Lesson 21
Related Rates
October 9, 2013
Lesson 21 Related Rates
1 / 10
Homework 20 Questions?
Lesson 21 Related Rates
2 / 10
Overview
This week we’re going to solve some word problems using increments and
implicit derivatives.
Recall that if f is a function of x, then
∆f ≈ f 0 (x)∆x .
We use increments when we are trying to “estimate using calculus.”
Lesson 21 Related Rates
3 / 10
Overview
This week we’re going to solve some word problems using increments and
implicit derivatives.
Recall that if f is a function of x, then
∆f ≈ f 0 (x)∆x .
We use increments when we are trying to “estimate using calculus.”
Last week we also talked about finding the derivative of both sides of an
equation. For example, if y is a function of x, and
x 2 + xy + y 2 = 10,
then,
2x + y + x
dy
dy
+ 2y
= 0.
dx
dx
Lesson 21 Related Rates
3 / 10
Example 1 (HW #1–3)
When a steel factory buys x tons of iron from Australia and y tons of iron
from Brazil, they can produce
I = 15x 2 + 3xy + 10y 2
tons of steel. Currently they buy 100 tons from Australia and 60 tons from
Brazil. A shortage reduces the amount they can buy from Australia by 10
tons; estimate the amount of additional iron needed from Brazil to keep
production constant.
Lesson 21 Related Rates
4 / 10
Example 1 (HW #1–3)
When a steel factory buys x tons of iron from Australia and y tons of iron
from Brazil, they can produce
I = 15x 2 + 3xy + 10y 2
tons of steel. Currently they buy 100 tons from Australia and 60 tons from
Brazil. A shortage reduces the amount they can buy from Australia by 10
tons; estimate the amount of additional iron needed from Brazil to keep
production constant.
We want to find ∆y given ∆x = −10, x = 100, y = 60, and I constant.
Recall
dy
∆y ≈
∆x.
dx
Lesson 21 Related Rates
4 / 10
Example 1 (HW #1–3)
When a steel factory buys x tons of iron from Australia and y tons of iron
from Brazil, they can produce
I = 15x 2 + 3xy + 10y 2
tons of steel. Currently they buy 100 tons from Australia and 60 tons from
Brazil. A shortage reduces the amount they can buy from Australia by 10
tons; estimate the amount of additional iron needed from Brazil to keep
production constant.
We want to find ∆y given ∆x = −10, x = 100, y = 60, and I constant.
Recall
dy
∆y ≈
∆x.
dx
To find
dy
, we need an implicit derivative.
dx
Lesson 21 Related Rates
4 / 10
Example 1, con’t
From the previous slide,
I = 15x 2 + 3xy + 10y 2 .
Lesson 21 Related Rates
5 / 10
Example 1, con’t
From the previous slide,
I = 15x 2 + 3xy + 10y 2 .
Since I is assumed to be constant, we can take the derivative of both sides:
0 = 30x + 3y + 3x
Lesson 21 Related Rates
dy
dy
+ 20y
dx
dx
5 / 10
Example 1, con’t
From the previous slide,
I = 15x 2 + 3xy + 10y 2 .
Since I is assumed to be constant, we can take the derivative of both sides:
0 = 30x + 3y + 3x
Solving for
dy
, we get
dx
dy
dy
+ 20y
dx
dx
dy
30x + 3y
=−
.
dx
3x + 20y
Lesson 21 Related Rates
5 / 10
Example 1, con’t
From the previous slide,
I = 15x 2 + 3xy + 10y 2 .
Since I is assumed to be constant, we can take the derivative of both sides:
0 = 30x + 3y + 3x
Solving for
dy
, we get
dx
Then
∆y ≈ −
dy
dy
+ 20y
dx
dx
dy
30x + 3y
=−
.
dx
3x + 20y
30(100) + 3(60)
(−10) = 21.2 tons.
3(100) + 20(60)
Lesson 21 Related Rates
5 / 10
Example 2 (HW #1–3)
Two people working at a factory can check x = 1.2 and y = 1.3 thousand
parts per hour, respectively. Their combined rate is:
R=
xy
.
x +y
The first employee is replaced with a new hire, who checks 200 fewer parts
per hour. Estimate how many more parts per hour the second employee
needs to check to keep their combined rate the same as before.
Lesson 21 Related Rates
6 / 10
Example 2 (HW #1–3)
Two people working at a factory can check x = 1.2 and y = 1.3 thousand
parts per hour, respectively. Their combined rate is:
R=
xy
.
x +y
The first employee is replaced with a new hire, who checks 200 fewer parts
per hour. Estimate how many more parts per hour the second employee
needs to check to keep their combined rate the same as before.
dy
dy
Again, we seek ∆y ≈
∆x. To find
, we take an implicit derivative of
dx
dx
the equation, using the fact that R is constant.
Lesson 21 Related Rates
6 / 10
Example 2 (HW #1–3)
Two people working at a factory can check x = 1.2 and y = 1.3 thousand
parts per hour, respectively. Their combined rate is:
R=
xy
.
x +y
The first employee is replaced with a new hire, who checks 200 fewer parts
per hour. Estimate how many more parts per hour the second employee
needs to check to keep their combined rate the same as before.
dy
dy
Again, we seek ∆y ≈
∆x. To find
, we take an implicit derivative of
dx
dx
the equation, using the fact that R is constant.
0=
(y + x dy
dx )(x + y ) − (1 +
(x + y )2
Lesson 21 Related Rates
6 / 10
dy
dx )(xy )
Example 2, con’t
From the previous slide, we had
0=
(y + x dy
dx )(x + y ) − (1 +
(x + y )2
Lesson 21 Related Rates
dy
dx )(xy )
7 / 10
.
Example 2, con’t
From the previous slide, we had
0=
We can solve for
dy
:
dx
(y + x dy
dx )(x + y ) − (1 +
(x + y )2
dy
dx )(xy )
dy
y2
=− 2
dx
x
Lesson 21 Related Rates
7 / 10
.
Example 2, con’t
From the previous slide, we had
0=
We can solve for
dy
:
dx
(y + x dy
dx )(x + y ) − (1 +
(x + y )2
dy
dx )(xy )
.
dy
y2
=− 2
dx
x
Then
∆y ≈ −
(1.3)2
(−0.2) = 0.235 = 235 more parts per hour .
(1.2)2
Lesson 21 Related Rates
7 / 10
Example 3 (HW #4–6)
The amount of trash generated by a community of p thousand people is
approximately W (p) = p 2 + 20p + 50 tons. If the population is currently
20,000 and is increasing at a rate of 3,000 per year, at what rate is the
amount of trash increasing?
Lesson 21 Related Rates
8 / 10
Example 3 (HW #4–6)
The amount of trash generated by a community of p thousand people is
approximately W (p) = p 2 + 20p + 50 tons. If the population is currently
20,000 and is increasing at a rate of 3,000 per year, at what rate is the
amount of trash increasing?
The key to this problem is viewing p as a function of time, p(t) = 3t + 20.
Lesson 21 Related Rates
8 / 10
Example 3 (HW #4–6)
The amount of trash generated by a community of p thousand people is
approximately W (p) = p 2 + 20p + 50 tons. If the population is currently
20,000 and is increasing at a rate of 3,000 per year, at what rate is the
amount of trash increasing?
The key to this problem is viewing p as a function of time, p(t) = 3t + 20.
By the chain rule,
dW
dW dp
=
dt
dp dt
= (2p + 20)
dp
dt
dp
= 3, so
dt
dW
(0) = 180 tons per year.
dt
When t = 0, p = 20 and
Lesson 21 Related Rates
8 / 10
Example 4 (HW #4–6)
Kleiber’s law says that the metabolic rate q of an animal is roughly M 3/4 ,
where M is the animal’s mass. A kitten’s growth rate during the first four
months is approximately linear:
3
1
t+
120
4
where M is in kilograms and t is in days. At what rate is the metabolic
rate of the kitten increasing on day 100?
M(t) ≈
Lesson 21 Related Rates
9 / 10
Example 4 (HW #4–6)
Kleiber’s law says that the metabolic rate q of an animal is roughly M 3/4 ,
where M is the animal’s mass. A kitten’s growth rate during the first four
months is approximately linear:
3
1
t+
120
4
where M is in kilograms and t is in days. At what rate is the metabolic
rate of the kitten increasing on day 100?
M(t) ≈
Again, by the chain rule,
dq
dq dM
3
1
=
= M −1/4 ·
.
dt
dM dt
4
120
When t = 100, M(100) ≈ 1.583, so
dq
(100) ≈ 0.75 · (1.583)−0.25 · (0.008) = 0.005.
dt
Lesson 21 Related Rates
9 / 10
Review
1. Finish Homework 21.
2. HW #5 is similar to Example 2.6.6 in the book, only using A = πr 2
instead of the volume of a cylinder.
3. Read p.177–180 on related rates for Friday.
Lesson 21 Related Rates
10 / 10