The University of Sydney
School of Mathematics and Statistics
Solutions to Tutorial for Week 2
MATH1011: Applications of Calculus
Semester 1, 2014
Web Page: http://www.maths.usyd.edu.au/u/UG/JM/MATH1011/
Lecturer: Clinton Boys, Ross Ogilvie and George Papadopoulos
Questions to complete BEFORE the tutorial
1. Differentiate the following
(a) sin(2x)
Solution: 2 cos(2x).
(b) cos(4x)
Solution: −4 sin(4x).
(c) sin(4x3 + 1)
Solution: 12x2 cos(4x3 + 1).
(d) cos(7x2 + x)
Solution: −(14x + 1) sin(7x2 + x).
(e)
sin x
cos x
Solution:
cos2 x+sin2 x
cos2 x
= sec2 x.
(f) x2 sin(4x + 3)
Solution: 2x sin(4x + 3) + 4x2 cos(4x + 3)
2. Plot, using the same axes, the graphs of
y = sin(x)
y = sin x +
π
2
.
Write down the period, amplitude, vertical shift and horizontal shift of each graph and
describe the relationship between the graphs.
Solution: The graph of the first function is solid, and the second dashed.
y
1
−2π
−π
π
2π
x
−1
Both functions have period 2π and amplitude 1. The graph the second function is the
result of moving the graph of the first π2 units to the left.
1
3. Plot, using the same axes, the graphs of
y = sin x
y = 2 sin x.
Write down the period, amplitude, vertical shift and horizontal shift of each graph and
describe the relationship between the graphs.
Solution: The graph of the first function is solid, and the second dashed.
y
2
1
−2π
−π
π
2π
x
−1
−2
Both functions have period 2π. The first has amplitude 1, the second 2. The graph of
y = 2 sin x is obtained from that of y = sin x by moving each point on the graph either
up or down so as to double its distance from the x-axis. In other words, there is a vertical
stretching by a factor of 2.
Tutorial questions
4. Plot, using the same axes, the graphs of
y = sin(x)
y = 3 + sin (2x) .
Write down the period, amplitude, vertical shift and horizontal shift of each graph and
describe the relationship between the graphs.
Solution: The graph of the first function is solid, and the second dashed.
y
4
3
2
1
−2π
−π
π
2π
x
−1
y = 3+sin 2x has period π, amplitude 1 and mean value 3. Thus the graph of y = 3+sin 2x
is the result of horizontally squashing the graph of y = sin x (halving the period) and
then moving it 3 units in the positive y-direction (that is, upwards).
2
5.
(a) Find an equation for a sinusoidal function, f (t), with amplitude 5 and period 3π.
Solution:
f (t)
5
−3π
3π
t
−5
The graph for f is the result of stretching the graph y = sin t horizontally by a
factor of 32 and vertically by a factor of 5. Thus
2t
.
3
In fact, the start point and mean value of the graph are not specified, so f can be
any function of the form
f (t) = 5 sin
2(t − θ)
3
2(t − θ)
or f (t) = A + 5 cos
3
for some numbers A and θ.
f (t) = A + 5 sin
(b) Find an equation for a sinusoidal function, f (t), with amplitude 1 and period π6 .
Solution:
f (t)
1
−2π
−π
π
2π
t
−1
We can stretch the graph y = sin t horizontally by a factor of 1/12 to obtain
f (t) = sin 12t.
In fact, the start point and mean value of the graph are not specified, so f can be
any function of the form
f (t) = A + sin 12(t − θ)
or f (t) = A + cos 12(t − θ)
for some numbers A and θ.
2
.
9
Solution: The ordinary sine curve has period 2π, so we need to scale it horizontally by a factor of 2π/ 92 = 9π. Thus one such function is
(c) Find an equation for a sinusoidal function, f , with amplitude 8 and period
f (t) = 8 sin 9πt.
3
In fact, the start point and mean value of the graph are not specified, so f can be
any function of the form
f (t) = A + 8 sin 9π(t − θ)
or f (t) = A + 8 cos 9π(t − θ)
for some numbers A and θ.
6. Sketch the graph of y = 8 sin (3πx) + 2.
Solution:
f (t)
10
2
−3
−2
−1
1
2
3
t
−6
(The t and f (t) scales are unequal here, to make it easier to see the shape of the curve.)
7. Plot, using the same axes, the graphs of
y = cos(x)
y = 2 cos (3x − π) − 1.
Write down the period, amplitude, vertical shift and horizontal shift of each graph and
describe the relationship between the graphs.
Solution: The graph of the first function is solid, and the second dashed.
y
1
−2π
−π
π
−1
−2
−3
4
2π
x
y = 2(cos 3x − π) − 1 = 2 cos 3(x − π3 ) -1 has period 2π/3, amplitude 2, mean value −1
and horizontal shift π/3. Squash cos x horizontally by a factor of 3 (i.e. stretch by a
factor of 1/3) and then move the graph one unit downwards and π/3 units to the right.
8. The sinusoidal function f is illustrated below. The sinusoidal function g has mean value 0
and is such that g(0) = 0, and is related to the function f in the following manner: the
amplitude of g is twice the amplitude of f and the period of g is three times that of f .
Find an equation for the function g.
f (t)
5
4
3
2
1
−1
1
2
3
t
Solution: We see from the graph that f (t) has the amplitude 2, mean value 3 and
period 1. Thus, f (t) can be given by the formula
f (t) = 2 sin 2πt + 3.
The amplitude of g(t) is therefore 4, its mean value is 0 and period is 3. So, the equation
of g(t) is
2πt
.
g(t) = ±4 sin
3
9. Using the addition law sin(A + B) = sin(A) cos(B) + cos(A) sin(B) and the fact that
sin(−θ) = − sin(θ) and cos(−θ) = cos(θ), show that
sin(A − B) = sin(A) cos(B) − cos(A) sin(B).
Solution:
Using the identity sin(A + B) = sin(A) cos(B) + cos(A) sin(B), we may write
sin(A − B) = sin(A + (−B)) = sin(A) cos(−B) + cos(A) sin(−B)
= sin(A) cos(B) − cos(A) sin(B)
as required.
10. Using the addition law cos(x + y) = cos(x) cos(y) − sin(x) sin(y), prove that cos(2x) =
cos2 x − sin2 x. Hence prove the formulae
1
sin2 x = (1 − cos 2x) and
2
5
1
cos2 x = (cos 2x + 1).
2
Solution:
cos(2x) = cos(x + x) = cos x cos x − sin x sin x
= cos2 x − sin2 x.
Now, using the Pythagorean identity sin2 x + cos2 x = 1, we can arrange the above to get
cos(2x) = cos2 x − (1 − cos2 x) = 2 cos2 x − 1 giving
1
cos2 x = (cos 2x + 1).
2
Or on the other hand, we can cos(2x) = 1 − sin2 x − sin2 x = 1 − 2 sin2 x, giving
1
sin2 x = (1 − cos 2x).
2
6