StudentAidUnit 11: Trigonometric Functions and their Graphs — Exercise 11 3
11th Class Mathematics · Unit 11: Trigonometric Functions and their Graphs
1.Find the maximum and minimum values of the function: (i) 3 - sin 3x (ii) 3 + sin 2x (iii) \frac{1}{2} + \sin(5x + \pi) (iv) \frac{3}{2} + \cos\left(x - \frac{\pi}{4}\right) (v) 1 - 3\cos 2x (vi) 1 + 2\sin\left(x + \frac{\pi}{6}\right) (vii) \frac{1}{10 - 2\sin 3x} (viii) \frac{1}{7 + 3\cos(-2x)} (ix) \frac{1}{5 - 3\cos(3x-1)}
Given
(i) Function
f(x) = 3 - sin 3x
Parameters
a = 3, b = -1
Formula
Maximum when sin 3x = 1
M = a + |b| = 3 + |-1| = 4
Minimum when sin 3x = -1
m = a - |b| = 3 - |-1| = 2
Given
(ii) Function
f(x) = 3 + sin 2x
Parameters
a = 3, b = 1
Formula
Maximum when sin 2x = 1
M = a + |b| = 3 + |1| = 4
Minimum when sin 2x = -1
m = a - |b| = 3 - |1| = 2
Given
(iii) Function
f(x) = 12 + sin(5x + pi)
Parameters
a = 12, b = 1
Formula
Maximum when sin(5x + π) = 1
M = a + |b| = 12 + |1| = 32
Minimum when sin(5x + π) = -1
m = a - |b| = 12 - |1| = -12
Given
(iv) Function
f(x) = 32 + cosleft(x - pi4right)
Parameters
a = 32, b = 1
Formula
Maximum when cos(x - π/4) = 1
M = a + |b| = 32 + |1| = 52
Minimum when cos(x - π/4) = -1
m = a - |b| = 32 - |1| = 12
Given
(v) Function
f(x) = 1 - 3cos 2x
Parameters
a = 1, b = -3
Formula
Maximum when cos 2x = -1
M = a + |b| = 1 + |-3| = 4
Minimum when cos 2x = 1
m = a - |b| = 1 - |-3| = -2
Given
(vi) Function
f(x) = 1 + 2sinleft(x + pi6right)
Parameters
a = 1, b = 2
Formula
Maximum when sin(x + π/6) = 1
M = a + |b| = 1 + |2| = 3
Minimum when sin(x + π/6) = -1
m = a - |b| = 1 - |2| = -1
Given
(vii) Function
g(x) = 110 - 2sin 3x
Find range of denominator
10 - 2sin 3x ranges from 10-2(1)=8 to 10-2(-1)=12
Formula
Maximum of g(x) when denominator is minimum
M = 18
Minimum of g(x) when denominator is maximum
m = 112
Given
(viii) Function
g(x) = 17 + 3cos(-2x)
Note: cos(-2x) = cos(2x)
g(x) = 17 + 3cos(2x)
Find range of denominator
7 + 3cos(2x) ranges from 7-3=4 to 7+3=10
Formula
Maximum of g(x) when denominator is minimum
M = 14
Minimum of g(x) when denominator is maximum
m = 110
Given
(ix) Function
g(x) = 15 - 3cos(3x-1)
Find range of denominator
5 - 3cos(3x-1) ranges from 5-3=2 to 5+3=8
Formula
Maximum of g(x) when denominator is minimum
M = 12
Minimum of g(x) when denominator is maximum
m = 18
(i) Maximum = 4, Minimum = 2quad (ii) Maximum = 4, Minimum = 2quad (iii) Maximum = 32, Minimum = -12quad (iv) Maximum = 52, Minimum = 12quad (v) Maximum = 4, Minimum = -2quad (vi) Maximum = 3, Minimum = -1quad (vii) Maximum = 18, Minimum = 112quad (viii) Maximum = 14, Minimum = 110quad (ix) Maximum = 12, Minimum = 18
2.(a) Find the maximum and minimum temperature during the day. (b) Find the temperature at t = 9 hours (9:00 a.m.). Given: T(t) = \frac{13}{2}\sin\left(\frac{\pi}{6}t - \frac{\pi}{9}\right) + 15, where t is the time in hours with t = 0 corresponding to midnight.
Given
Temperature equation
T(t) = 132sinleft(pi6t - pi9right) + 15
(a) Maximum and minimum temperature
Tmax = a + |b| = 15 + 132 = 432 = 21.5°
Minimum temperature
Tmin = a - |b| = 15 - 132 = 172 = 8.5°
(b) Put t = 9 in equation
T(9) = 132sinleft(pi · 96 - pi9right) + 15
Simplify angle
= 132sinleft(3pi2 - pi9right) + 15 = 132left[-cospi9right] + 15
Numerical evaluation
T(9) = 132(-0.9397) + 15 ≈ -6.11 + 15 = 8.89°
Tmax = 21.5°, Tmin = 8.5°; T(9) ≈ 8.89°
3.A man on the top of 100m high light house is in the line with two ships on the same side of it, whose angles of depression from the man are 17° and 19° respectively. Find the distance between the ships.
Given
Height of light house
h = 100 m
Angles of depression
17° and 19°
Distance to second ship from light house foot
tan 19° = 100BC ⇒ BC = 100tan 19° = 1000.3443 ≈ 290.42 m
Distance to first ship
tan 17° = 100x + 290.42 ⇒ x + 290.42 = 100tan 17° = 327.085
Distance between ships
x = 327.085 - 290.42 = 36.665 m
Distance between ships = 36.66 m
4.P and Q are two points in line with a tree. If the distance between P and Q is 30m and the angles of elevation of the top of the tree at P and Q be 12° and 15° respectively, find the height of the tree.
Given
Distance PQ
PQ = 30 m
Angles of elevation
12° at P, 15° at Q
In triangle QRS
tan 15° = hy ⇒ h = ytan 15°
In triangle PRS
tan 12° = h30 + y ⇒ h = (30 + y)tan 12°
Equating and solving
ytan 15° = (30+y)tan 12° ⇒ y(tan 15° - tan 12°) = 30tan 12°
Calculation
y(0.0553) = 6.372 ⇒ y = 115.34 m
Height of tree
h = ytan 15° = 115.34 × 0.268 = 30.91 m
h = 30.91 m
5.(a) Model an equation that represent the height h(t) of a rider on the Ferris wheel at any given time t. (b) Find the maximum height of a rider. (c) Find the height of the rider from the ground after 35 seconds. Given: A giant Ferris wheel has a diameter of 60 feet. The lowest point of the wheel is located 6 feet above the ground. The wheel completes one full revolution every 80 seconds.
Given
Diameter
d = 60 ft, so radius r = 30 ft
Lowest point height
6 ft above ground
Period
T = 80 seconds, so c = 2pi80 = pi40
Center height
a = 6 + 30 = 36 ft
(a) Model equation (negative cosine for lowest start)
h(t) = -30cosleft(pi t40right) + 36
(b) Maximum height
hmax = a + |b| = 36 + |30| = 66 ft
(c) Height at t = 35 seconds
h(35) = -30cosleft(35pi40right) + 36 = -30cosleft(7pi8right) + 36
Calculation
cos(7pi/8) ≈ -0.924 ⇒ h(35) ≈ -30(-0.924) + 36 ≈ 63.72 ft
h(t) = -30cosleft(pi t40right) + 36; hmax = 66 ft; h(35) ≈ 63.72 ft
6.(a) What is the maximum height reached by the swing seat? (b) What is the minimum height reached by the swing seat? (c) How long does it take for the swing to complete one full back-and-forth motion (period)? (d) At what times does the swing seat first reach a height of 2.12 meters? Given: h(t) = 1.5 + 1.2\sin(3\pi t)
Given
Height function
h(t) = 1.5 + 1.2sin(3pi t) (in meters, t in seconds)
(a) Maximum height
hmax = 1.5 + 1.2(1) = 2.7 m
(b) Minimum height
hmin = 1.5 + 1.2(-1) = 0.3 m
(c) Period of sin(3πt)
Period = 2pi3pi = 23 second
(d) Solve h(t) = 2.12
2.12 = 1.5 + 1.2sin(3pi t) ⇒ sin(3pi t) = 0.517
Find angle
3pi t = sin-1(0.517) ≈ 0.543 ⇒ t ≈ 0.057 s
hmax = 2.7 m; hmin = 0.3 m; Period = 23 s; t ≈ 0.05 s
7.A carnival ride consists of a vertical wheel with a diameter of 40 feet. The center of the wheel is 28 feet above the ground. The wheel rotates at a constant speed and takes 120 seconds to make one complete revolution. Model an equation that describes the height h(t) of a rider on the wheel as a function of time t. How high is the rider from the ground after 90 seconds? At what times will the rider be 36 feet above the ground?
Given
Diameter
d = 40 ft, so radius A = 20 ft
Center height
D = 28 ft
Period
T = 120 s, so B = 2pi120 = pi60
Height function (starts at top, C = 0)
h(t) = -20cosleft(pi t60right) + 28
Height after 90 seconds
h(90) = -20cosleft(3pi2right) + 28 = -20(0) + 28 = 28 ft
Solve for h(t) = 36
36 = -20cosleft(pi t60right) + 28 ⇒ cosleft(pi t60right) = -0.4
First solution
pi t60 = cos-1(-0.4) ≈ 1.982 ⇒ t ≈ 37.87 s
Second solution within period
t = 120 - 37.87 ≈ 82.13 s
h(t) = -20cosleft(pi t60right) + 28; h(90) = 28 ft; t ≈ 37.87 s or 82.13 s
8.(a) Find the temperature at t = 9 hours. (b) At what time the temperature will be maximum? (c) Calculate the maximum temperature. Given: T = 64 + 8\sin\left(\frac{\pi}{12}(t-8)\right), where t is the time in hours. The temperature oscillates 8 degrees above and below an average temperature of 64 degrees.
Given
Temperature equation
T = 64 + 8sinleft(pi12(t-8)right)
(a) Temperature at t = 9 hours
T(9) = 64 + 8sinleft(pi12right) = 64 + 8(0.259) ≈ 66.07°F
(b) Maximum temperature occurs when sin = 1
pi12(t-8) = pi2 ⇒ t - 8 = 6 ⇒ t = 14 hours (2 p.m.)
(c) Maximum temperature
Tmax = 64 + 8(1) = 72°F
T(9) ≈ 66.07°F; tmax = 14 hours; Tmax = 72°F
9.(a) Find the population of a city at t = 7 months. (b) Find the maximum population. Given: P(t) = 70000 + 10000\cos\left(\frac{\pi t}{6} - \frac{\pi}{2}\right), where t is the time in months with t = 0 corresponding to January 1st.
Given
Population function
P(t) = 70000 + 10000cosleft(pi t6 - pi2right)
(a) Population at t = 7 months
P(7) = 70000 + 10000cosleft(7pi6 - pi2right) = 70000 + 10000cosleft(2pi3right)
Evaluate cosine
cos(2pi/3) = -1/2 ⇒ P(7) = 70000 + 10000(-1/2) = 70000 - 5000 = 65000
(b) Maximum population when cos = 1
Pmax = 70000 + 10000(1) = 80000
P(7) = 65000; Pmax = 80000