Fall 2004 Math 151
Exam 2A: Solutions
c
Mon, 01/Nov
2004,
Art Belmonte
9. (a) Repeatedly differentiate until a pattern is evident.
f (x) = sin x
f 0 (x) = cos x
f 00 (x) = − sin x
f 000 (x) = − cos x
1
1. (c) As x → ∞, we see that 3−x = x → 0 since the
3
numerator is fixed and the denominator increases without
bound.
f (4) (x) = sin x
We see that the derivatives repeat in a cycle of four. Hence
f (28) (x) = sin x. More precisely, f (k) = f (k mod 4) where
k mod 4 signifies remainder upon division by 4. Because
28 mod 4 = 0, we conclude that
2. (d) The natural logarithm is defined for all positive real
arguments; i.e., for |x| > 0 or x 6= 0.
3. (e) Isolate x step by step.
log2 (2x + 3) =
2x + 3 =
2x
x
=
=
f (28) (x) = f (0) (x) = f (x) = sin x.
3
23 = 8
10. (c) At x =
5
5/2
f 0 (x) =
L(x) =
=
1
=
f 0 (g(2))
1
f 0 (1)
=
4
2
5x + 3x =
q
1
.
8
s=
x=1
ds
dt
f 0 (x) = 2x tan x + x 2 sec2 x
h
i
(cos t)2 , t
v(t) = r0 (t)
=
[2 (cos t) (− sin t) , 1] = [− sin 2t, 1]
=
[−2 cos 2t, 0]
−2 cos π3 , 0 = [−1, 0]
when t =
−1/2 d x 1 2
x +1
2x
2
dt
2x
,
p
x2 + 1
since d x/dt = 2 from the statement of the problem.
12. (d) Let w = 2−x (to give it a name). Then
2−x + 2−x = w + w = 2w = 21 2−x = 21−x .
8. (c) The derivative of position is velocity. In turn the
derivative of velocity is acceleration.
=
=
=
√
7. (b) Now f (x) = 2 e x = 2 (e x )1/2 = 2e x/2 , whence
1/2 √
= ex .
f 0 (x) = 2e x/2 12 = e x/2 = e x
r(t)
1/2
.
(x − 0)2 + (0 − 1)2 = x 2 + 1
Differentiation with respect to time t yields
6. (b) Use the product rule to differentiate f (x) = x 2 tan x.
=
√
√
√
f ( π) + f 0 ( π)(x − π)
√ √
0−2 π x − π
√
2π − 2 π x.
11. (e) The distance between (x, 0) and (0, 1) is
1
=
sin x 2 = sin π = 0,
√
cos x 2 · (2x) = −2 π.
Therefore, the desired linear approximation is
5. (a) Since f (1) = 2 and g = f −1 , we have g(2) = 1 and thus
a(t) = v0 (t) = r00 (t)
π, we have
f (x) =
4. (a) The √
range of f −1 is the domain of f . Since we are given
f (x) = 1 − x, we require 1 − x ≥ 0 or 1 ≥ x. This is the
interval (−∞, 1].
g 0 (2) =
√
13. (c) Geometrically, Newton’s method amounts to following
tangent lines to where they intersect the x-axis. With the
point P(2, 2) on the graph of f and the slope of the tangent
line thereat between 3 and 5, we have the following diagram.
It is clear from the picture that the next Newton iterate lies in
the interval 1 < x < 2 or (1, 2). No computation is required!
π
6.
1
(ii) Therefore,
Solution: rise/run by counting
4
3
d y/dt
dy
(1 − t) e−t
=
= 2
.
−1/3
dx
d x/dt
3 (t + 1)
slope 3
2
(iii) The slope m of the desired tangent line is obtained by
plugging t = 0 into the expression for d y/d x. This
yields m = 32 . Hence an equation of the tangent line is
y − 0 = 32 (x − 1) or y = 32 x − 32 .
y
1
0
−1
slope 5
−2
16. Given f (x) = eax sin bx, we have
−3
−4
−1
0
1
2
x
3
4
5
=
f 00 (x)
=
=
Alternatively, take an analytical approach. Since the graph of
y = f (x) passes through (2, 2), we have f (2) = 2. Using a
slope of f 0 (2) = 3, the next Newton iterate is
x2 = x1 −
f 0 (x)
aeax sin bx + eax (b cos bx) = eax (a sin bx + b cos bx)
aeax (a sin bx + b cos bx) + eax ab cos bx − b2 sin bx
eax a 2 − b2 sin bx + 2ab cos bx
17. Let y be the height of the rocket and θ the camera’s angle of
elevation. Here is a diagram followed by the relevant
computations involved in this related rates problem.
2
4
f (x 1 )
= 2 − = ≈ 1.33 ∈ (1, 2) .
0
f (x 1 )
3
3
Similarly, using a slope of f 0 (2) = 5 gives
2
2 1/2
(y + 4000 )
2
8
f (x 1 )
= 2 − = = 1.6 ∈ (1, 2) .
x2 = x1 − 0
f (x 1 )
5
5
y
14. Implicitly differentiate the equation
2
x 2 + y 2 = 4 x 2 − y 2 + 3x + 7
θ
4000
with respect to x, then solve for d y/d x.
dy
dy
= 4 2x − 2y
+3
2 x 2 + y 2 2x + 2y
dx
dx
dy
3
dy
x 2 + y2 x + y
=
2x − 2y
+
dx
dx
4
dy
3
2y + x 2 + y 2 y
= 2x + − x 2 + y 2 x
dx
4
2x + 34 − x 2 + y 2 x
dy
=
dx
2y + x 2 + y 2 y
19
4
2
3
=
=
=
camera angle is changing at
−
7
40
4
12
125
√ 4000
30002 +40002
= 45 . The
= 0.096 rad/s or 5.5◦ /s.
18. The linear approximation to f at a = 1 is L(x) = 2x + 3.
Thus √
f (1) = L(1) = 5 and f 0 (1) = L 0 (1) = 2. We √
are given
H = f or H (x) = ( f (x))1/2 . At x = 1, H (x) = 5 and
H 0(x) = 12 ( f (x))−1/2 f 0 (x) = 12 √1 (2) = √1 . Hence
5
5
the linear approximation to H at a = 1 is
− 21
3
= 4 =−
7
4
L H (x) = H (1) + H 0 (1)(x − 1) =
(i) The parametric derivatives are
=
dθ
dt
dθ
dt
dθ
dt
y
4000
d y/dt
4000
d y/dt
(cos θ )2
4000
600 4 2
12
= 0.096
=
4000 5
125
since for y = 3000, we have cos θ =
15. We are given x = (t + 1)2/3 and y = te−t . The point
(x, y) = (1, 0) corresponds to t = 0.
dx
dt
dy
dt
=
sec2 θ ·
Finally, plug in (x, y) = (2, 1).
4 3 − 10
dy
= 4
=
dx
7
tan θ
(t + 1)−1/3
= (1) e−t + te−t (−1) = (1 − t) e−t .
2
√
1
5 + √ (x − 1) .
5