MATHEMATICS 100, Section 103
Midterm #2 Solutions
1. a) Differentiate the equation: if x sin y = sin(xy) then sin y + x cos y · y 0 = cos(xy) ·
(xy 0 + y). Plug in x = 2, y = π: 0 + 2 · (−1)y 0 = 1 · (2y 0 + π), −2y 0 = 2y 0 + π, y 0 = −π/4.
b) If f (x) = (ln x)sin x then ln f (x) = sin x ln(ln x). Differentiate the last equation:
f 0 (x)
1
1
sin x
= cos x ln(ln x) + sin x ·
· = cos x ln(ln x) +
,
f (x)
ln x x
x ln x
sin x sin x f 0 (x) = f (x) cos x ln(ln x) +
= (ln x)sin x cos x ln(ln x) +
.
x ln x
x ln x
c) Let f (x) =
x
, then
x2 + 4
f 0 (x) =
x2 + 4 − 2x2
4 − x2
1(x2 + 4) − x(2x)
=
=
.
(x2 + 4)2
(x2 + 4)2
(x2 + 4)2
Thus f 0 (0) = 0 when 4 − x2 = 0, x = ±2. Since x = −2 is not in [0, 4], the only critical
point in [0, 4] is x = 2. We evaluate
f (0) = 0, f (2) =
1
4
1
2
= , f (4) =
= .
4+4
4
16 + 4
5
The maximum value is 1/4 (at x = 2), and the minimum value is 0 (at x = 0).
2. a) We have ex ≈ 1 + x +
e−0.2 ≈ 1 − 0.2 +
x2
(you can either remember it or derive it here). Hence
2
(−0.2)2
= 0.8 + 0.02 = 0.82.
2
M
(0.2)3 , where M is a constant such that |(ex )000 | ≤ M for x
3! x 000
between −0.2 and 0. We have (e ) = ex and |ex | = ex ≤ 1 for all x ≤ 0, hence we can
1
1
take M = 1. Thus the error can be estimated by (0.2)3 =
.
3!
150
b) The error is bounded by
c) If f (x) = 2x then f 0 (x) = 2x ln 2, f 00 (x) = 2x (ln 2)2 , f 000 (x) = 2x (ln 2)3 . Hence the third
order Maclaurin polynomial is
f (x) ≈ 1 + (ln 2)x +
(ln 2)2 2 (ln 2)3 3
x +
x .
2!
3!
3. Let y(t) denote the number of cells after t hours. Then y(t) = 200ekt for some k. We
know that y(6) = 240, hence
200e6k = 240, e6k =
6
6
1
6
240
= , 6k = ln( ), k = ln( ).
200
5
5
6
5
We wish to find t such that y(t) = 400:
200ekt = 400, ekt = 2, kt = ln 2, t =
ln 2
=
k
1
6
ln 2
6 ln 2
6 = ln 6 − ln 5 .
ln( 5 )
4. The water in the container has the shape of an inverted cone with height h cm and
radius h cm at the top, where h is measured to be 60 cm. The volume of the water is
V (h) = πh3 /3. The error in computing it based on the measured value of h is bounded
by V 0 (60)dh, where dh = 2. We have V 0 (h) = 3πh2 /3 = πh2 , so V 0 (60) = 602 π = 3600π.
Hence the error is at most 3600π · 2 = 7200π cm3 .
5. Let d(t) denote the distance between the boy and the ball, h(t) – the height of the
ball, x(t) – the distance the boy has run from the starting point, where all distances are
measured in meters and t is measured in seconds. Then d2 = h2 + x2 . Differentiate in t:
2dd0 = 2hh0 + 2xx0 , dd0 = hh0 + xx0 . We have h(t) = 28t − 16t2 , h0 (t) = 28 − 32t, x(t) = 4t,
x0 (t) = 4. At t = 1, we have h(1) = 28 − 16 = 12, h0 (1) = 28 − 32 = −4, x(1) = 4,
x0 (1) = 4, and
p
√
√
√
d(1) = 122 + 42 = 144 + 16 = 160 = 4 10.
Hence
√
8
32
4 10d0 (1) = 12 · (−4) + 4 · 4 = −48 + 16 = −32, d0 (1) = − √ = − √ .
4 10
10
Since d0 (1) < 0, the distance is decreasing.