Solutions
Math 105a
Exam 2
1. (a)
dy
dx
=
1
q
1
2x
2x = p
2
1
(x2 )
x4
dy
(b) To nd
given ey + ln y 2 = xy + x, it is necessary to implicitly dierentiate both sides with
dx
respect to x.
dy
1
dy
dy
+ 2 2y = y + x
dx
y
dx
dx
dy
2
ey +
x =y+1
dx
y
ey
dy
dx
=
ey
y+1
+ y2
+1
dy
dx
or simplify to get
x
=
y2 + y
yey
+2
xy
2. (a) Apply l'H^opital's rule since the limit leads to the indeterminate form (I.F.) 0=0.
lim
!0
tan()
L0 H
=
lim
!0
1
cos ( )
2
1
= lim
!0 cos ()
2
=
cos2 (0)
=
(b) Apply l'H^opital's rule (twice) since the limit leads to the I.F. 1=1.
ex + x L0 H
e x + 1 L0 H
ex
=
=
lim
lim
=1
lim
2
x!1
x!1 2x
x!1 2
x
3. Is the following argument valid?
1
1
1
1
Consider lim+
. Because lim+ = 1 and lim+
=1
x
sin x
x !0
x !0 x
x!0 sin x
it must surely be true that the limit in question has the value 1 1 = 0.
The argument is not valid. Recall from our study of limits (Theorem 2.1) that
lim (f (x) + g (x)) = lim f (x) + lim g (x)
x
!a
x
!a
x
!a
both limits on the right hand side of the equation exist. For our example, the individual limits
1
lim
and lim+
are innite, i.e., do not exist. Therefore, we cannot break up the (original)
x!0+ x
x!0 sin x
limit of a sum into the sum of individual limits. To nd a legitimate method to evaluate the limit start
1
1
by rewriting
using algebra, and then reevaluate the limit by applying l'H^opital's rule.
x
sin x
1
1
sin x x L0 H
cos x 1
sin x
L0 H
=
=
lim+
= lim+
lim+
lim
= 0.
x
sin x
x sin x
sin x
x !0
x!0 sin x + x cos x
x!0+ 2 cos x
x!0
only if
1
4. (a) The local linearization of f (x) = arctan(x) near x = 1 is given by f (x) f (1) + f 0 (1)(x
1
1
Note: f (1) = arctan 1 =
since tan(=4) = 1; and f 0 (x) =
so f 0 (1) = .
2
4
1+x
2
1
Therefore, f (x) + (x 1).
4 2
1
1).
(b) To approximate arctan(1:01), let x = 1:01
1
then f (1:01) = arctan(1:01) + (1:01
4 2
1) 0:7904.
5. A kite 100 feet above the ground moves horizontally at a speed of 8 ft/sec. At what rate is the angle
between the string and the horizontal changing when 200 feet of string have been let out. Be sure to
express your answer using the appropriate units.
dx
d
We are given
= 8 ft/sec, but need to nd .
dt
dt
Using trigonometry we can establish a relationship between the base x of the right triangle and the
angle .
100
tan =
x
Implicitly dierentiating both sides with respect to t gives:
1
d = 100
dx
cos2 dt
x2 dt
We can use the Pythagorean theorem to nd x and trigonometrypto nd cos when z = 12. In particp
p
3
ular, if z = 200 ft then x = 2002 1002 = 100 3 and cos =
.
2
Therefore,
d
dt
=
100 cos dx
2
x2
dt
=
100
p 2
3
2
p
2
100 3
=
1
radians/sec.
50
In other words, the angle between the string and the horizontal is changing at a rate of
1
radians/sec.
50
6. Suppose the function f is continuous at the point P when x = c. The graph of f has
a vertical tangent at P
if lim f 0 (x) and lim+ f 0 (x) are either both +1 or both
x
!c
x
!c
1.
if lim f 0 (x) and lim+ f 0 (x) are both innite with opposite signs (one +1
x! c
x! c
and the other 1).
a cusp at P
If f (x) = 3x2=3 , then f 0 (x) =
p32x .
To determine if the graph of f has a vertical tangent or cusp
at x = 0 we need to look at the behavior of f 0 (x) as x ! 0 and x ! 0+ . In particular,
2
p32x = 1
lim f 0 (x) = lim
x
!0
!0
x
and lim+ f 0 (x) = lim+
x
!0
x
!0
p32x = 1
The graph of f (x) = 3x2=3 has a cusp at x = 0 since lim f 0 (x) and lim+ f 0 (x) are both innite with
x !c
x!c
opposite signs.
7. A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of
the window is 12 feet, nd the dimensions of the rectangle that will produce the largest area for the
window.
Goal: to maximize the area of the window. Let x represent the width of each side of the triangle and
the width of the rectangle. Let h represent the height of the triangle. Let y represent the height of the
rectangle. To nd a formula for the area of the window we need to add the area of the triangle to the
1
area of the rectangle. Therefore, we need to maximize A = xh + xy . The perimeter (outer edge) of
2
the window, given by P = 3x + 2y , is xed at 12 feet. To simplify the formula for area we need to nd
expressions for both y and h in terms of x.
3
Solve for y using the equation for perimeter: 12 = 3x + 2y
=)
y=6
x
2
s
r
p
2
1
3 2
3
Solve for h using the Pythagorean theorem: h = x2
x
=
x =
x
2
4
2
Substituting these values into the formula for area gives
p
1
A= x
2
Dierentiating gives
dA
dx
dA
dx
= 0 when 6 +
p
p
=6+
!
3
x +x 6
2
3
x = 6x +
2
=
3 6 2
x
4
p
3 6
x
2
3 6
x=0
2
=)
x=
p
3 6
< 0 for all values of x.
2
Therefore the area of the window is maximized when x =
Furthermore,
d2 A
dx2
p
3
12
p
3
The height of the rectangular portion of the window is y = 6
3
6
.
12
2:81 feet.
6
p
3
12
6 3
p
= p
2
3 6
3
18
6
1:78 feet.
8. Consider the function f with f 0 (x) =
4(x + 1)
p
3 3 x2
p
(a) f 0 (x) is undened when 3 3 x2 = 0
f 0 (x) = 0
when 4(x + 1) = 0
=)
and
=)
f 00 (x) =
4(x 2)
p
9 3 x5
x=0
x=
1
Use a sign chart to see that f 0 (x) > 0 for 1 < x < 0 and x > 0, and f 0 (x) < 0 for x < 1. There-
fore, f is increasing on ( 1; 0) and (0; 1). The graph of f is decreasing on ( 1; 1). In other
words, the graph of f has a local minimum at x = 1. However, there is no local extreme at x = 0.
p
(b) f 00 (x) is undened when 9 3 x5 = 0
f 00 (x) = 0 when 4(x 2) = 0
=)
Use a sign chart to see that f 00 (x)
=)
x=0
x=2
0 for x < 0 and x > 2, and f 00 (x) < 0 for 0 < x < 2.
Therefore, f is concave up on ( 1; 0) and (2; 1). The graph of f is concave down on (0; 2). In
other words, the graph of f has inection points at x = 0 and x = 2.
>
p
(c) The graph of f passing through ( 1; 3), (0; 0), (1; 5), and (2; 6 3 2) is shown below.
9. Consider f (x) = x2=3 on [ 1; 1].
(a) The graph of f is provided below.
f (1) f ( 1)
1 1
(b) Notice that
=
= 0, but there is no c in the interval ( 1; 1) such that f 0 (c) = 0.
1 ( 1)
2
Does this contradict the Mean Value Theorem?
No, this does not contradict the Mean Value Theorem. The function f (x) = x2=3 is not dieren2
tiable on ( 1:1). In particular, f 0 (x) = 1=3 is undened at x = 0. We can see this graphically
3x
by noticing that the graph of f has a cusp at x = 0. In other words, the function f does not
satisfy the conditions of the Mean Value Theorem since the derivative does not exist at x = 0.
Therefore, the conclusion of the Mean Value Theorem does not apply to f (x) = x2=3 on [ 1; 1].
4