Fall 2005 Math 151
Exam 2A: Solutions
c
Mon, 31/Oct
2005,
Art Belmonte
8. (e) The limiting expression is a decaying exponential.
Therefore, as x → −∞, we see that e x−1 → 0.
9. (a) Use log properties. Be sure to check your answer.
log4 x + log4 x 2
= 6
1. (d) Now f 0 (x) = 4e4x − 15e3x + 3e x + cos x, whence
f 0 (0) = 4 − 15 + 3 + 1 = −7.
2. (b) We have
= 6
log4 x
= 2
x
F 0 (x) =
F 0 (2) =
f 0 (g (x)) g 0 (x)
f 0 (g (2)) g 0 (2)
10. (c) By inspection, f (1) = 4. Thus g (4) = 1. Accordingly,
g 0 (4) =
.
=
sin θ
= 1.
3. (a) Use the fact that lim
θ →0 θ
3x (1 + cos x)
sin 4x
x→0
=
lim
=
=
lim
=
3x
4x
(1 + cos x)
lim
3
4
(1 + cos x)
x→0
sin 4x
13. (b) The acceleration is a (t) = v0 (t) = [6t − 8, 4].
Therefore, a (2) = [12 − 8, 4] = [4, 4].
5. (d) We have
14. Use product and chain rules, as applicable.
dy dx
2x
(y) + x 2
dt
dt
(4 (−2)) (3) + (4) (4)
16 − 24 = −8.
(a) Given f (x) =
f 0 (x)
6. (d) Solve y = f (x) for x.
y = f (x) =
Thus f −1 (x) =
yx + 5y
(y − 2) x
=
=
x
=
3 (−1 − 1 − 1 + 1)2 (3 + 2 + 1)
3 (−2)2 (6) = 72.
(sin 3x + tan 4x)2
√
x 2 +3x+4
= exp
x 2 + 3x + 4
1/2 ,
15. Implicitly differentiate y 5 + 3x 2 y 3 + x 3 + 5 = 0 with
respect to x, then substitute the data (x, y) = (2, −1).
4 dy
3
2
2 dy
5y
+ (6x) y + 3x
+ 3x 2 =
3y
dx
dx
dy
dy
+ −12 + 36
+ 12 =
Substitute data: 5
dx
dx
41y 0 =
7. (c) When t = −1, we have
=
=
(sin 3x + tan 4x) (−2 sin 2x) − (cos 2x) 3 cos 3x + 4 sec2 4x
1/2 1 −1/2 x 2 + 3x + 4
(2x + 3)
x 2 + 3x + 4
2
√
2
(2x + 3) e x +3x+4
0
p
or f (x) =
.
2 x 2 + 3x + 4
f 0 (x) = exp
1 − 5x
.
x −2
2 3t 2 − 2t + 1
3 t3 − t2 + t + 1
=
cos 2x
, we have
sin 3x + tan 4x
(b) Given f (x) = e
we have
2x + 1
x +5
2x + 1
1 − 5y
1 − 5y
y−2
=
1
1
= .
7
5x 4 + 2 x=1
Thus v (2) = [12 − 16,√
8 − 3] = [−4,
√5]. The speed is the
magnitude of velocity: 16 + 25 = 41, a scalar.
d
(x cos 2x) = (1) (cos 2x)+(x) (−2 sin 2x) = cos 2x −2x sin 2x.
dx
=
1
f 0 (1)
12. (e) Velocity, the derivative of position, is a vector.
i
h
v (t) = r0 (t) = 3t 2 − 8t, 4t − 3
4x
3
3
(2) =
4
2
4. (c) The product rule gives
d 2 x y
=
dt
=
1
f 0 (g (4))
11. (a) Given position r (t) = t 3 − 4t 2 + 2, 2t 2 − 3t , we have
r (2) = [8 − 16 + 2, 8 − 6] = [−6, 2].
sin 4x
4x
x→0
= 42 = 16
This checks out via substitution into the original equation.
f 0 (5) g 0 (2)
(11) (4) = 44.
=
=
h 0 (t)
3 log4 x
At stated point: y
The tangent line is a horizontal line: y = −1.
.
1
0
=
0
0
0
0
.
16. Let f (x) = x 5 + x 3 + 1. Then f 0 (x) = 5x 4 + 3x 2 .
Given x 1 = −1, Newton’s method yields
x2
=
=
f (x 1 )
f 0 (x 1 )
1
7
−1
= −1 + = − = −0.875.
−1 −
5+3
8
8
x1 −
√
1
17. The quadratic approximation of f (x) = 4 x = x 4 near 16 is
0
00
f (16)
f (16)
f (16)
+
Q (x) =
(x − 16)1 +
(x − 16)2 .
0!
1!
2!
3 x −7/4 . Thus
Now f 0 (x) = 14 x −3/4 and f 00 (x) = − 16
3
1
(x − 16) −
(x − 16)2
32
2 · 24 · 27
3
1
2+
(x − 16) −
(x − 16)2 .
32
4096
Q (x) =
2+
=
18. Here is a diagram depicting the situation.
balloon
z
boy
y
x
• Place the boy at the origin of the x y-plane as he passes
under the balloon. At time t = 0, the balloon is at the
point (0, 45). For t ≥ 0, the boy’s position is x = 15t
and the balloon’s position is y = 45 + 5t. Note that
d x/dt = 15 and d y/dt = 5 for all t.
• Let z be the distance (through the air) between the
balloon and the boy. By the Pythagorean Theorem,
z 2 = x 2 + y 2 . Differentiate with respect to t.
2z
dz
dt
=
dz
dt
=
dx
dy
+ 2y
dt
dt
dy
dx
+y
x
dt
dt
z
2x
• When t = 3 seconds, we have x = 45, y = 60,
dy
dx
dt = 15, dt = 5, and
p
p
z = 452 + 602 = 15 32 + 42 = 75.
Therefore,
dz
dt
=
=
=
(45) (15) + (60) (5)
75
65
15 (45 + 20)
=
15 (5)
5
13 ft/sec.
2