MATH 100, HOMEWORK 5 SOLUTIONS
Section 3.9, #24: This was done in class (check your notes!)
Section 3.9, #38: If the lighthouse is at a point Q, and if the light beam
meets the shore at R, then P QR is a right triangle with |P Q| = 3 and the
right angle at P . Let x denote the oriented distance between P and R, and
θ – the oriented angle between QP and QR, so that x = 3 tan θ. Then
x0 (t) = 3 sec2 θ · θ0 (t). We know that the light beam makes 4 full revolutions
per minute, so that θ0 (t) = 8π (measured in radians per minute). Also, when
x = 1, from the triangle P QR we have
√
√
√
10
,
|QR| = 1 + 9 = 10, sec θ =
3
hence x0 (t) = 3 ·
10
80π
· 8π =
≈ 83.77 km/min.
9
3
Course Notes 1, #4: If f (x) = tan x then f (0) = 0,
f 0 (x) = sec2 x, f 0 (0) = 1,
f 00 (x) = 2 sec x(sec x tan x) = 2 sec2 x tan x, f 00 (0) = 0,
f 000 (x) = 4 sec x(sec x tan x) tan x + 2 sec2 x sec2 x
= 4 sec2 x tan2 x + 2 sec4 x, f 000 (0) = −2,
f (4) (x) = 4(2 sec x)(sec x tan x) tan2 x + 4 sec2 x(2 tan x) sec2 x
+2(4 sec3 x)(sec x tan x)
= 8 sec2 x tan3 x + 16 sec4 x tan x, f (4) (0) = 0,
f (5 )(x) = 16 sec x(sec x tan x) tan3 x + 8 sec2 x(3 tan2 x) sec2 x
+16(4 sec3 x)(sec x tan x) tan x + 16 sec4 x sec2 x, f (5) (0) = 16.
Hence the Maclaurin polynomial is
tan x ≈ x +
−2 3 16 5
1
2
x + x = x − x3 + x5 .
3!
5!
3
15
Course Notes 2, #2: The error in the n-th order approximation is bounded
M
by
(0.25)n+1 , where M is a constant such that |(cos x)(n+1) | ≤ M .
(n + 1)!
1
Since (cos x)(n+1) is either ± cos x or ± sin x for all n, we can take M = 1.
1
(0.25)n+1 < 10−5 , i.e.
We then need to find n such that
(n + 1)!
1
1
< 10−5 , (n + 1)!4n+1 > 105 .
n+1
(n + 1)! 4
We have 4!44 = 6144 and 5!45 = 122, 880, so that we should take n + 1 = 5,
n = 4.
Section 11.10, #10: If f (x) = xex then f (0) = 0 and
f 0 (x) = xex + ex , f 0 (0) = 1,
f 00 (x) = xex + ex + ex = xex + 2ex , f 00 (0) = 2,
f 000 (x) = xex + ex + 2ex = xex + 3ex , f 000 (0) = 3,
...,
f (n) (x) = xex + nex , f (n) (0) = n,
hence the Maclaurin series is
0+x+
2 2 3 3
n
1
1
x + x + . . . + xn + . . . = x + x2 + x3 + . . . +
xn + . . .
2!
3!
n!
2!
(n − 1)!
2