Lesson 21
Approximation
By Increments
Related Rates
Section 2.6 (cont.)
March 3rd, 2014
Lesson 21
Approximation
By Increments
Related Rates
In this lesson we will combine implicit differentiation with
approximation by increments in order to answer certain
questions.
We will also introduce the idea of related rates and solve
various real world problems with this concept.
Lesson 21
Example
Approximation
By Increments
Related Rates
The output at a certain plant is Q = 6x 2 + 3xy + 4y 2 units per
day, where x is the number of hours of skilled labor used and y
is the number of hours of unskilled labor used. Currently, 60
hours of skilled labor and 80 hours of unskilled labor are used
each day. Use calculus to estimate the change in unskilled
labor that should be made to offset a 1-hour increase in skilled
labor so that the output will be maintained at its current level.
When x = 60 and y = 80, we have
Q = 6(60)2 + 3(60)(80) + 4(80)2 = 61, 600.
Thus the equation becomes 61, 600 = 6x 2 + 3xy + 4y 2 , and so
y is defined implicitly as a function of x.
Lesson 21
Approximation
By Increments
Related Rates
The question says to ”use calculus to estimate the change in
unskilled labor”. Since y measures the hours of unskilled labor,
we must estimate ∆y . The phrase ”use calculus” means that
we should use our method of approximation by increments.
∆y ≈
dy
|x=x0 ,y =y0 ∆x
dx
To compute dy
dx , we use implicit differentiation. Remember that
the problem asked us to estimate the change in y based on a
certain change in x ”so that the output will be maintained at
its current level”. This means that Q is a constant, and
therefore dQ
dx = 0.
Lesson 21
dy
dy
+ 8y
dx
dx
dy
⇒ −12x − 3y = (3x + 8y )
dx
dy
−12x − 3y
⇒
=
dx
3x + 8y
Q = 6x 2 + 3xy + 4y 2 ⇒ 0 = 12x + 3y + 3x
Approximation
By Increments
Related Rates
According to the statement, the initial value of x is 60 and the
initial value of y is 80, i.e. x0 = 60, y0 = 80. Also, we are
offsetting a 1-hour increase in skilled labor, so ∆x = 1.
−12(60) − 3(80)
dy
∆y ≈
|x=x0 ,y =y0 ∆x =
(1)
dx
3(60) + 8(80)
48
=−
hours
41
Lesson 21
Example
Approximation
By Increments
Related Rates
At a certain factory, output Q is related to inputs u and v by
the equation
3u + 4v
Q = 5u 2 +
(u + v )2
If the current levels of input are u = 5 and v = 10, use calculus
to estimate the change in input v that should be made to
offset a decrease of 0.7 unit in input u so that output will be
maintained at its current level.
We must use quotient rule when differentiating here.
0 = 10u +
(u + v )2 (3 + 4 dv
du ) − (3u + 4v )[2(u + v )(1 +
(u + v )4
dv
du )]
Lesson 21
Approximation
By Increments
Related Rates
We could solve this equation for dv
du , but doing so would be
cumbersome. Since we are interested in the value of dv
du when
u = 5 and v = 10, we can also just plug those values in for u
and v and then solve for dv
du :
0 = 10(5) +
(15)2 (3 + 4 dv
du ) − 2(55)(15)(1 +
(15)4
= 50 + (15)
(15)(3 + 4 dv
du ) − 110(1 +
(15)4
dv
45 + 60 dv
du − 110 − 110 du
= 50 +
(15)3
65
50
dv
= 50 −
−
3
3
(15)
(15)
du
dv
du )
dv
du )
!
Lesson 21
Thus
Approximation
By Increments
dv
du
3
65
65
3
= − (15)
50 ( (15)3 − 50) = − 50 + (15) =
Therefore
Related Rates
∆v ≈
dv
∆u
du u=5,v =10
=
33, 737
(−0.7)
100
=−
236, 159
1000
33,737
10 .
Lesson 21
Approximation
By Increments
Related Rates
A related rates problem is one where we are given an equation,
told the rates of change of some of the variables, and asked to
find the rate of change of the remaining variable (thus relating
the rates). We will use the following method for solving these
problems.
Solving Related Rates Problems
(1) Draw a figure (if appropriate) and assign variables.
(2) Find a formula relating the variables.
(3) Use implicit differentiation to find how the rates are
related.
(4) Substitute any given numerical information into the
equation in step (3) to find the desired rate of change.
Lesson 21
Example
Approximation
By Increments
Related Rates
Boyle’s law states that when gas is compressed at a constant
temperature, the pressure P and volume V of a given sample
satisfy the equation PV = C , where C is a constant. Suppose
that at a certain time the volume is 40 in3 , the pressure is 70
lb/in2 , and the volume is increasing at a rate of 12 in3 /sec.
How fast is the pressure changing at this instant? Is it
increasing or decreasing?
Both P and V are functions of time t, and we are asked to find
how fast the pressure P is changing. So we must compute dP
dt .
Step 1: There is no picture to draw here.
Step 2: We are given the equation PV = C .
Lesson 21
Step 3: We now use implicit differentiation to relate
dV
dt .
Approximation
By Increments
Related Rates
PV = C
⇒
dV
dP
V +P
=0
dt
dt
⇒
dP
dt
dP
P
=−
dt
V
and
dV
dt
Step 4: The problem asks us to find how fast the pressure is
changing when the volume is 40 in3 , the pressure is 70 lb/in2 ,
and the volume is increasing at a rate of 12 in3 /sec, i.e. when
V = 40, P = 70, and dV
dt = 12. We plug these values into the
equation in step (3):
70
lb
dP
= − (12) = −21
/sec
dt
40
in2
Thus the pressure is decreasing.
Lesson 21
Example
Approximation
By Increments
Related Rates
The basal metabolic rate is the rate of heat produced by an
animal per unit time. Observations indicate that the basal
metabolic rate of a warm-blooded animal of mass w kilograms
(kg) is given by
M = 70w 3/4 kilocalories per day
a. Find the rate of change of the metabolic rate of an 80-kg
cougar that is gaining mass at the rate of 0.8 kg per day.
b. Find the rate of change of the metabolic rate of a 50-kg
ostrich that is losing mass at the rate of 0.5 kg per day.
Lesson 21
Approximation
By Increments
Related Rates
Again we are given the equation, and drawing a picture won’t
help, so we skip straight to differentiating.
105
dw
dM
=
M = 70w 3/4 ⇒
dt
2w 1/4 dt
105
2(80)1/4
(0.8) =
105
2(50)1/4
−9.87 kCal
day
(−0.5) =
a.
dM
dt
=
b.
dM
dt
=
105
√ (4)
445 5
=
21
√
4
5
105
√ (− 1 )
2
2 4 50
≈ 14.04 kCal
day
=−
105
√
4 4 50
≈
Lesson 21
Approximation
By Increments
Related Rates
Example
At a certain factory, output is given by Q = 60K 1/3 L2/3 units,
where K is the capital investment (in thousands of dollars) and
L is the size of the labor force, measured in worker-hours. If the
output is kept constant, at what rate is the capital investment
changing at a time when K = 8, L = 1, 000, and L is increasing
at the rate of 25 worker-hours per week?
Lesson 21
Approximation
By Increments
Related Rates
Q = 60K
1/3 2/3
L
1/3
dK
dL
K
+ 40
⇒ 0 = 20
dt
L
dt
2/3
1/3
dK
dL
L
K
⇒ 20
= −40
K
dt
L
dt
K dL
dK
⇒
= −2
dt
L dt
L
K
2/3
Since K = 8, L = 1, 000, and dL
dt = 25, we have
dK
8
400
= −2
(25) = −
= −0.4 = −$400
dt
1000
1000