Week #1 - Introduction to Functions
Section 1.5
From “Calculus, Single Variable” by Hughes-Hallett, Gleason, McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
SUGGESTED PROBLEMS
For Exercises 1-9, draw the angle using a ray through the origin, and determine whether
the sine, cosine, and tangent of that angle are positive, negative, zero, or undefined.
1.
3π
2
Graph below:
From the quadrant, sin is negative, cos is zero and tan = sin / cos is undefined.
3.
π
Graph below:
4
From quadrant, all sin, cos, and tan are positive.
5.
π
Graph below:
6
1
From quadrant, all sin, cos, and tan are positive.
Given that sin(π/12) = 0.259, and cos(π/5) = 0.809, compute (without using the trigonometric functions on your calculator) the quantities in Exercises 10-12. You may want to
draw a picture showing the angles involved, and then check your answers on a calculator.
10. cos(−π/5)
From the symmetry of y = cos(x) across the y axis, we can see that cosine is unaffected
when we change the sign of the angle: that cos(θ) = cos(−θ). We can therefore say that
cos(−π/5) = cos(π/5) = 0.809.
11. sin(π/5)
We can find this by using the trigonometric identity sin2 (θ) + cos2 (θ) = 1.
p
p
sin(π/5) = 1 − cos2 (π/5) ≃ 1 − 0.8092 ≃ 0.588
We know that we should be taking the positive root because π/5 is in the first quadrant
(where sin(θ) is always positive).
12. cos(π/12)
By the trig identity sin2 (θ) + cos2 (θ) = 1,
q
p
cos(π/12) = 1 − sin2 (π/12) ≃ 1 − 0.2592 ≃ 0.966
13. Consider the function y = 5 + cos(3x).
(a) What is its amplitude?
(b) What is its period?
(c) Sketch its graph.
(a) amplitude = coefficient of cosine = 1 (implicitly)
(b) period = 2π/coeff of x = 2π/3
5
4
y
6
(c) Graph below
−π/2−π/3−π/6 0 π/6 π/3 π/22π/35π/6 π
x
2
Find the period and amplitude in Exercises 14-17.
14. y = 7 sin(3t) Period = 2π/3. Amplitude = 7.
15. z = 3 cos(u/4) + 5 Period = 8π. Amplitude = 3.
16. w = 8 − 4 sin(2x + π) Period = pi. Amplitude = 4.
17. r = 0.1 sin(π t) + 2 Period = 2. Amplitude = 0.1.
For Exercises 19-28, find a possible formula for each graph.
21.
−4 sin(2x)
25.
2 − sin(x)
In Exercises 29-33, find a solution to the equation if possible. Give the answer in exact
form and in decimal form.
29. 2 = 5 sin(3x)
One solution is x = 13 sin−1 (2/5) ≃ 0.137. (Make sure your calculator is in radian mode to
get this value.) Because sin is periodic, this is only one of an infinite number of solutions.
31. 8 = 4 tan(5x)
One solution is x = 15 tan−1 (2) ≃ 0.2214. Because tan is periodic, this is only one of an
infinite number of solutions, separated by π/5.
33. 8 = 4 sin(5x)
This has so solution because sin(5x) = 2 is not possible for any x. The range of sin(5x)
is only [−1, 1], so can’t reach a value of 2.
36. What is the difference between sin x2 , sin2 x, and sin(sin(x))? Express each of the three
as a composition. (Note: sin2 xis another way of writing (sin(x))2 .)
If f (x) = x2 and g(x) = sin(x), then
• sin x2 - First take square of x, then take the sine. (= g(f (x)))
• sin2 x - equal to (sin(x))2 , take sine first, then square the result. (= f (g(x)))
• sin(sin(x)) - take sine of x, then take sine of that result. (= g(g(x)))
QUIZ PREPARATION PROBLEMS
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34. A compact disk spins at a rate of 200 to 500 revolutions per minute. What are the
equivalent rates measured in radians per second?
Using the fact that 1 revolution = 2π radians, and 1 minute = 60 seconds, we have
200rev/min = 200 · 2πrad/min
1
= 200 · 2π rad/sec
60
(200)(6.283)
≃
60
≃ 20.94radians per second
Similarly, 500 rpm is equivalent to 52.36 radians per second.
35. When a car’s engine makes less than about 200 revolutions per minute, it stalls. What is
the period of the rotation of the engine when it is about to stall?
200 rotations per minute →
→
1
minutes per rotation
200
60
3
=
seconds per rotation
200
10
The period would be 0.3 seconds.
38. The Bay of Fundy in Canada has the largest tides in the world. The difference between
low and high water levels is 15 meters (nearly 50 feet). At a particular point the depth of
the water, y meters, is given as a function of time, t, in hours since midnight by
y = D + A cos(B(t − C))
(a) What is the physical meaning of D?
(b) What is the value of A?
(c) What is the value of B? Assume the time between successive high tides is 12.4 hours.
(d) What is the physical meaning of C?
(a) D = the average depth of the water.
(b) A = the amplitude = 15/2 = 7.5
(c) Period = 12.4 hours.Thus, (B)(12.4) = 2π so, B = 2π/12.4 ≈ 0.507
(d) C is the time of high tide.
42. A baseball hit at an angle of θ to the horizontal with initial velocity v0 has horizontal
range, R, given by
R=
v02
sin(2θ)
g
4
Here g is the acceleration due to gravity. Sketch R as a function of θ for 0 ≤ θ ≤ π/2.
What angle gives the maximum range? What is the maximum range?
0
R
v0^2/g
The function R has period π, so its graph is as shown in the figure below. The maximum
ν2
value of the range is g0 and it occurs when θ = π/4.
π/4
0
π/2
θ
43. A population of animals oscillates sinusoidally between a low of 700 on January 1 and a
high of 900 on July 1.
(a) Graph the population against time.
(b) Find a formula for the population as a function of time, t, in months since the start
of the year.
800
700
Population
900
(a) Graph below
Jan 1
Jul 1
Jan 1
Time
(b) Since t is measured in months from Jan 1st, the desired period is 12 months. As a
result, the coefficient inside our trig function should be 2π/12 = π6 .
The total variation in the population cycle is from 700 to 900. This means that the
central value is 800, and that the amplitude is 100.
Since the curve should start at its lowest value (Jan 1 is when t = 0), any of the
following two formulas would work.
π
P = 800 − 100 cos( t)
6
or
π
P = 800 + 100 sin( (t − 3))
6
5
or
π
P = 800 + 100 sin( t − π/2))
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45. The point P is rotating around a circle of radius 5 shown in Figure 1.52. The angle , in
radians, is given as a function of time, t, by the graph in Figure 1.53.
(b) Describe in words the motion of the point P on the circle.
Figure 1.52
Figure 1.53
(b) Firstly, you should note that 6.3 ∼ 2π. In other words, θ is ranging from 0 to 2π,
or over a full circle.
From t = 0 to t = 2.5, the angle θ is increasing until it reaches 2π, so the point
P moves from the point (5, 0) counter-clockwise until it completes a full circle at
t = 2.5, arriving at (5, 0) again, but from below. At t = 2.5, the point stops, and
then begins to move back in the opposite direction (clockwise now) around the circle
until it returns to its original starting point at (5, 0) from above, arriving at t = 5.
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