We’ll start by going over the problem from last week:
Problem 0.1.
Find the second derivative d
2 y dx 2 at an arbitrary place on the curve y 2 = x 3 + 5 x
−
6 .
We have that
2 y dy dx
We can differentiate again to get:
= 3 x 2 + 5 .
2 dy dx
2
+ 2 y d 2 y dx 2
= 6 x and solve for d
2 y dx 2
: d 2 y dx 2
=
6 x
−
2
2 y dy dx
Now, we plug in the expression for dy dx to get
2 d 2 y dx 2
=
6 x
−
2
2 y
3 x
2
+5
2 y
2
.
The point is that you can use implicit differentiation to find second derivatives too.
Now, we’re ready to discuss the first applied topic of the course: related rates.
The basic idea is the following: often you are given one rate of change and wish to find another:
Example 0.2.
Someone is pouring water into a conical tank. Suppose the person pours water in at 1 cm 3 /sec . At a certain point in time, the water has reached a height of 5 cm and a radius of 3 cm . How quickly is the water level rising?
To do this problem, we need to find a relationship between first step is to find a relationship between V and h : dV dt and dh dt
. The
V =
1
3
πr 2 h.
This is a start, but how is h related to r ? Use similar triangles: r =
Therefore,
V =
3
25
πh 3
3 h
5
.
1

so dv dt
=
9
25
πh 2 dh dt
.
Now, plug in h = 5 and dh dt
= 1.
This is a related rates problem- we can find the rate of change of a quantity using another rate of change.
Example 0.3.
A rectangle is being “stretched out” with constant area. Suppose the rectangle is 3 cm -by1 cm and the longer side is increasing at a rate of 1 cm/s .
How quickly is the shorter side shrinking?
Have A = ℓ
· w = 3. Thus w =
3 ℓ and
We plug in ℓ = 3 and dw dt dℓ dt
= 1 to get
=
−
3 ℓ 2 dℓ dt dw dt
=
−
1
3 cm/s.
Let’s do a more complicated example where the function isn’t a polynomial:
Example 0.4.
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
R =
1
1
R
1
+ 1
R
2 how quickly is the equivalent resistance changing?
To solve this problem, we take a time derivative and notice that R
2 a constant.
= 5Ω is
R =
1 x
1
+ 1
5 so dR dt
=
−
1 x
+
1 s
−
2
· which is equal to
1 /x 2
(1 /x + 1 / 5) 2 dx dt
Now, we plug in x = 10 and dx dt
=
−
2 to get
1 x 2
1 / 100
(1 / 10 + 1 / 5) 2
· −
2 which comes out to −
9
2 Ω / min.
· dx dt
2

Example 0.5.
You’re running at 5 km/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 2 = x 2 + y 2 by Pythagoras. So by differentiating:
2 D dD dt
= 2 x dx dt so dD dt
=
Now, use D = 50 and x =
√
2499 to get x
D dx dt
√
2499
10 km/hr .
3