Calculus
Homework # 6
Homework # 6
8.8 # 11: (3 pts, p. 626)
(a) Approximate f by a Taylor polynomial with degree n at the number a.
(b) Use Taylor’s Inequality to estimate the accuracy of the approximation f (x) ≈ Tn (x) when
x lies in the given interval.
(c) (Not required) Check your result in part (b) by graphing |Rn (x)|.
f (x) =
√
x,
a = 4,
4 ≤ x ≤ 4.2
n = 2,
Part (a): Here is a table of derivatives and values at a = 4:
Derivative #
0
1
2
3
Derivative
√
x
1 −1/2
2x
− 14 x−3/2
3 −5/2
8x
Value at x = 4
2
1
4
1
− 32
3
128
The Taylor polynomial T2 (x) for this function at x = 4 is:
T2 (x) = 2 + 14 (x − 4) −
1
64 (x
− 4)2
Part (b): The Taylor Inequality is:
|Rn | ≤
M
|x − a|n+1
(n + 1)!
All that is needed is to find an M such that M ≥ |f (n) (x)| for all n > 2. From the table above,
f (3) (x) = 38 x−5/2 and successive derivatives will have an absolute value less than this because
each will have x raised to a higher magnitude negative exponent. Therefore, a good choice for
3
. Also, the value of this derivative will be at its maximum when x = 4 rather than
M is 256
when x = 4.2. However, the x value to use in the Taylor Inequality is the value x = 4.2 to
get an estimate of the error which takes into account the part of the interval where the Taylor
polynomial will fit the function f (x) the least well. Therefore, the Taylor Inequality shows that:
|R2 | ≤
3
256
3!
|4.2 − 4|3
Using x = 4.2 to get a high error estimate
3
256 · 3! · 53
1
1
|R2 | ≤
=
≈ 1.5625 × 10−5
512 · 125
64000
|R2 | ≤
Simplifying
Part (c): This is not required. Here is a graph of the
√ absolute value of the difference
between the Taylor polynomial and the function f (x) = x . The two values are very close
(the difference is close to zero)√when x ≈ 4. From an actual calculation, the value of T2 (x) is
1
≈ 1.52 × 10−5 of x when x = 4.2.
within 65992
1
Calculus
Homework # 6
0.0003
0.00025
0.0002
y
0.00015
0.0001
5e-05
0
3.4 3.6 3.8 4 4.2 4.4 4.6
x
9.1 # 22: (3 pts, p. 639) Describe in words the region of R3 represented by the equations or
inequalities:
y = −2
This equation describes a vertically oriented plane in R3 that is perpendicular to the y-axis
and/or parallel to the x-z plane with y = −2.
9.2 # 26: (4 pts, p. 647) Find the magnitude of the resultant force and the angle it makes
with the positive x-axis:
y
20 lb
45°
0
x
30°
16 lb
2
Calculus
Homework # 6
These two forces are:
F1 = 20 cos 45◦ i + 20 sin 45◦ j
F2 = 16 cos 330◦ i + 16 sin 330◦ j
Their sum is:
F1 + F2 = (20 cos 45◦ + 16 cos 330◦ )i + (20 sin 45◦ + 16 sin 330◦ )j
The magnitide of this vector is in absolute terms and then a numerical approximation:
√
∥F1 + F2 ∥ = (20 cos 45◦ + 16 cos 330◦ )2 + (20 sin 45◦ + 16 sin 330◦ )2
√ √
√
√
= (10 2 + 8 3 )2 + (10 2 − 8)2
No real simplification possible
√
Continuing numerically
≈ 27.99854208432 + 6.14213562372
≈ 28.6643365328 lbs
The magnitude
The angle that this vector makes with the positive x-axis (will be in quadrant I) is:
20 sin 45◦ + 16 sin 330◦
20 cos 45◦ + 16 cos 330◦
√
−1 10 2 + 8
√
√
= tan
10 2 + 8 3
≈ 12.3731703995◦
angle = tan−1
Again, no real simplification possible
Continuing numerically
y
20 lb
28.7 lb
12.4°
45°
0
x
30°
16 lb
3