Week #4 - Interpreting Derivatives, Local Linearity, Marginal Rates
Section 3.9
Some material from “Calculus, Single and MultiVariable” by Hughes-Hallett, Gleason,
McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
TEST PREPARATION PROBLEMS
12. (a) To sketch this you would need to compute the y values for f (x) for several points.
(b) The local linearization (or tangent line) to f (x) at x = 2 will be defined by
f (x) = x3 − 3x2 + 3x + 1
at x = 2, f (2) = 23 − 3 · 22 + 3(2) + 1 = 3
f ′ (x) = 3x2 − 6x + 3
f ′ (2) = 3(2)2 − 6(2) + 3 = 3
The tangent line is therefore
y = 3(x − 2) + 3
(c) See diagram.
13. (a)
so at x =
π
,
4
f (x) = cos(x)
π 1
=√
f
4
2
The tangent line at x = π/4 is therefore
π
1
−1 +√
y = √ x−
4
2
2
1
f ′ (x) = − sin(x)
π
1
= −√
f′
4
2
(b) See the graph below.
The tangent line will produce overestimates of f (x) near x = π/4.
(c) The maximum error will occur at the ends of the interval
x ∈ [0, π/2].
−1 π
1
At x = 0 ,the error is cos(0) − √ 0 −
+√
≈ −0.262
4
2
2
−1 π π 1
At x = π/2 ,the error is cos(π/2) − √
+√
−
≈ −0.152
4
2 2
2
Since error is usually measured in absolute values, and the -0.262 value is largest in
absolute value, we can say
|Error| ≤ 0.262 for x ∈ [0, π/2]
14. (a)
so at x = 0,
f (x) = (1 + x)k
f ′ (x) = k(1 + x)k−1
f (0) = 1k = 1
f ′ (0) = k(1k−1 ) = k
so the tangent line at x = 0 will be
y = k(x − 0) + 1 or y = 1 + kx
√
(b) As an estimate for the square root of 1.1, we could note that 1.1 = (1 + 0.1)1/2 .
This matches exactly the form of f (0.1) if we choose k = 12 . From our linearization
above,
1
f (0.1) ≈ 1 + (0.1) = 1.05
2
√
so yes, a good approximation for 1.1 is 1.05. (Calculator gives the value of ≈ 1.0488.
(c) Since the square root function is concave down,
√ the tangent line will produce overestimates
of the actual function. The actual value of 1.1 will be slightly less than 1.05.
15. The value of a is the x value where the linearization is done, or where we draw the line
tangent to the graph a = 1.
f (a) is the y value at the linearization point. However, this is not shown on the graph.
We can find it, though, by working backwards from the point/slope linearization formula.
y = 2x − 1 = f ′ (1)(x − 1) + f (a)
2
Since f ′ (1) must equal 2, 2x − 1 = 2(x − 1) + f (a), or f (a) = +1.
The approximation will produce underestimates of f (x), because the tangent line lies
below the actual curve.
f (1.2) ≈ 2(1.2) − 1 = 1.4
16. This question essentially asks for the one iteration of Newton’s method. To start, we
re-arrange the terms so they are all on the left.
x
x − 2} = 0
|e +{z
f (x)
The question specifies that we use the tangent line at x = 0 to approximate the solution,
so
and at x = 0,
f (x) = ex + x − 2
f ′ (x) = ex + 1
f (0) = e0 + 0 − 2 = −1
f ′ (0) = e0 + 1 = 2
The tangent line at x = 0 is g(x) = 2x − 1, and this function has its root at 0 = 2x − 1
or x = 21 . Thus an (approximate) solution to the equation given would be x = 1/2.
17. Similar to #16, we define f (x) = x + ln(1 + x) − 0.2. The question specifies that we use
the tangent line at x = 0 to approximate the solution, so
f ′ (x) = 1 +
f (x) = x + ln(1 + x) − 0.2
and at x = 0,
f (0) = 0 + ln(1) − 0.2 = −0.2
1
1+x
f ′ (0) = 1 +
1
=2
1
The tangent line at x = 0 is g(x) = 2x − 0.2, and this function has its root at 0 = 2x − 0.2
or x = 0.1. Thus an (approximate) solution to the equation given would be x = 0.1.
(With several more steps of Newton’s method, we could find a more accurate solution
≈ 0.102)
19.
r
f (T ) = 331.3
so at T = 0o C,
T
1+
273.15

1
q 1
f ′ (T ) = 331.3
2
1+
f ′ (0) =
f (0) = 331.3
T
273.15
331.3
≈ 0.606
(2)(273.15)


1
273.15
Thus the speed of sounds for air temperatures around the water freezing mark (0o C) is
f (T ) ≈ 0.606(T − 0) + 331.3, or f (t) ≈ 0.606T + 331.3 m/s
21. (a)
T = 2π
s
dT
1 1 1
π
= 2π p
=√
dl
2 l/g g
l·g
l
g
3
The trick in this question is to re-write the derivative in terms of T :
√ !
T
dT
2l
π
1 2π l
=
×
=√
=
√
dl
2l
2l
g
2l
l·g
| {z }
T
Taking our limit definition of the derivative as
∆T
dT
≈
,
dl
∆l
T
∆T
≈
∆l
2l
T
so ∆T ≈ ∆l
2l
as stated in the problem.
(b) If there is a 2% change in l, then
∆T ≈
or
∆l
= 0.02, or ∆l = 0.02l:
l
T
(0.02l) = 0.01T
2l
∆T
≈ 0.01 (formula for relative change)
T
so the period will increase by roughly 1% if we increase the length by 2%.
23. Since f has a positive second derivative, so it is concave up. This means that the graph
of f (x) lies above its tangent line. Whether f is increasing or decreasing, we can see in
the diagrams below that f (1+ ∆x) ≥ f (1)+ f ′ (1)∆x. i.e. the actual value of the function
to either side of x = 1 will be above the linear approximation at the same point.
25. Note that the sine function here uses radians as measure, even though θ is measured in
π
term inside the sine function).
degrees (you can see the conversion occurring due to the
90
≈ 16, 398 meters.
(a) If θ = 20, the range = f (20) = 22510 sin π·20
90
4
(b) To find the a linear approximation, we need the rate of change of f with respect to
θ:
πθ
π
df
= 25510 cos
×
dθ
90
90
πθ
≈ 890.5 cos
meters distance per degree elevation
90
At θ = 20, f (20) ≈ 16, 398
f ′ (20) ≈ 682.2
so an approximation to f near θ = 20 would be
f (θ) ≈ 682.2(θ − 20) + 16, 398 meters
(c) The true range for θ = 21o is f (21) = 17, 070 meters.
The linear approximation gives f (21) ≈ 17, 080 meters, so the two estimates are quite
close.
26. (a) The time in the air is obtained by using θ = 20 in t(θ).
t(20) ≈ 34.9 seconds.
(b)
π πθ
′
cos
t (θ) = 102
180
180
πθ
= 1.78 cos
seconds
180
For the angle θ = 20,
t(20) = 34.9 seconds
t′ (20) = 1.673 seconds / degree of inclination
so an approximation to t near θ = 20 would be
t(θ) ≈ 1.673(θ − 20) + 34.9 seconds
(c) The true time-of-flight for θ = 21o is t(21) = 36.6 seconds.
The linear approximation gives t(21) ≈ 36.6 seconds, so the approximation agrees
with the exact value to three significant figures.
27. (a) The peak altitude is obtained by using θ = 20 in h(θ).
h(20) ≈ 1942 meters.
(b)
π πθ
πθ
′
sin
cos
h (θ) = 12755(2)
180
180
180
πθ
πθ
= 445.2 sin
cos
180
180
5
For the angle θ = 20,
h(20) = 1942 meters
h′ (20) = 143 meters / degree of inclination
so an approximation to h near θ = 20 would be
h(θ) ≈ 143(θ − 20) + 1942 meters
(c) The true highest point for θ = 21o is h(21) = 1638 meters.
The linear approximation gives h(21) ≈ 1635 meters, which is very close to the true
value, though a little low.
33. ex has a tangent line/local linearization near x = 0 of y = x + 1 (slope 1, point (0, 1)).
Multiplying this approximation by itself, we get (ex )(ex ) or e2x ≈ (x + 1)(x + 1) =
x2 + 2x + 1
To compare with the actual linearization of g(x) = e2x , we find its derivative and value
at x = 0,
g(x) = e2x
g(0) = 1
g′ (x) = 2e2x
g′ (0) = 2
so a linearization of g(x) = e2x near x = 0 is y = 2(x − 0) + 1 or
y = 2x + 1
Note that his is the same as the approximation we obtained before, except that our
product version had an additional term, x2 .
These approximations give the same straight-line estimate of the function, but I would
expect the first (multiplication) version to be more accurate because it contains more
information (the squared term that the pure linear approximation was missing).
We will see more of this idea in Taylor polynomials and Taylor series.
−1
34. (a) Let f (x) = 1/(1 + x). Then f ′ (x) = (1+x)
2.
′
At x = 0, f (0) = 1 and f (0) = −1.
So near x = 0, f (x) ≈ −1x + 1 = −x + 1
(b) For small x values (i.e. x near zero), we can approximate 1/(1 + x) with 1 − x.
Replace the variable x with y (because the name doesn’t matter),
1/(1 + y) ≈ 1 − y
If we choose y small but equal to x2 , then
1
≈ 1 − x2
1 + x2
(c) The linearization of 1/(1 + x2 ) is the linear part of 1 − x2 , or just 1. Since the
d1
/(1 + x2 )
derivative at x = 0 is the coefficient for x in the linear part, this means dx
at x = 0 must equal zero.
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