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```Differentials
∆

When we first started to talk about derivatives, we said that ∆ becomes
when the change in x and change in y become very small.
dy can be considered a very small change in y.
dx can be considered a very small change in x.
Let y  f  x  be a differentiable function.
The differential d x is an independent variable.
The differential dy is:
dy  f   x  dx
dy =

Related Rates
If we are pumping air into a balloon, both the volume and the radius of
the balloon are increasing and their rates of increase are related to each
other.
However, it is much easier to measure directly the rate of increase of the
volume than the rate of increase of the radius.
In this section, we will learn:
How to compute the rate of change of one quantity
in terms of that of another quantity.
In a related-rates problem, the idea is to compute the rate of change of one
quantity in terms of the rate of change of another quantity—which may be
more easily measured.
The procedure is to find an equation that relates the two quantities and
then use the Chain Rule to differentiate both sides with respect to time.
Example 1: Consider a sphere of radius 10cm.
If the radius changes 0.1cm (a very small
amount) how much does the volume change?
= 0.1
V 
4
r
=?
3
3
dV  4 r dr
2
dV  4 10cm   0.1cm
2
dV  40 cm
3
The volume would change by approximately 40 cm 3
Example 2: Now, suppose that the radius is changing at an instantaneous
rate of 0.1 cm/sec. At what rate is the sphere growing when r = 10 cm?

= 0.1/ sec

V 
4
r

=?

3
3
dV
 4 r
2
dt
dV
dt
dV
dt
dr
dt
 4 10cm 
 40
cm
2
cm 

  0.1

sec


3
sec
The sphere is growing at a rate of 40 cm 3 / sec
Example 3: Air is being pumped into a spherical balloon so that its
volume increases at a rate of 100 cm3/s. How fast is the radius of the
balloon increasing when the diameter is 50 cm?
V 
4
r
3
3
dr
dt

1
4 (25)
2
100 
1
25
The radius of the balloon is increasing
at the rate of 1/(25π) ≈ 0.0127 cm/s.
Example 4: Water is draining from a cylindrical tank at 3 liters/second.
How fast is the surface dropping?
dV
L
 3
dt
  3000
V r h
r
dt
dh
dt
2
r is constant
dh
dt
3000

3
sec
sec
2
dV
cm
cm
3
sec
r
2
dh
dt
=?
Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.

Example 4: Hot Air Balloon Problem
@  

d
4
dt
 0.14
m in
h
How fast is the balloon rising?

h
tan  
500ft
500
d
sec 
2
dt
 

sec


4

 
2
2

1
dh
500 dt
2
 0.14  
1
dh
500 dt
 0.14   500 
dh
dt
Example 5: Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the trucks changing 6 minutes later?
@ 6 min x= 4 mi & y = 3 mi
from initial position
z2 = x2 + y2
@ 6 min z = 5 mi

2
= 2
+ 2

5
= 4 ∙ 40 + 3 ∙ 30

= 50 /ℎ

=
Example 6: Car A is traveling west at 50 mi/h and car B is traveling north
at 60 mi/h. Both are headed for the intersection of the two roads.
At what rate are the cars approaching each other when car A is 0.3 mi
and car B is 0.4 mi from the intersection?
dx / dt = –50 mi/h & dy / dt = –60 mi/h.
both x & y are decreasing

=?

z2 =
x2 +
y2

⇒ 2
= 2
+ 2

When x = 0.3 mi & y = 0.4 mi, z = 0.5 mi.
```