Math 1102: Calculus I (Math/Sci majors)
MWF 3pm, Fulton Hall 230
Homework 4 Solutions
Please write neatly, and show all work. Caution: An answer with no
work is wrong!
Problem A. Use Weierstrass’ (,δ)-definition of lim f (x) = L to show
x→a
that the sum of two continuous functions is continuous. (Recall that
we showed rigorously in class that the limit of a sum is the sum of the
limits, provided each exist.)
Solution A. Let f and g be the two continuous functions in question.
We assume that they have a common domain of definition (otherwise
the sum of the functions makes no sense – who wrote this question?),
and we will write f + g for the function x 7→ f (x) + g(x). By definition
of continuity, the function f + g is continuous if
lim (f + g)(x) = (f + g)(a)
x→a
at each point a where f + g is defined. At each such a, continuity of f
and g imply that both
lim f (x) = f (a),
x→a
and
lim g(x) = g(a).
x→a
We’ve seen in class (rigorously, using Weierstrass’ definition) that when
two summand limits exist, the limit of a sum is the sum of limits. Thus
lim (f + g)(x) = lim f (x) + g(x)
x→a
x→a
= lim f (x) + lim g(x)
x→a
x→a
= f (a) + g(a) = (f + g)(a),
so that f + g is continuous at each point where it is defined.
Problem B. Ignore Weierstrass’ definition, and evaluate the following
limits:
1 − cos x
(1) lim
x→0
x2
sin(2h)
(2) lim
h→0
h
Hints:
(1) Look at how we evaluated lim
h→0
cos h − 1
in class.
h
sin h
.
h→0 h
(2) Compare with lim
Solution B.
(1): We have
1 − cos x
1 − cos x 1 + cos x
= lim
·
2
x→0
x→0
x
x2
1 + cos x
1 − cos2 x
1
= lim
·
x→0
x2
1 + cos x
2
sin x
1
= lim
·
x→0 x2
1 + cos x
sin x sin x
1
= lim
·
·
.
x→0 x
x
1 + cos x
The limit of a product is the product of limits, when the latter exist.
sin x
We’ve seen in class that lim
= 1. The last function, (1 + cos x)−1 ,
x→0 x
is the reciprocal of a function that has a non-zero limit as x goes to 0,
so it exists as well. We have
1 − cos x
1
1
lim
=1·1·
= .
2
x→0
x
1+1
2
lim
(2): Let t = 2h, so that h = t/2. As h → 0, t does as well, and we have
sin t
sin t
sin(2h)
= lim
= 2 lim
= 2.
t→0 t/2
t→0 t
h→0
h
lim
Problem C. Find the equation of the line tangent to the graph of
y = log(tan x) when x = π/4.
Solution C. We’ve seen already that
1
and
(log x)0 = ,
x
1
.
(tan x)0 =
cos2 x
In order to compute the tangent line, we compute the derivative using
the chain rule:
1
1
cos x
1
1
y0 =
·
=
·
=
.
2
2
tan x cos x
sin x cos x
sin x cos x
When x = π/4, this evaluates to
y 0 |x=π/4 =
√1
2
1
= 2.
· √12
At x = π/4, the point on the graph of y = log(tan x) has y-coordinate
y = log(tan π4 ) = log 1 = 0. The equation of a line through π4 , 0 with
slope 2 is given by
π
y =2 x−
.
4
Problem D. Suppose an isosceles triangle has base angle θ, and that
the perimeter of the triangle is the fixed length 2. In this case, the area
of the triangle, A, is a function of the angle θ. Find the derivative of
the area with respect to θ, dA/dθ, when the triangle is equilateral.
Solution D. Suppose the length of the two equal length sides is x, and
the base is given by y. In this case, the perimeter is 2x + y, and we
have
2x + y = 2.
On the other hand, drawing the altitude of the triangle that meets the
base of the isosceles triangle, we see a right triangle with an angle θ,
hypotenuse x, and side adjacent to the angle y/2. Trigonometry tells
us that
y/2
cos θ =
,
x
or
y = 2x cos θ.
Plugging in to the perimeter equation, we find
2x + 2x cos θ = 2,
and solving for x gives
1
.
1 + cos θ
The area of the triangle is half of its base times its height. The base
is given by y, and the same right triangle above tells us the height is
given by x sin θ, so we obtain
xy sin θ
A=
2
2
= x sin θ cos θ
sin θ cos θ
=
(1 + cos θ)2
x=
Now that we have A as a function of θ alone, we may compute the
d
derivative directly. (We use primes below as shorthand for dθ
).
0
dA
(sin θ cos θ)0 (1 + cos θ)2 − sin θ cos θ ((1 + cos θ)2 )
=
dθ
(1 + cos θ)4
(cos2 θ − sin2 θ)(1 + cos θ)2 − sin θ cos θ · 2(1 + cos θ) · (− sin θ)
=
(1 + cos θ)4
(cos2 θ − (1 − cos2 θ))(1 + cos θ)2 + 2 cos θ(1 + cos θ)(1 − cos2 θ)
=
(1 + cos θ)4
(2 cos2 θ − 1)(1 + cos θ)2 + 2 cos θ(1 − cos θ)(1 + cos θ)2
=
(1 + cos θ)4
2 cos θ − 1
2 cos2 θ − 1 + 2 cos θ(1 − cos θ)
=
=
2
(1 + cos θ)
(1 + cos θ)2
When the triangle is equilateral, all of the angles are π/3, so we may
compute:
2 cos π3 − 1
dA =
= 0,
dθ θ=π/3 (1 + cos π3 )2
since cos π/3 = 1/2.
Problem E. If n is an integer, show that
cos((n + 1)x) = 2 cos x cos(nx) − cos((n − 1)x).
Solution E. We use the sum formula for cos, and the fact that cos is
an even function while sin is odd, expanding the right-hand side:
2 cos x cos(nx) − cos((n − 1)x) = 2 cos x cos(nx) − cos(nx − x)
= 2 cos x cos(nx) − (cos(nx) cos(−x) − sin(nx) sin(−x))
= 2 cos x cos(nx) − (cos(nx) cos x + sin(nx) sin x)
= 2 cos x cos(nx) − cos(nx) cos x − sin(nx) sin x
= cos x cos(nx) − sin(nx) sin x
= cos(x + nx) = cos((n + 1)x).
Note that n did not have to be an integer in the above analysis. (Who
wrote these questions?)
5
Problem F. Find the values of x at which the tangent line to the
graph of y = cos x + sin x is horizontal.
Solution F. The tangent line is horizontal when its slope is 0. The
slope of the tangent at x is also the value of the derivative at x. We
compute directly that the derivative is given by
y 0 = − sin x + cos x = cos x − sin x.
Thus the tangent line is horizontal exactly when cos x = sin x. Squaring
both sides, and replacing cos2 x with 1 − sin2 x, we find
√
sin x = ±1/ 2.
This occurs for x = π/4 + πn/2, for any integer n. However, cos x =
sin x in particular implies that cos and sin have the same sign, so we
only look at the first and third quadrants. Thus when x = π/4 + πn,
for any integer n, we have cos x = sin x, and the tangent line to
y = sin x + cos x is horizontal.
Problem G. Let f (x) = sin(2x). What is f (101) (x)?
Solution G. We compute directly, using the chain rule repeatedly:
f 0 (x) = 2 cos(2x)
f (2) (x) = −4 sin(2x)
f (3) (x) = −8 cos(2x)
f (4) (x) = 16 sin(2x)
Thus f (4) (x) = 16f (x), and f (4n) (x) = 16n f (x), for any integer n. This
implies that f (100) (x) = 1625 f (x), so that
f (101) (x) = (1625 f (x))0 = 1625 f 0 (x) = 2 · 1625 cos(2x) = 2101 cos(2x).
Problem H. Let f (x) = ex sin x. What is f (95) (x)?
Solution H. We compute directly, using the product rule repeatedly:
f 0 (x) = ex sin x + ex cos x
f (2) (x) = ex sin x + ex cos x + ex cos x − ex sin x
= 2ex cos x
f (3) (x) = 2(ex cos x − ex sin x)
f (4) (x) = 2(ex cos x − ex sin x − ex sin x − ex cos x)
= 2(−2ex sin x) = −4ex sin x.
Thus f (4) (x) = −4f (x), and f (4n) = (−4)n f (x), for any integer n. This
implies that f (92) (x) = (−4)23 f (x), so that
(3)
f (95) (x) = (−4)23 f (x)
= (−4)23 f (3) (x)
= (−4)23 · 2(ex cos x − ex sin x) = −247 ex (cos x − sin x)
Problem I. Suppose that cos−1 x is defined so that its range is in the
interval [π, 2π]. Compute the derivative
d
cos−1 x .
dx
Solution I. By definition, we have cos(cos−1 x) = x. Taking the derivative of each side, we obtain
− sin(cos−1 x) · (cos−1 x)0 = 1,
so that
1
.
− sin(cos−1 x)
√
Since sin2 x + cos2 x = 1, we have sin x = ± 1 − cos2 x. Thus
p
√
sin(cos−1 x) = ± 1 − cos2 (cos−1 x) = ± 1 − x2 .
(cos−1 x)0 =
Since the range of cos−1 is in the interval [π, 2π], the value
of sin(cos−1 x)
√
is less than or equal to 0, so that sin(cos−1 x) = − 1 − x2 . We conclude
1
1
(cos−1 x)0 =
=√
.
−1
− sin(cos x)
1 − x2
Problem J. Define the one-sided limit lim+ f (x).
x→a
Solution J. Suppose there exists L so that the following holds: For
any > 0, there exists δ > 0 so that, for all 0 < x − a < δ, we have
|f (x) − L| < . In that case, lim+ f (x) exists and is equal to L.
x→a
Problem K. A particle moves along the x-axis so that its position
3
at time t is given by x(t) = t2 − t2 . Find the intervals on which the
x-coordinate is increasing/decreasing. Find the acceleration of the particle at time t = 1/2.
2
Solution K. We have the derivative x0 (t) = 2t − 3t2 , and the second
derivative x00 (t) = 2 − 3t. Factoring x0 (t), we see
t
x0 (t) = (4 − 3t).
2
0
Thus x (t) is equal to 0 when t = 0 or 4/3. For t < 0, t/2 < 0 and
4 − 3t > 0, so that x0 (t) < 0. When 0 < t < 4/3, t/2 > 0 and
4 − 3t > 0, so that x0 (t) > 0. When t > 4/3, t/2 > 0 and 4 − 3t < 0, so
that x0 (t) < 0. Thus x(t) is increasing for t between 0 and 4/3, while
x(t) is decreasing for t < 0 or t > 4/3.
Finally, the acceleration is given by x00 (1/2) = 2 − 3 ·
1
2
= 12 .