Integrals 215 problem 02.mws

1. Trigonometric integrals and AM radio.

This activity is based on How to Tune a Radio by Clark Benson from the University of Arizona and The National Security Agency, found in Applications of Calculus, Philip Straffin, Editor. Published by the Mathematical Association of America, MAA Notes, Vol. 29, 1993.

(a) What is an amplitude modulated signal, or AM signal? The idea is as follows. Consider the function f(t) = sin(n*t) for a kind of big n . For example plot f(t) = sin(20*t) over [0, 2*pi] .

> plot(sin(20*t),t=0..2*Pi):

You should observe a high frequency (small period, period = Pi/10 ) wave. Let's think of this as a carrier wave, that is a wave that can be used to transport other signals. To do this we consider functions like f(t) = g(t)*sin(n*t) where g(t) is slowly varying in comparison to sin(n*t) . Do this for the functions below.

Submission (a):

Plot t*sin(20*t) , t^2*sin(20*t) , and sin(t)*sin(20*t) over [0, 2*Pi] . Explain how you can find t , t^2 , and sin(t) hidden in the graphs.

Submission worksheet:


(b1) A useful trigonometric integral. First we will need the fact that int(cos(n*t),t = 0 .. Pi) = 0 , n <> 0 .

Submission (b1):

Plot cos(n*t) on [0, pi] for n = 1, 2 and 3 . Argue from the graphs that int(cos(n*t),t = 0 .. Pi) = 0 . Confirm this for yourself with a paper and pencil calculuation of the integral.

Submission worksheet:


(b2) Other useful trigonometric integrals. We will also need the facts that int(sin(n*t)*sin(m*t),t = 0 .. Pi) = 0 and int(cos(n*t)*cos(m*t),t = 0 .. Pi) = 0 , for n <> m along with int([sin(n*t)]^2,t = 0 .. Pi) = Pi/2 and int([cos(n*t)]^2,t = 0 .. Pi) = Pi/2 for n = 1, 2, 3 , ...

Submission (b2):

Using the half-angle identities or the product to sum trigonometric identities along with the result from (b1) above convince yourself of each of these identities. Also use Maple to evaluate int(sin(3*t)*sin(4*t),t = 0 .. Pi) and int(cos(20*t)*cos(12*t),t = 0 .. Pi) , and int([sin(144*t)]^2,t = 0 .. Pi) .

(If you don't know the identities and have no clue where to find them, try using a search engine (I like google) on the www.)

Submission worksheet:


(c) Using these ideas to broadcast and RECEIVE signals. Suppose we have 3 radio stations, WAAA, WBBB and WCCC all broadcasting some signal, and each is broadcasting a signal that we call A(t) , B(t) and C(t) respectively. Our radios at home then receive the superimposed signal, that is A(t)+B(t)+C(t) and how is it exactly that we are able to tune in to say a talk show on WAAA which is A(t) , or some music on WBBB, which is B(t) or the Brewers game on WCCC, which is C(t) . The surprising answer is given by trigonometric integrals. We begin by assigning each of WAAA, WBBB and WCCC their own, unique carrying signal. Suppose WAAA's is sin(8*t) , WBBB's is sin(16*t) and WCCC's is sin(24*t) . Now we require that each send their signals as a sum of sin functions (not as restrictive a condition as it sounds) with much lower frequencies than their carrying frequenies. Let's consider a specific example:

A(t) = 3*sin(t)-2*sin(2*t) , B(t) = 5*sin(t)+6*sin(2*t) , C(t) = 4*sin(t)+7*sin(2*t)+3*sin(3*t) .

Then WAAA broadcasts (3*sin(t)-2*sin(2*t))*sin(8*t) , WBBB broadcasts (5*sin(t)+6*sin(2*t))*sin(16*t) and WCCC broadcasts (4*sin(t)+7*sin(2*t)+3*sin(3*t))*sin(24*t) .

Submission (c):

Submit the t hree separate graphs. The first should be A(t) and A(t)*sin(8*t) over [0, 2*Pi] , the second B(t) and B(t)*sin(16*t) over [0, 2*Pi] , and the third C(t) and C(t)*sin(24*t) over [0, 2*Pi] . What do you observe?

Submission worksheet:


(d) More with the example. The superimposed signal our radio receives is:

f(t) = (3*sin(t)-2*sin(2*t))*sin(8*t)+(5*sin(t)+6*s...

Recall, the idea is that we know that each station will send Sum(a[i]*sin(i*t),i = 1 .. 3) , but that we don't know what the a[i] are. Here's the way to get them. Let's focus on WAAA. We would like to determine, a[1] , a[2] and a[3] . To get a[i] we simply multiply f(t) by (-4/Pi)*cos((8+i)*t) , where recall 8 was the carrying frequency assigned to WAAA. Now integrate from 0 to Pi . Let's let Maple to this and find out if we get a[1] = 3 , a[2] = -2 and a[3] = 0 .

> f := t -> (3*sin(t)-2*sin(2*t))*sin(8*t) + (5*sin(t)+6*sin(2*t))*sin(16*t) + (4*sin(t)+7*sin(2*t)+3*sin(3*t))*sin(24*t):

> (-4/Pi)*Int(f(t)*cos(9*t),t=0..Pi);


> (-4/Pi)*int(f(t)*cos(9*t),t=0..Pi);


Wow, you might not want to try that one at home! What changes when we want to find a[2]? The answer is not much, we simply do everything as above except we use cos(10*t) instead (10 = 8 + 2).

> (-4/Pi)*int(f(t)*cos(10*t),t=0..Pi);


Finally, to get a[3]:

> (-4/Pi)*int(f(t)*cos(11*t),t=0..Pi);


So, we now know, that WAAA broadcast 3*sin(t)-2*sin(2*t)+0*sin(3*t) , which is music to the ears.

Submission (d):

For the above f(t), retrieve the broadcasts from WBBB and WCCC. Suppose WBBB's signal was

B(t) = 5*sin(t)+6*sin(2*t)+7*sin(3*t) . Write the new form of f(t) . What integral would you calculate to recover the coefficient a[3] in this signal. Get Maple to perform this integration too.

Submission worksheet:


(e) Where do we use the trigonometric identities? Consider WAAA's signal. Recall that we multiplied f(t) by cos(9*t) , cos(10*t) and cos(11*t) respectively to retrieve all the necessary coefficients. Why did we decide on these particular cosine expressions? The answer is because of trigonometric identities, and our observations in (b2) above. To see more clearly how cos(9*t) does its job (retrieves a[1] for WAAA), look again at

> (-4/Pi)*Int(f(t)*cos(9*t),t=0..Pi);


Notice that if we distribute all the products we have lots of products of sine terms with cos(9*t) , 7 to be exact. But each of these products can be rewritten.

Submission (e):

Rewrite the integrand above using the trigonometric identity sin(A)*sin(B) = 1/2*(cos(A-B)-cos(A+B)) , and then use our results from (b2) to argue that value of this integral is 3. Also explain why we multiplied by cos(9*t) to do this. Could we have multiplied by something else to do the same thing?

Submission worksheet:


2. Inverse trig functions and integrals.

Later on in the course, we will be computing integrals using trigonometric substitution. Actually, we have used trigonometric substitutions in the first course in calculus, but not those which involve inverse trigonometric functions. To compute Int(sqrt(1-x^2),x) , one can make the substitution x = sin(theta) , which converts the integral into Int(cos(theta)^2,theta) (valid when 0 <= cos(theta) ). Then one can use the trigonometric identity cos(theta)^2 = (1+cos(2*theta))/2 to convert this integral into Int((1+cos(2*theta))/2,theta) , which then can be evaluated directly, giving the indefinite integral 1/2*theta+1/4*sin(2*theta) . But now we need to convert this expression into one involving the variable x . Since theta = arcsin(x) , the first term in this expression is just 1/2*arcsin(x) . To convert the second term into an expression involving x, we use the fact that sin(2*theta) = 2*sin(theta)*cos(theta) , and the identity cos(theta) = sqrt(1-sin(theta)^2) (valid when 0 <= sin(theta) ), to finally obtain the result

Int(sqrt(1-x^2),x) = x*sqrt(1-x^2)/2+arcsin(x)/2 .

Note that this expression is valid when theta lies in the first quadrant, that is, when x is between 0 and 1. Of course, Maple can evaluate this indefinite integral very simply.

> int(sqrt(1-x^2),x);


We can use the computation above to evaluate the area under the curve y = sqrt(1-x^2) , over the domain [0,1], using the fundamental theorem of calculus.

> f:=x->1/2*x*sqrt(1-x^2)+1/2*arcsin(x);f(1)-f(0);

f := proc (x) options operator, arrow; 1/2*x*sqrt(1...


This is no surprise, since it represents the area of a quarter circle of radius one. Maple can evaluate this integral for us directly

as well.

> int(sqrt(1-x^2),x=0..1);



The objective of this exercise is to compute the area under the curve y = 1/(1+x^2) , on the interval [0,1].

(a) Use the derivatives of inverse trigonometric functions to determine an indefinite integral of this function. Then use the fundamental theorem of calculus to evaluate the area.

(b) Compute this area directly by having Maple evaluate a certain definite integral.

Submission worksheet: