Note that the volume of the entire container is 1000
3 cm . If
the water has height h, the “empty space” has volume equal to the base times
the height squared- the base will have area equal to (10 − h)2 , and the height is
3
(10 − h), so the volume comes out to (10−h)
. Therefore, the volume of water
3
in the tank is
1000 (10 − h)3
V =
−
3)
3
. Taking the derivative, we get
dV
dh
= (10 − h)2
dt
dt
so
dh
dV /dt
.
=
dt
(10 − h)2
In other words, there is an inverse-square dependence of dh
dt on the quantity
dV
cm3
10 − h. We can now plug in the value dt = 5 s and h = 3cm to get that
3
/s
5
f racdhdt = 5cm
49cm2 = 49 cm/s.
Now, try to explain in your own words why you think there was an inverse
square dependence on the quantity 10 − h.
Example 0.2. Consider two resistors connected in parallel. One resistor is
variable, and the other resistor has constant resistance 5Ω. The resistance of
the variable resistor is decreasing at a rate of 2Ω/min. Given that the equivalent
resistance of the circuit is
1
R= 1
1
R1 + R2
1
how quickly is the equivalent resistance changing?
To solve this problem, we take a time derivative and notice that R2 = 5Ω is
a constant.
R=
1
x
1
+
1
5
so
−2 dx
dR
1 1
1
=−
+
· 2 ·
dt
x s
x
dt
which is equal to
dx
1/x2
2
(1/x + 1/5) dt
Now, we plug in x = 10 and
dx
dt
= −2 to get
1/100
· −2
(1/10 + 1/5)2
which comes out to
−2
9 Ω/min.
Example 0.3. You’re running at 5km/hr. You run on a road that is 1 meter
away from a tree. You run past the tree. How quickly is your distance from the
tree changing when you’re 50 meters away from the tree?
We have
D = x2 + y 2
by Pythagoras. So by differentiating:
2D
so
Now, use D = 50 and x =
√
dD
dx
= 2x
dt
dt
dD x dx
=
dt
D dt
2499 to get
2
√
2499
10 km/hr.