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course Mth 173
10/7 19:05
Text_09Text problem 1.5.10
Given function y = 7sin(3t)
We can see that the amplitude of y = 7sin(3t) is 7 because the factor of 7 stretches the oscillation up tp 7 and down to -7.
The period of the given sine function is given by 2pi / |3| = 2pi / 3
Thus the amplitude is 7 and the period is 2pi / 3
Text problem 1.5.12
Given function w = 8 - 4sin(2x + pi) , the factor 8 and pi just shift the graph in the along the y axis and the x axis respectively, they do not change or contribute to the period or the amplitude of the graph. Thus the amplitude of the given sine curve is given by |-4| = 4.
The period for the same sine curve = 2pi / |2| = 2pi / 2 = pi
Thus the amplitude is 4 and the period is pi.
Text problem 1.5.13
Given function r = .1sin(pi*t) + 2, similar to the previous question, the factor 2 just shifts the graph by 2 units in the positive y direction.
Thus the amplitude for the given sine function would be |.1| = .1
The period for the same sine function = 2pi / |pi| = 2*pi / pi = 2
Thus the amplitude of the sine function is .1 and the period of the sine function is 2.
Text problem 1.5.17
The graph looks like an upside-down cosine graph with amplitude 8. So the function f(t) = -8cos(Bx). The period of the function is calculated as 2pi / B. The graph performs one oscillation in time t = 20pi, thus the period of the function is 20pi. Thus 2pi / B = 20pi , B = 1/10 = 0.1. thus the formula of the graph will be f(t) = -8cos(.1x)
Text problem 1.5.21
The graph looks like a proper sine function with the amplitude 3. So the function f(t) = 3sin(Bx). The period of the function is calculated as 2pi / B. the graph performs one oscillation between time t = 0 and t = 18 units. thus the period of the function is 18 units. Thus 2pi / B = 18, B = pi/9. Thus the formula of the graph will be f(t) = 3sin( pi/9 * x)
Text problem 1.5.25
To calculate sin ( pi / 5 ) , given cos( pi / 5 ) = 0.809
We know that (cos(x))^2 + (sin(x))^2 = 1
Thus (cos(pi / 5))^2 + (sin( pi / 5))^2 = 1
0.809 ^ 2 + (sin( pi / 5))^2 = 1
0.654481 + (sin( pi / 5))^2 = 1
Thus (sin( pi / 5))^2 = 0.345519
sin( pi / 5) = sqrt(0.345519) = 0.588 approx
Text problem 1.5.27
Given function 2 = 5sin(3x)
Thus rearranging we get 0.4 = sin(3x)
Thus sin^-1(0.4) = 3x
Thus in exact form x = (1/3)* sin^-1(0.4)
In decimal form we obtain the value of x as
x = (1/3) * ( 0.412 ) = 0.1370
Text problem 1.5.28
Given equation 1 = 8cos(2x + 1) - 3
Thus 4 = 8cos(2x + 1)
Further solving gives ½ = cos(2x + 1)
cos^-1(½) = 2x + 1
thus 2x = cos^-1( ½ ) - 1
and x = [co^1( ½ ) - 1 ] / 2
Since cos(pi/3) = ½ thus cos(2x + 1) = cos(pi/3)
Thus 2x + 1 = pi/3
2x = pi/3 - 1 = ( pi - 3 ) / 3
Thus x = ( pi - 3 ) / 6 = 0.0234 approx
Thus the value of the function in exact form is [cos^-1( ½ ) - 1 ] / 2 and in decimal form is 0.0236(approx)
Text problem 1.5.30
Given equation 1 = 8tan(2x + 1) - 3
Thus 4 = 8tan(2x + 1)
Rearranging the expression we get ½ = tan(2x + 1)
Thus tan^-1( ½ ) = 2x + 1
Rearranging we get the value of x as
[tan^-1( ½ ) - 1] / 2 = x , the answer in the exact form.
The actual value can be calculated by substituting the value of tan^-1( ½ )
tan^-1( ½ ) = 0.464
thus at approx. value of x = ( 0.464 - 1 ) / 2 = -0.268
Text problem 1.5.32
Given 3 function graphs f(t) , g(t) and h(t) and 3 formulas,
(a) y = 2cos(t - pi/2) , (b) y = 2cost (c) y = 2cos( t + pi/2 )
The value of (a) at t = o is 0, (b) at t = 0 is 2 and (c) at t = 0 is 0
Thus we see that only (b) values to 2 at t = 0, and out of the 3 graphs only f(t) passes through (0, 2 ). Thus the function f(t) corresponds to formula (b)
Next at t = pi/2+
That value of (a) at pi/2 , y = 2cos( pi/2 - pi/2) = 2cos(0) = 2
And the value of (c) at pi/2, y = 2cos( pi/2 + pi/2 ) = 2cos(pi) = -2
The value of the graph g(t) is -2 at time t = pi/2 , and the value of the graph h(t) is 2 at time t = pi/2
Thus (a) corresponds to h(t) and (c) corresponds to g(t)
Thus (a) - h(t), (b) - f(t) and (c) - g(t).
Text problem 1.5.38
The smallest depth = 5.5 feet and the largest depth = 8.5 feet.
Thus the average depth = (5.5 feet + 8.5 feet) / 2 = 7 feet
Thus we see that the depth increase 7 feet to 8.5 feet by 1.5 feet and lowest to 5.5 feet by 1.5 feet. Thus the function oscillates between 5.5 feet and 8.5 feet with the base remaining at 7 feet. Thus the amplitude of the function is 1.5 feet and the function is to shifted to 7 feet along the positive y axis from the x axis for a normal sine - cosine function. The period of the function = 6 hours.
If the function is y = asin(bt), the period is given by 2pi / |b|, thus the value of b for the period of 6 hours is calculated as
6 hours = 2pi / |b| , |b| = pi / 3, thus b = pi/3 or -pi/3
Thus the possible functions are
y = 1.5sin( pi/3 * t) + 7 feet or y = 1.5sin( -pi/3 * t) + 7 feet
the function may be cosine too
y = 1.5cos( pi/3 * t ) + 7 feet , the negative b for cos is same as positive b as cos(-b) = cos(b)
Text problem 1.5.42
As seen from the graph we see that the maximum temperature is approximately 90℉. And the lowest temperature is 60℉. The average temperature is (60℉ + 90℉) / 2 = 75℉. Thus we know that the temperature increases from 75℉ to 90℉ and then decreases to 75℉, which further decreases to 60℉ and then increases back to 75℉, in 12 months. Thus the function has an amplitude of 15℉ and it completes one oscillation in 12 months. Since the function has an average temperature of 75℉, the function is to forward shifted along the positive y axis by 75 units, the function looks like an upside down cosine function, thus the function can be written as y = -acos(bx) + 75 ℉. Here a is the amplitude which is 15℉ and the period of the function is 2*pi / b = 12, thus b = pi / 6. Thus the trigonometric function will be y = H = -15cos(pi/6 * x) + 75℉, where y is temperature and x is the month.
Text problem 1.5.45
Given that the car’s engine makes 200 revolutions per minute. That is 200 revolutions in 60 seconds. The period of the engine is defined as the time taken by the engine to complete 1 revolution. Thus the time taken by the engine to complete one revolution = 60 / 200 seconds = 0.3 revolutions / second. Thus the period of the car engine when it is about to stall is 0.3 revolutions/second.
Text problem 1.5.47
The approximate period of moon’s revolution around the earth is 27.3 days. That is the moon completes one revolution around the earth in 27.3 days, which is the period of revolution of moon around the earth.
Text problem 1.5.52
The area of the trapezoidal cross-section can be found by cutting the trapezoidal into various segments. Initial the rectangle of the trapezoidal of width w and height h is a part of the trapezoid. The remaining part of the trapezoid forms 2 similar right angle triangle. The base of the triangle can be found by using the θ. The height of the right angle triangle will be h. the base of the right angle triangle is calculated as tan θ = height / base = h / base of the right angle triangle.
Thus the base of the right angle triangle = h / tan θ = h cot θ
The area of the triangle, thus will be ( ½ ) * base * height = ( ½ ) * h cot θ *h = ( ½ ) * cot θ * h^2 .
There are 2 such triangles which are similar, thus the total area of the triangle = cot θ* h^2.
Thus the total area of the trapezoid = area of the rectangle + area of the total triangle
= h*w + cot θ * h^2
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