MATH 13200
Midterm 1 Practice Answers
January 27, 2016
1. Since cos θ and sin θ are continuous at θ = 0, we have
lim cos α = cos α = 1
lim sin α = sin α = 0
α→0
α→0
Using the change of coordinates α = θ − θ0 , and the addition formula, we see that
lim cos θ = lim cos(α + θ0 )
α→0
θ→θ0
= lim cos α cos θ0 − sin α sin θ0
α→0
= lim cos α cos θ0 − lim sin α sin θ0
α→0
α→0
= cos θ0 lim cos α − sin θ0 lim sin α
α→0
α→0
= cos θ · 1 − sin θ · 0
= cos θ0
Thus, by the definition of continuity, cos θ is continuous at θ = θ0 .
2. Using the quotient rule, we have
1
Dx (sec x) = Dx
cos x
0 · cos x − 1 · Dx (cos x)
=
(cos x)2
sin x
=
cos2 x
sin x
1
=
·
cos x cos x
= tan x · sec x
3. By the monotonicity theorem, f (x) = cot x is increasing when f 0 (x) > 0 and decreasing when
f 0 (x) < 0. We know that
Dx cot x = − csc2 x
Since the square of a number is always positive, Dx cot x is negative whenever it is defined.
The x-values in [−π, 3π] where csc x is undefined (i.e. where the y-value of Px is 0) are
−π, 0, π, 2π, 3π. Thus, cot x is increasing on
(−π, 0) ∪ (0, π) ∪ (π, 2π) ∪ (2π, 3π)
and decreasing nowhere.
4. (a) The concavity theorem says that f (x) is concave up when f 00 (x) is increasing, and is
concave down when f 00 (x) is decreasing. We have
f 0 (x) = −
3
x4
f 00 (x) =
12
x5
Since x5 is greater than 0 for x > 0, and less than 0 for x < 0, we see that f (x) is concave
up for (0, 3] and concave down for [3, 0).
(b) The only point at which f changes from concave up to concave down is at 0. However, f
is not continuous at 0. Thus, 0 is not an inflection point, and hence f has no inflection
points.
5. (a) f (c) is the slope of the secant line through (0, 0) and the point on the graph of f (x) at
x = c.
(b) We would set it to equal 1 at 0, since
lim
x→0
sin x
=1
x
(c) Since the derivative is the limit of the slopes of the secant lines, part (b) means that the
derivative at 0 is 1. I.e., sin0 (0) = 1.
6. We are trying to maximize
A = (2x) · y
From the definition of sin and cos, we see that
x = cos θ
y = sin θ
Thus,
A = 2 cos θ sin θ
Taking the derivative with respect to θ, we see
A0 = 2(cos2 θ − sin2 θ)
Apply the Pythagorean identity:
A0 = 2(2 cos2 θ − 1)
To maximize A, we must look for the critical points of A as a function of θ. In particular, we
look for when A0 is undefined, when A0 = 0 and at the endpoints.
We see that A0 is always defined. Furthermore, it’s easy to see that at the endpoints A = 0, so
we can exclude those points. So we know the maximum must be when
A0 = (2 cos2 θ − 1) = 0
Solving for cos θ shows that
1
cos θ = √
2
√
Since there is exactly one θ-value in the interval [0, π/2] such that such that cos θ = 1/ 2, we
conclude that
π
θ=
4
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