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MATH 1914 Homework 9 Solutions
Section 3.5 Problems
6. Use the guidelines of this section to sketch the curve.
y = 2x − tan x,
−π/2 < x < π/2
y 0 = 2 − sec2 x
y 00 = −2 sec x(sec x tan x)
= −2 sec2 x tan x
Critical points are the values of x so that y 0 = 0
Or
2 − sec2 x = 0
sec2 x = 2
1
=2
cos2 x
1
cos2 x =
2
1
cos x = √
2
π π
x=− ,
4 4
Where the solutions are over the interval −π/2 < x < π/2. By either plugging test values or
considering the fact that cos x increases on the interval (− π2 , 0) and decreases on the interval
(0, π2 ), the signs for y 0 is as follows:
x
y0
(− π2 , − π4 ) (− π4 , π4 ) ( π4 , π2 )
−
+
−
The function achieves local minimum at x = − π4 , y = − π2 + 1, and local maximum at
x = π4 , y = π2 − 1
For the concavity, set y 00 = 0. This implies either sec x = 0 or tan x = 0. Since sec x = cos1 x
can never be zero, the only solution comes from tan x = 0, or x = 0. The table for concavity
1
is as follows:
x
y 00
(− π2 , 0) (0, π2 )
+
−
Before drawing the graph, the function y has the following vertical assymptotes: when
x → − π2 + , tan x → −∞ and when x → π2 − , tan x → ∞. Also, notice when replacing x with
−x, the y values also gets negated. This implies that the function is symmetric about the
origin. The graph is as follows.
8 Use the guidelines of this section to sketch the curve. In guideline D find an equation
of the slant asymptote.
8.
2x2
y=
x−1
First perform long division to simplify the calculation:
y = 2x + 2 +
2
x−1
2
(x − 1)2
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y 00 =
(x − 1)3
y0 = 2 −
2
(*)
The critical points are where
2−
2
=0
(x − 1)2
(x − 1)2 = 1
x − 1 = ±1
x = 0, 2
The derivative is also undefined at x = 1. The table for the first derivative is as follows:
x
y0
(−∞, 0) (0, 1) (1, 2) (2, ∞)
+
−
−
+
with local minimum at (0, 0) and local maximum at (2, 8).
Similarly, inflection points are the x-values such that y 00 = 0, which is never satisfied for
all values of x. However, y 00 is undefined at x = 1. The table for concavity is as follows:
x
y 00
(−∞, 1) (1, ∞)
−
+
In addition, the funciton also has a vertical asymptote at x = 1. y → −∞ as x → 1−
and y → +∞ as x → 1+ . For the slant assymptote, notice in the (*) expression of y, the
2
term x−1
has limit 0 as x → ∞. Therefore the slant assymptote is y = 2x + 2. The graph is
as follows.
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Section 3.7 Problems
11. Find two positive numbers whose product is 200 and whose sum is a minimum.
Let the two numbers be x and y. Since xy = 200, y =
function of x, the sum can be written as
s(x) = x +
200
.
x
To minimize the sum as a
200
x
Its extreme is achieved at the critical point, when s0 (x) = 1 − 200
= 0 or equivalently
x2
√
x = 200. Using√either the first or second derivative test one can conclude this is a local
minimum. x = y 200.
16. Find the dimensions of the isosceles triangle of largest area that can be inscribed in
a circle of radius r.
Let θ be the angle OAB as drawn in the graph. Let a(θ) be the area of the isoscele triangle
ACD as a function of the angle θ. Since the radius of the circle is r,
OA = OD = OC = r
AB = OA · cos θ = r cos θ
OB = OA · sin θ = r sin θ
Therefore the area can be written as
1
a(θ) = AD · CB
2
1
= · 2AB · (CO + OB)
2
= r cos θ(r + r sin θ)
= r2 cos θ(1 + sin θ)
where r is a constant. To maximize the area is to find the critical point of this function,
or
a0 (θ) = r2 ((− sin θ)(1 + sin θ) + cos2 θ)
= r2 (− sin θ − sin2 θ + cos2 θ)
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To solve the equation a0 (θ) = 0, rewrite the second factor as
− sin θ − sin2 θ + (1 − sin2 θ) = 0
−2 sin2 θ − sin θ + 1 = 0
This is equivalent to solving the equation −2u2 −u+1 = 0 when u = sin θ. The quadratic
formula yields u = 21 or u = −1 (discard). Note this is when θ = π6 = 30◦ .
To intepret the result, notice the angle AOB equals 90◦ − 30◦ = 60◦ . Consequently, the
angle DOB equals 60◦ and the total angle AOD equals 120◦ . The surrounding two angles
DOC and AOC are equal based the symmetry, and each should be 120◦ . This implies that
ACD is in fact an equilateral triangle.
18. Two vertical poles at P and Q are secured by a rope PRQ going from the top of the
first pole to a point R on the ground between the poles and then to the top of the second
pole as in the figure. Show that the shortest length of such a rope occurs when θ1 = θ2 .
Let x be the distance between M and R as shown in the graph. Let H be the fixed height
of the corridor and L be the distance between the two poles P and Q. The length of the rope
s(x) can be written as a function in x:
s(x) = P R + QR
p
√
= H 2 + x2 + (L − x)2 + H 2
Its minimum is achieved at is critical point where s0 (x) = 0:
2x
−2(L − x)
s0 (x) = √
+p
=0
2 H 2 + x2
(L − x)2 + H 2
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This is equivalent to
2(L − x)
2x
√
=p
2
2
2 H +x
(L − x)2 + H 2
(L − x)
x
√
=p
2 H 2 + x2
(L − x)2 + H 2
Without solving for actual values of x, this equation is the exact statement for
RN
MR
=
PR
RQ
cos θ1 = cos θ2
Since the cosine function is one-to-one on the interval (0, π2 ), the minimum is achieved
when θ1 = θ2 .
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