Solution: APPM 1350
Final Exam (150 pts)
Fall 2013
1. The following problems are not related:
(a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist:
√
arcsin(x)
(i) lim e−x x
(ii) lim
(iii) lim (sin x)x
x→∞
x→0
x
x→0+
(b) (5 pts) Find and classify all relative extrema of f (x) =
x5 − 5x
.
5
(c) (10 pts) A right triangle is formed in the first quadrant by both the x-axis and y-axis and a line
through the point (1, 2). Find the vertices of the triangle so that its area is a minimized.
[Hint: the slope of the hypotenuse should be the same whether slope is found using the y-intercept and
the point (1,2) or the x-intercept and the point (1,2).]
Solution:
(a)(i) Note that, lim e
x→∞
−x √
√
x = lim
x→∞
x
1
√ = 0.
x
L0 H
= lim
ex
x→∞
2ex
(a)(ii) Applying L’Hospitals Rule yields,
arcsin(x) L0 H
= lim
x→0
x→0
x
lim
√ 1
1−x2
1
1
= lim √
= 1
x→0
1 − x2
(a)(iii) If we let y = limx→0+ (sin x)x , then, using continuity,
0·∞
ln y = lim ln(sin(x)x ) = lim x ln(sin(x)) = lim
x→0+
x→0+
x→0+
ln(sin x) L0 H
−x2
cot x
= lim −1 = lim
1/x
x→0+ tan(x)
x→0+
x2
Finally, applying L’Hospitals Rule again yields
ln(y) = lim
x→0+
so ln(y) = 0 implies y = e0 = 1, that is
−x2 L0 H
−2x
=0
= lim
tan(x)
x→0+ sec2 (x)
lim (sin x)x = 1. Note that we could also do this limit directly
x→0+
x)
using continuity by rewriting the limit as lim (sin x)x = lim eln(sin(x)
x→0+
x→0+
= lim ex ln(sin(x)) .
x→0+
5x4 − 5
(b) Taking the derivative yields f 0 (x) =
= x4 − 1, so the critical points are x = ±1. Now we need
5
to classify the critical points. Using, for example the First Derivative Test, we have
1st Derivative Test for f (x):
%
&
|
-1
%
|
1
Therefore, we have a relative maximum at (−1, 45 ) , and relative minimum at (1, − 45 ) . (Note could also
use the 2nd Derivative Test.)
(c) Assume the x-intercept and y-intercept of the line are (0, y) and (x, 0) respectively.
Note that we must have x > 1, since if x = 1 then the line through (x, 0) and (1, 2) is vertical and we
don’t get a triangle, and if x < 1 then the line through (x, 0) and (1, 2) has a negative y-intercept and we
don’t get a triangle solely in the first quadrant.
y−2
0−2
Then the slope of the hypotenuse using the two intercepts is
=
. This produces the relationship
0−1
x−1
2
.
y =2+
x−1
1
2
So we wish to minimize the area A(x, y) = xy, subject to y = 2 +
and x > 1. Substituting, yields
2
x−1
2
x
1
)=x+
A(x) = x(2 +
2
x−1
x−1
Then differentiating and finding critical points yields,
A0 (x) = 1 +
(x − 1) − x
1
=1−
=0
2
(x − 1)
(x − 1)2
Thus (x − 1)2 = 1 implies x − 1 = ±1 and so the critical points are x = 0, 2. But x = 0 is not in the
2
domain x > 1 so we may ignore it. Note that A00 (x) =
, and by the 2nd Derivative test A00 (2) > 0,
(x − 1)3
so we have a relative minimum at x = 2 (we may also use the First Derivative Test to determine this).
Since x = 2 is the only critical point however, it gives an absolute minimum. Thus, select x = 2 then
y = 4 and A = 4 and the vertices that yield the minimum area are (0, 0), (2, 0), (0, 4).
2. The following problems are not related:
(a) (15 pts, 5 pts ea.) Find dy/dx if:
−1 (x)
(i) y = ex+tan
√
(ii) y = x2 cosh( x)
(iii) xey − tanh(y) = 10, 690
(b) (7 pts) Consider the function f (x) = 2x2 − 4x + 1 on the interval [0, 2].
(i) (3 pts) Does the function f (x) satisfy the hypotheses of the Mean Value Theorem on the given
interval?
(ii) (4 pts) If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean
Value Theorem. If it does not satisfy the hypotheses, write DNE.
(c) (8 pts) A kite 100 feet above the ground moves horizontally at a speed of 8 ft/s. At what rate is the
angle between the string and the horizontal decreasing when 200 ft of string has been let out?
Solution:
2
dy
1
x+tan−1 (x)
x+tan−1 (x) 2 + x
(a)(i) Using the Chain Rule, we have
=e
1+
= e
dx
1 + x2
1 + x2
(a)(ii) Using the Product Rule and Chain Rule, we have
√ √
√
√
√
d
1
x3/2
x2 cosh( x) = 2x cosh( x) + x2 sinh( x) √ = 2x cosh( x) +
sinh( x).
dx
2
2 x
(a)(iii) Using implicit differentiation yields
ey + xey y 0 − sech2 (y)y 0 = 0 =⇒ y 0 = −
ey
ey
=
xey − sech2 (y)
sech2 (y) − xey
(b)(i) Yes. The function is a polynomial, so it is continuous on [0, 2] and differentiable on (0, 2).
(b)(ii) We see that c = 1, since
f 0 (c) =
f (b) − f (a)
1−1
⇐⇒ 4c − 4 =
= 0 ⇐⇒ 4c = 4 ⇐⇒ c = 1
b−a
2−0
which is in (0, 2).
(c) Let θ denote the angle of the kite string with the ground, let x denote the horizontal distance of the
kite from the person holding the string and let y be the amount of string let out. Then
cot(θ) =
Now,
x
dx
dθ
dθ
sin2 (θ) dx
⇒ x = 100 cot(θ) ⇒
= −100 csc2 (θ)
⇒
=−
100
dt
dt
dt
100 dt
dx
100
1
= 8ft/s and when y = 200 we have sin(θ) =
= , so
dt
200
2
sin2 (θ) dx
(1/2)2
1
dθ
=−
=−
· 8 = − rad/s
dt
100 dt
100
50
so the angle is decreasing at a rate of
1
rad/s.
50
3. The following problems are not related:
(a) (15 pts, 5 pts ea.) Evaluate the integrals:
Z
Z 2 1/x3
1
e
dx
(ii)
dx
(i)
4
2
x
x +4
1
(b) (5 pts) Find h0 in terms of f 0 and g 0 if h(x) =
Z
(iii)
√
e2x
dx
1 − e4x
f (x)g(x)
.
f (x) + g(x)
(c) (10 pts) The Ideal Gas Law relating the pressure, volume, and temperature of a gas is given by
P V = nRT where R is the ideal gas constant. Assuming that the current pressure of the gas, P , is
10 atmospheric pressure units (atm) and assuming that n, R, and T are held fixed, use differentials to
estimate the relative change in the volume, V , if the pressure increases by 0.1 atms.
Solution:
(a)(i) Let u =
3
1
then du = − 4 dx and
3
x
x
Z
Z
Z 2 1/x3
e
1 1/8 u
1 1 u
1 u 1
1
e du =
e du =
dx = −
e
= (e − e1/8 )
4
x
3 1
3 1/8
3
3
1
1/8
(a)(ii) Factoring out a 4 from the denominator yields
Z
Z
Z
dx
1
dx
1
dx
=
=
2
2
x
x +4
4
4
(x/2)2 + 1
4 +1
dx
, and so we have
2
Z
Z
1
dx
2
du
1
1
−1
−1 x
=
=
tan
(u)
+
C
=
tan
+C
4
(x/2)2 + 1
4
u2 + 1
2
2
2
let u = x/2, then du =
(a)(iii) Let u = e2x , then du = 2e2x dx and
Z
e2x dx
1
√
=
4x
2
1−e
Z
√
du
1
1
= sin−1 (u) + C = sin−1 (e2x ) + C
2
2
2
1−u
(b) Use the Quotient Rule and Product Rule,
h0 =
f f 0 g + f 2 g 0 + f 0 g 2 + f gg 0 − f f 0 g − f gg 0
f 2g0 + f 0g2
(f 0 g + f g 0 )(f + g) − f g(f 0 + g 0 )
=
=
(f + g)2
(f + g)2
(f + g)2
(c) We have
P V = nRT
Then
⇒
V =
nRT
P
⇒
dV = −
nRT
dP
P2
dP
− nRT
dV
1
1
P2
= − dP = − (0.1) = −0.01
=
nRT
V
P
10
P
4. The following problems are not related:
(a) (15 pts, 5 pts ea.) Find dy/dx given:
(i) y = (sin x)
x
2x
(ii) y =
1 + 2x
u(x)
(iii) y = e
Z
g(x)
cos(f (t)) dt
where u(x) =
0
(b) (5 pts) In your blue book clearly sketch the graph of a function f (x) that satisfies the following
properties (label any extrema, inflection points or asymptotes):
• f (0) = 0, f (−x) = f (x)
• lim f (x) = −∞,
x→1−
lim f (x) = +∞
x→−1−
• lim f (x) = 2
x→∞
• f 0 (x) > 0 if x < 0
• f 00 (x) < 0 if |x| < 1
(c) (10 pts) A bacteria culture initially contains 106 cells and its population grows at rate proportional to
its size. After an hour the population has increased to 420. Find an expression for the number of bacteria
after t hours.
Solution:
(a)(i) Using logarithmic differentiation yields,
ln(y) = ln[sin(x)x ] =⇒ ln(y) = x ln[sin(x)] =⇒
y0
cos(x)
= ln[sin(x)] + x ·
y
sin(x)
and so y 0 = (sin(x))x ln(sin(x)) + x cot(x) .
(a)(ii) Applying the Quotient Rule yields,
dy
2x ln(2) · (1 + 2x ) − 2x · 2x ln(2)
2x ln(2)
=
=
dx
(1 + 2x )2
(1 + 2x )2
(a)(iii) First note that, dy/dx = eu(x) u0 (x) by the Chain Rule, and applying the Fundamental Theorem
of Calculus to determine u0 (x), we get
dy
= eu(x) u0 (x) = eu(x) · cos(f (g(x))) · g 0 (x) = eu(x) g 0 (x) cos(f (g(x)))
dx
(b) There are vertical asymptotes at x = ±1, a horizontal asymptote at y = 2 and a local maximum at
(0, 0), the graph could, for example, look like:
5
2.5
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2.5
-5
dy
= ky for some
dt
kt
constant k, and so y(t) = y0 e where y0 = y(0) = 106. Now note that y(1) = 420 implies
420
420
k
k
=⇒ k = ln
420 = y(1) = 106e =⇒ e =
106
106
420
ln
t
420 t
106
and so y(t) = 106e
= 106
.
106
(c) Let y(t) represent the population of the bacteria in cells at time t hours, then
5. The following problems are not related:
(a) (15 pts, 5 pts ea.) Evaluate the integrals:
Z 4 p
sin(x)
(i)
dx
16 − x2 +
1 + x2
−4
Z
1+x
dx
1 + x2
√
1
(b) (5 pts) Use the Squeeze Theorem to show lim x sin
= 0.
x
x→0+
(ii)
(c) (10 pts) The equation of motion of a particle is s(t) =
Z
(iii)
dx
x ln(xe )
t3 t2
− − 6t ft
3
2
(i ) Find the velocity at time t.
(ii ) When is the particle moving in the positive direction?
(iii ) Find the acceleration after 3 seconds.
(iv ) Find the total distance traveled during the first 4 seconds.
Solution:
sin(x)
(a)(i) First note that y =
is an odd function, and using geometry to calculate the first integral we
1 + x2
get,
Z 4 p
Z 4p
Z 4
Z 4p
sin(x)
sin(x)
π · 42
2
2
2 dx + 0 =
16 − x +
dx
=
16
−
x
dx
+
dx
=
16
−
x
= 8π
2
1 + x2
2
−4
−4
−4 1 + x
−4
(a)(ii) Breaking up the integral yields,
Z
Z
Z
1+x
1
x
dx =
dx +
dx
2
2
1+x
1+x
1 + x2
Now, in the second integral, let u = 1 + x2 then du = 2x dx, and
Z
Z
Z
1
1
x
1
1
−1
u du = tan−1 (x)+ ln |u|+C = tan−1 (x) + ln(1 + x2 ) + C
dx+
dx = tan (x)+
1 + x2
1 + x2
2
2
2
Z
(a)(iii) First note that
Z
dx
1
=
e
x ln(x )
e
1
dx
=
e
x ln(x )
e
Z
Z
dx
dx
, now let u = ln(x) then du =
and so
x ln(x)
x
dx
1
=
x ln(x)
e
Z
du
1
1
= ln |u| + C = ln |ln(x)| + C
u
e
e
Alternately, one could also use the substitution u = ln(xe ).
√
√
(b) Note that −1 ≤ sin x1 ≤ 1 implies − x ≤ x sin
√
1
lim x sin
= 0 by the Squeeze Theorem.
+
x
x→0
1
x
≤
√
√
x and lim ± x = 0 implies that
x→0+
(c)(i) Here v(t) = s0 (t) = t2 − t − 6.
(c)(ii) Note, v(t) = t2 − t − 6 = (t − 3)(t + 2) and v(t) > 0 when t > 3 , so the particle is moving in a
positive direction after exactly 3 seconds.
(c)(iii) Note that a(t) = v 0 (t) = 2t − 1 and so a(3) = 6 − 1 = 5 seconds.
(c)(iv) The total distance in the first 4 seconds is
Z
4
Z
3
|v(t)|dt =
0
Z
−v(t) dt +
0
4
Z
2
Z
−(t − t − 6)dt +
v(t) dt =
3
3
0
2
t
4
(t2 − t − 6)dt
3
3 3
4 3
3
t
4
42
3
32
t3 t2
− − 6t +
− − 6t =
−
− 6(4) − 2
−
− 6(3)
= −
3
2
3
2
3
2
3
2
0
3
3
3
4
42
3
32
10
49
−
− 6(4) − 2
−
− 6(3) =
+ 13 = ft.
3
2
3
2
3
3