Solutions to Math 105a Exam 2 Review
cos x
1. (a) f 0 (x) =
sin x
1
2
2 =
cos2 (ln(1 + 2 )) 1 + 2 (1 + 2 ) cos2 (ln(1 + 2 ))
(ln 2)2sin y cos2 y + 2sin y sin y 2sin y (ln 2) cos2 y + sin y
0
h (y ) =
=
cos2 y
cos2 y
cos f 0 ( ) =
1 + sin2 1
g 0 (z ) = 2zearctan z + (1 + z 2 )earctan z = (2z + 1)earctan z
1 + z2
Apply l'H^opital's rule since the limit leads to the indeterminate form (I.F.) 0=0.
(b) g 0 () =
(c)
(d)
(e)
2. (a)
1
( sin x) = lim sin x = 0
ln(cos x)
= lim cos x
x!0 sin x
x !0
x!0 cos2 x
cos x
(b) Method 1: Rewrite the expression using sine and cosine:
lim
sec x
= lim
1 + tan x x!(=2)
lim
x!(=2)
1
cos x
sin x
cos x
1+
=
lim
x!(=2)
Method 2: Since the original limit leads to the I.F.
1
=1
cos x + sin x
1=1 we can apply l'H^opital's rule.
sin x
sec x
cos2 x
= lim
= lim sin x = 1
x!(=2) cos12 x
x!(=2) 1 + tan x
x!(=2)
(c) Apply l'H^opital's rule since the limit leads to the I.F. 1=1:
lim
1
ln x
1
p
= lim x1 = lim p = 0
x!1 2 x
x!1 p
x!1 x
x
lim
(d) The original limit leads to the I.F. 1 0, in order to apply l'H^opital's rule (twice in this case),
it is necessary to rewrite the expression so that the limit yields the I.F. 1=1.
x2
2x
2
lim x2 e x = lim x = lim x = lim x = 0
x!1
(e)
x+4
lim+
x! 2 x 2
3. Consider x2 =
x!1 e
x!1 e
x!1 e
=1
y2
y2
1
. Implicitly dierentiate both sides with respect to x.
dy
dy
dy
1) 2y dx
y 2 2y dx
2y dx
2x =
=
(y 2 1)2
(y 2 1)2
dy
x(y 2 1)2
Therefore,
=
(y 2
dx
y
p
4. (a) The local linearization of f (x) = 3 1 + x near x = 0 is given by f (x) f (0) + f 0 (0)(x 0).
1
1
Note: f (0) = 1 and f 0 (x) = p
so f 0 (0) = .
3
2
3
3 (1 + x)
1
Therefore, f (x) 1 + x.
3
p3
p
1
(b) To approximate :95, let x = 0:05, then 3 :95 = f ( 0:05) 1 + ( 0:05) = 0:9833.
3
1
(c) The graph of f is concave down for x > 1, so the tangent line at x = 0 will sit above the
graph of f over that same interval. Therefore,
the local linearization at x = 0:05 will be an
p
overestimate for the actual value of 3 :95.
5. Consider the curve given by x ln y + y 3 = 1 + ln x.
dy
(a) Implicitly dierentiate both sides with respect to x, solve for , and then substitute in the
dx
x and y -values at (1; 1).
ln y +
x
y
dy
dy
1
dx
+ 3y 2 =
dx
x
dy
dx
=)
ln y
2
+
y 3y
1
= xx
ln 1
1
=
2
4
+ 3(1)
(b) The local linearization of the curve near x = 1 is the equation of the tangent line passing
1
through (1; 1). The slope of the tangent at (1; 1) is . Therefore, the local linearization is
4
1
y 1 + (x 1).
4
1
(c) When x = 1:1, y 1 + (1:1 1) = 1:025.
4
6. An example of a slanted ellipse is given by the equation x2 + xy + y 2 = 3.
At (1; 1),
dy
dx
=
1
1
1
1
dy
(a) Implicitly dierentiate both sides with respect to x, solve for , and then substitute in the
dx
p
x and y -values at ( 3; 0).
dy
dy
dy
2x y
2x + y + x + 2y =0
=)
=
dx
dx
dx
x + 2y
p
p
2 3 0
= 2
At ( 3; 0),
= p
3+0
(b) A line is normal to a curve at a point if it is perpendicular to the curve's tangent at that
point. For a given line ofpslope m 6= 0, the slope of a line perpendicular line it is 1=m.
Therefore, at the point ( 3; 0) the slope of any line perpendicular
to the tangent is 1=2.
p
1
3
Therefore, the equation of the normal line is y = x
.
2
2
(c)
dy
= 0 for
dx
dy
=0
dx
dy
dx
any points on the ellipse with a horizontal tangent.
when 2x
y
=0
=)
y
2
= 2x.
Substitute y = 2x into the original equation for the ellipse and solve for x.
x2 + x(
2x) + ( 2x)2 = 3
()
()
3x2 = 3
1
x=
Substitute the x-values into y = 2x to nd the corresponding values for y . The ordered
pairs of the points with horizontal tangents are ( 1; 2) and (1; 2).
7. A stone dropped into a still pond sends out a circular ripple. The radius is increasing at a rate of
3 ft/sec. After ten seconds, what is the rate of change of the area?
We need to nd
t = 10
dA
.
dt
Area A = r2 , dierentiating with respect to t gives
seconds, the radius is r = 10(3) = 30 feet. Therefore,
dA
dt
dA
dt
= 2r
dr
.
dt
At time
= 2 (30)(3) = 180 ft2 /sec.
8. A 15 foot ladder is leaning against a house when its base starts to slide away. By the time the
base is 12 feet from the house, the base is moving at a rate of 7 ft/sec.
dx
dy
(a) We are given
= 7 ft/sec, but need to nd . Using the Pythagorean Theorem we can
dt
dt
establish a relationship between the sides of the right triangle.
x2 + y 2 = 152
Dierentiating both sides with respect to t and solving for
2x
dx
dt
+ 2y
dy
dt
=)
=0
p
dy
dt
=
x
y
dx
dt
dy
dt
gives:
12
(7) 9:33 ft/sec.
9
In other words, the top of the ladder is sliding down the wall at 9.33 ft/sec.
When x = 12 then y =
152
122 = 9. Therefore,
dx
d
dy
dt
=
(b) We are given
= 7 ft/sec, but need to nd . Using trigonometry we can establish a
dt
dt
relationship between the base of the right triangle and the angle .
x
x
sin =
=)
= arcsin
15
15
Dierentiating both sides with respect to t gives:
d
dx
1
= q
dt
x 2 dt
15 1
15
3
d
dt
When x = 12 then y = 9, therefore,
=
7
q
15 1
12 2
=
15
7
= 0:78 radians/sec.
9
In other words, the angle between the ladder and the wall is changing at a rate of 0.78
radians/sec.
9. Let f (x) = x3 + ax2 + bx + c, where a, b, and c are some constants. Then f 0 (x) = 3x2 + 2ax + b
and f 00 (x) = 6x + 2a.
(a) Since the graph of f has an inection point at x = 3, then f 00 ( 3) = 0
6( 3) + 2a = 0
=)
a=9
Therefore, a = 9, but there are no restrictions on the values of b and c.
(b) From (a), an inection point at x = 3 implies a = 9 therefore, f 0 (x) = 3x2 + 18x + b. Since
the graph of f has a local maximum at x = 4, then f 0 ( 4) = 0.
3( 4)2 + 18( 4) + b = 0
=)
b = 24
Therefore, a = 9 and b = 24, but there is no restriction on the value of c.
(c) Since a = 9 and b = 24, f (x) = x3 +9x2 +24x + c, f 0 (x) = 3x2 +18x +24, and f 00 (x) = 6x +18.
To nd critical points:
f 0 (x) = 3x2 + 18x + 24 = 3(x + 4)(x + 2) = 0
=)
x=
2; 4
Use the Second Derivative Test to nd local extrema:
f 00 (
f 00 (
4) < 0, therefore the graph of f has a local maximum at x = 4.
2) > 0, therefore the graph of f has a local minimum at x = 2.
g (x)
x!1 f (x)
10. (a) A function f (x) dominates g (x) as x ! 1 if lim
(b) lim
x!1
ln(100x)
x!1
g (x) = x2=3
x2=3
Therefore,
= lim
11. Consider f (x) = x2=3 (x
100 =
1
100x
2
x 1 =3
3
3
lim
x!1 2x2=3
= 0.
=0
dominates f (x) = ln(100x) as x ! 1.
5). Then f 0 (x) =
(a) f 0 (x) is undened when 3x1=3 = 0
5(x 2)
10(x + 1)
and f 00 (x) =
.
3x1=3
9x4=3
=)
f 0 (x) = 0 when 5(x 2) = 0
x = 0.
=)
x = 2.
0
(b) We can use a sign chart to see that f (x) < 0 for 0 < x < 2, and f 0 (x) > 0 for x < 0 and
x > 2. Therefore, f is increasing on ( 1; 0) and (2; 1). The graph of f is decreasing on
(0; 2).
(c) Inection points occur when the function changes concavity.
f 00 (x) is undened when 9x4=3
f 0 (x) = 0 when 10(x + 1) = 0
=0
=)
=)
x=
x = 0.
1.
Therefore, f has possible inection points at x = 0 and x = 1. We can use a sign chart
to see that f 00 (x) < 0 for x < 1, and f 00 (x) > 0 for 1 < x < 0 and x > 0. Therefore,
the graph of f changes from concave down to concave up at x = 1, and f does not change
concavity at x = 0. In other words, f has an inection point at x = 1.
(d) Using the information in part(b), the graph of f has a local maximum at x = 0 and a local
minimum at x = 2
4
(e) The graph of f is provided below.
(f) To nd the global extrema of f we need to evaluate f at the critical points and the endpoints
of the interval [ 1; 3].
f ( 1) = 6
3
f (3) = 2 9
f (0) = 0
3
f (2) = 3 4
p
p
Therefore, on the interval [ 1; 3], f has a global minimum of 6 at x = 1, and f has a
global maximum of 0 at x = 0.
(g) From above, we know that the maximum value of f on the interval [ 1; 3] is 0, while the
minimum value of f is 6. Therefore, 6 f (x) 0 on [ 1; 3].
12. The US Postal Service will accept a box for domestic shipment only if the sum of its length and
girth (distance around) does not exceed 108 inches. What dimensions will give a box with a square
end the largest possible volume?
We want to maximize the volume of the box. Let x represent the length of each side of the square
end. Let y represent the length of the box. We need to maximize V = x2 y . The girth given by
4x + y is xed at 108 inches. Therefore, y = 108 4x, substituting this expression for y into the
formula for volume gives V = x2 (108 4x) = 108x2 4x3 .
dV
dt
= 216x
12x2 = 12x(x
18) = 0 when x = 0; 18.
Given the context of the problem, x = 18 is the only appropriate critical value.
d2 V
At x = 18, 2 = 216
dt
inches.
24(18) < 0, therefore the volume of the box is maximized when x = 18
13. A silo (base not included) is to be constructed in the form of a cylinder with a hemisphere attached to the top. The cost of construction per square inch of surface area is twice as great for the
hemisphere as it is for the cylindrical sidewall, for which the cost is $25 per square foot. Determine
the dimensions to be used if the volume of the silo is xed at 5000 cubic feet and the cost of
construction is to be kept to a minimum.
Useful formulas:
Volume:
cylinder V = r2 h
Surface Area:
cylinder A = 2rh
4
sphere V = r3
3
sphere A = 4r2
5
We want to minimize cost C = 25(2rh) + 50(2r2 ).
2
The volume of the silo is given by V = r2 h + r3 is xed at 5000 cubic feet.
3
5000 32 r3 5000 2
2
= 2
r.
Therefore,
5000 = r2 h + r3
=)
h=
3
r2
r
3
Substituting this expression for h into the formula for cost gives
5000 2
250000 200 2
C = 50r
r
+ 100r2 =
+
r .
2
r
3
r
3
Solve
dV
dt
= 0 to nd critical points.
250000
r
750000 + 400r3
7500
=
= 0 when r = 3
.
r2
3r 2
4
d2 V
500000 400
The second derivative 2 =
+
> 0 for all values of r > 0. Therefore, cost is
r3
3
r dt
7500
minimized when r = 3
8:42. The dimensions of the silo are r 8:42 feet and h 16:84
4
feet.
dV
dt
400
+
r=
3
14. (a) Mean Value Theorem: if f is continuous on [a; b] and dierentiable at (a; b), then there exists
f (b) f (a)
a number c, with a < c < b, such that f 0 (c) =
.
b
a
(b) Below is a graph to providing a geometric interpretation of the Mean Value Theorem for the
function described in c. The slope of the tangent line at x = c is equal to the slope of the line
through points A and B.
(c) Since f is continuous on [0; 3] and dierentiable on (0; 3), f satises the conditions of the
Mean Value Theorem. Therefore, there exists a number c in the open interval (0; 3) such that
f (3) f (0)
15
f 0 (c) =
=
= 5. In particular, since f 0 (x) = x2 + 2,
3 0
3
p
c2 + 2 = 5 =) c = 3:
6