Worksheet 10 Answers (AC 1.8)
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School
Cornell University *
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Course
1110
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
4
Uploaded by CorporalSardine2209
Linear Approximation (Active Calculus
1.8) Solutions
Learning Objectives:
You should be able to...
Use the local linearization of a function to make predictions and approximate numerical values.
Describe the limitations of using linear approximation.
1. Thomas is roasting a 14 lb turkey as a test run for Thanksgiving. He starts roasting the turkey at noon.
At 1:00pm, he checks on the turkey using a fancy thermometer; he discovers that it has an internal
temperature of 35
.
6
◦
C,
1
and its temperature is rising at an instantaneous rate of 0
.
25
◦
C per minute.
(a) Approximate the temperature of the turkey at 1:06pm.
Answer
.
37
.
1
◦
C
at 1:06 pm.
(b) Let
I
(
t
) be the turkey’s internal temperature (in
◦
C)
t
minutes after noon. Use function notation
to express what you were told about the turkey.
Answer
.
We were told that
I
(60) = 35
.
6 and
I
′
(60) = 0
.
25.
(c) Here is a graph of
I
(
t
). Use a sketch to explain the approximation you made in
(a)
.
60
120
180
240
t
I
Answer
.
We were pretending that the graph of the turkey’s temperature was a line including
the point (60
, I
(60)) and having slope
I
′
(60). Instead of calculating
I
(66), we estimated using the
t
= 66 on the tangent line.
(d) Based on your sketch, was your approximation too high or too low?
Answer
.
Too high
.
1
According to the USDA, a turkey should be roasted to an internal temperature of 73
◦
C.
Adapted from materials created by the Cornell Active Learning in Math team and Harvard University’s Math Ma, 1a.
1/
4
Worksheet 10:
Linear Approximation (Active Calculus
1.8) Solutions
MATH 1110
(e) Would you be comfortable using the same method to predict the turkey’s temperature at 3 pm?
Why or why not?
Answer
.
No.
Although it seems pretty reasonable to say that the turkey’s temperature changes
at a roughly constant rate for 6 minutes, it seems pretty unlikely that it would change at a constant
rate for 2 hours.
2.
Interpreting rates of change.
Jasmin is interested in understanding how the ra-
dius of a balloon changes as it rises into the air.
She fills a spherical weather balloon with helium
and releases it from sea level (0 meters). As the
balloon rises, its radius can be modeled by a func-
tion
r
(
a
) where
a
is the altitude of the balloon in
meters (see figure to the right).
Note:
As the balloon rises from an altitude of
18
,
000 meters to an altitude of 25
,
000 meters, the
balloon expands from a radius of 2 meters to a
radius of 3 meters.
5000
10000
15000
20000
25000
1
1
.
5
2
2
.
5
3
3
.
5
Altitude (m)
Radius (m)
(a) What is the average rate of change of the balloon’s radius with respect to its altitude from
a
= 18
,
000
to
a
= 25
,
000? Include units in your answer and represent this rate of change on the graph above.
Answer
.
So, between 18,000 and 25,000 meters, the balloon’s radius is changing at an average of
1
7
,
000
meters of radius
meter of altitude
.
(b) After some careful calculations, the scientist determines that
r
(17
,
000) = 1
.
9 and
r
′
(17
,
000) =
1
10
,
000
. Use this data to estimate the radius of the balloon at 22
,
000 meters above sea level.
Answer
.
We expect the radius of the balloon to be around 2
.
4 meters
when the balloon is 22,000
meters above sea level.
(c) Is your answer from part
(b)
an overestimate or an underestimate of the radius of the balloon at
22
,
000 meters above sea level? Explain your reasoning.
Answer
.
The function
r
(
a
) is concave up and so the tangent line to
r
(
a
) at
a
= 17
,
000 is below
2/
4
Worksheet 10:
Linear Approximation (Active Calculus
1.8) Solutions
MATH 1110
the graph of the function. This means that the estimate from part (b) is an underestimate of the
radius.
(d) Estimate the altitude at which the balloon’s radius is 3
.
4 meters.
Answer
.
a
= 32
,
000
meters is the approximate altitude at which the balloon pops. (Using the
tangent line approximation which may not be particularly accurate.)
Linearization.
The (local) linearization
(or tangent line approximation
) of a function
f
(
x
) near
x
=
a
is
L
(
x
) =
f
(
a
) +
f
′
(
a
)(
x
−
a
)
Note:
By choosing an appropriate function
f
and an input value
a
, we can approximate complicated expres-
sions without using a calculator. That is, for
x
close to
a
, we can approximate
f
(
x
)
≈
L
(
x
).
3. Use linear approximation to estimate
√
23. Fully explain your reasoning. Compare your answer with an
estimate of
√
23 obtained by using a calculator.
Answer
.
To find a linear approximation, we can use the following steps:
(1) Choose an appropriate function
f
and a value
a
where you know
f
(
a
) precisely (and
a
is
close to the desired input).
(2) Find
f
(
a
) and
f
′
(
a
).
(3)
L
(
x
) =
f
(
a
) +
f
′
(
a
)(
x
−
a
) is the equation of the tangent line.
(4)
f
(
x
)
≈
L
(
x
) for values of
x
close to
a
.
In this case, the appropriate function is
f
(
x
) =
√
x
and
a
= 25.
Then we know
f
′
(
x
) =
1
2
√
x
and so
f
′
(25) =
1
10
. This means that the local linearization of
f
is
L
(
x
) = 5 +
1
10
(
x
−
25). Finally,
√
23 =
f
(23)
≈
L
(23) = 5 +
1
10
(23
−
25) = 4
.
8
and so
√
23
≈
4
.
8
.
A calculator returns
√
23
≈
4
.
7958315.
4. Approximate
1
4
.
9
using a local linearization. Fully explain your reasoning. Compare your answer with
an estimate of
1
4
.
9
obtained by using a calculator.
Answer
.
In this case, the appropriate function is
f
(
x
) =
1
x
and
a
= 5. Then we know
f
′
(
x
) =
−
1
x
2
and so
f
′
(5) =
−
1
25
. This means that the local linearization of
f
is
L
(
x
) = 0
.
2
−
1
25
(
x
−
5). Finally,
1
4
.
9
=
f
(4
.
9)
≈
L
(4
.
9) = 0
.
2
−
1
25
(4
.
9
−
5) = 0
.
2 + 0
.
004 = 0
.
204
3/
4
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Worksheet 10:
Linear Approximation (Active Calculus
1.8) Solutions
MATH 1110
and so
1
4
.
9
≈
0
.
204
.
A calculator returns
1
4
.
9
≈
0
.
2040816.
4/
4