Class Summary Page
for
MATH 162
Calculus with Applications, II
October, 1997
Go back to the first class summary page.
Wednesday, October 1. We made a segue from computing volumes of revolution to integration of rational functions (i.e., ratios of polynomials) by doing a volume problem that led to an integral that we didn't (yet) know how to integrate. We rotated the region between the curve y=sqrt((4+x-x^2)/(3+2*x-x^2))+1 and the lines y=1, x=0 and x=1/2 about the line y=1. (Note that one feature of this problem that was different from the examples we did on Monday was that the axis of rotation was not one of the coordinate axes.) We began by trying to determine what the curve y=sqrt((4+x-x^2)/(3+2*x-x^2))+1 looked like. It's not terribly important to have an exact sketch; in fact, we were content with noting that since we were only looking at x's near 0, this function would be approximately its value at x=0, namely sqrt(4/3)+1, which is a little larger than 2. We were then able to set up the integral for the volume, Int(Pi*((sqrt((4+x-x^2)/(3+2*x-x^2))+1)-1)^2,x-0..1/2), which simplifies to Pi*Int((4+x-x^2)/(3+2*x-x^2),x=0..1/2). At this point we stopped worrying about the definite integral and began concentrating on the indefinite integral, Int((x^2-x-4)/(x^2-2*x-3),x). We didn't see a good u-substitution to make. Hoping that we might be able to simplify the integrand by canceling a common factor in the integrand, we tried to factor each of these. The denominator is x^2-2*x-3=(x-3)(x+1). It wasn't immediately clear how to factor the numerator: x^2-x-4. We ended up using the quadratic formula to solve the equation x^2-x-4=0. The solutions to this equation are x=(1+sqrt(17))/2 and x=(1-sqrt(17))/2. Therefore, x^2-x-4 = (x-(1+sqrt(17))/2)*( x-(1-sqrt(17))/2). [Here's an important observation: if we had found that the solutions of the quadratic equation involved imaginary numbers (because there was a negative number in the square root of the quadratic formula), then we would say that this quadratic polynomial is irreducible -- because it cannot be "reduced" to being written as a product of two linear terms. For the time being, we will not worry about irreducible quadratic factors, but we'll eventually do so when we look at the second half of section 7.4.] So our plan to simplify the integrand by factoring didn't work. We next tried to get as much mileage as we could out of a certain analogy between numbers and function:
Integers (e.g., 4, 3) <---> Polynomials (e.g., x+1, x^2+5)
Fractions (e.g., 4/3, 3/4) <---> Rational Functions (e.g., (x+1)/(x^2+5), (x^2+5)/(x+1))
Proper Fractions (e.g., 3/4) <---> Proper Rational Functions (e.g., (x+1)/(x^2+5))
Improper Fractions (e.g., 4/3) <---> Improper Rational Functions (e.g., (x^2+5)/(x+1))
In trying to distinguish between proper and improper rational functions, we had to decide what it meant to say that a certain rational function was larger than another; instead of using their numerical size (which could be tricky -- what would you do when the graph of one of the functions crosses the graph of the other?), we decided to use the order of the polynomials: the rational function p(x)/q(x) is proper if the order of the polynomial p is less than that of q. Our analogy suggests that just as we can rewrite improper fractions as the sum of an integer and a proper fraction, we should try to rewrite an improper rational function as the sum of a polynomial and a proper rational function. We saw two ways to do this, using long division and adding and subtracting in the numerator. The first of these is guaranteed to eventually give you the decomposition that you want, while the second is faster if you see how to do it. For example, in this situation, the second method consists of writing (x^2-x-4)/(x^2-2*x-3) = (x^2-2*x-3+x-1)/(x^2-2*x-3) = (x^2-2*x-3)/(x^2-2*x-3) + (x-1)/(x^2-2*x-3) = 1 + (x-1)/(x^2-2*x-3). At this point we ran out of time. [To be continued ...]
Friday, October 3. We picked up where we left off on Wednesday. In order to compute the volume from last class, we now needed to evaluate Pi*(Int(1,x=0..1/2) + Int((x-1)/(x^2-2*x-3),x=0..1/2)). We were able to do the second integral by making the u-substitution u=x^2-2*x-3, since then du=2*(x-1)dx, and we already had a factor of x-1 in the numerator. The volume turned out to be (Pi/2)*(1+ln(5/4)) which is approximately 1.92, which compares reasonably well to the volume of an approximating cylinder with height 1/2 and radius 1. We then looked at three related indefinite integrals: Int((x+1)/(x^2-2*x-3),x)), Int((x-3)/(x^2-2*x-3),x)), and Int((x+3)/(x^2-2*x-3),x)). We were able to compute the first two of these by canceling the numerator with a factor in the denominator -- recall that on Wednesday we saw that x^2-2*x-3=(x-3)(x+1); we got ln|x-3|+c for the first integral, and ln|x+1|+c for the second. We needed a new idea for the third integral. It already was a proper rational function (so there was no point in trying to re-express it through a long division), we couldn't think of a good u-substitution, nor could we simplify the integrand by canceling common factors. Our new idea was to use partial fractions. We tried to rewrite (x+3)/(x^2-2*x-3) as A/(x-3) + B/(x+1). To solve for A and B, we put the right-hand side over a common denominator, and then tried to find A and B so that x+3 = B*(x+1)+A*(x-3) [= (A+B)*x + (B-3*A)]. Equating first the terms involving a single power of x, and then the terms involving no powers of x, we got the pair of equations: A+B=1 and B-3*A=3, which has the solution A=3/2 and B=-1/2. So (x+3)/(x^2-2*x-3) = (3/2)/(x-3) - (1/2)/(x+1), which means that Int((x+3)/(x^2-2*x-3),x)) = (3/2)*Int(1/(x-3),x)) - (1/2)*Int(1/(x+1),x)) = (3/2)*ln|x-3| - (1/2)*ln|x+1| + c. We ended by working an example involving a repeated linear factor in the denominator: Int(x/(x-1)^2,x) -- the previous example had two distinct linear factors. After some experimentation, we realized that the correct way to do the partial fraction decomposition was to try to find constants A and B so that x/(x-1)^2 = A/(x-1) + B/(x-1)^2. After the right hand side is put over the common denominator (x-1)^2, we see that we need to solve x=A*(x-1)+B [=A*x+B-A]. Equating first the terms involving a single power of x, and then the terms involving no powers of x, we got the pair of equations: A =1 and B-A=0 (since x=1*x+0), which has the solution A=B=1. Therefore Int(x/(x-1)^2,x) = Int(1/(x-1),x) + Int(1/(x-1)^2,x) = ln|x-1| - (x-1)^(-1) + c. [Note: Another (and simpler) way to do this problem is to make the u-substitution u=x-1 (with du=dx). In the numerator, the x becomes u+1, so the integral is Int((u+1)/u^2,u) = Int(u/u^2,u) + Int(1/u^2,u) = Int(1/u,u) + Int(1/u^2,u) = ln|u| -u^(-1) + c = ln|x-1| - (x-1)^(-1) + c.]
Monday, October 6. Today we began by working one of the homework problems on integration by parts: Int(1/((x+a)(x+b)),x) [Exercise 22 from Section 7.4]. We saw that if a is not equal to b, then we could do a partial fraction decomposition to get the answer. To get the answer in the case where a=b, we could do (at least) two things. One of these was to take the answer we had found in the case where a and b were assumed to be different, and then take the b-->a limit (which would require us to use L'Hospital's Rule). Another possibility was to just make the u-substitution u=x+a. As we saw, one thing that will not work is to try to do a partial fraction decomposition of 1/(x+a)^2 --- since this function is already in the partial fraction decomposition form [in this case, a constant over a repeated linear term]. We then went back to volume problems and computed the volume of a cored apple (we assumed it to be spherical) in two ways: by the method of disks, and by the method of shells. These methods were illustrated with real apples, and also with stacking rings (for the method of washers) and stacking cups (for the method of shells). An important part of using the method of cylindrical shells is computing the surface area of a cylinder. By cutting a paper cylinder down its side and rolling it flat, we saw that this surface area was the same as the area of a rectangle whose height was the height of the cylinder, and whose base was the circumference of the cylinder. We performed several checks of our answers, comparing the results of the two different methods, and also comparing the result when the radius of the core is zero to the known formula for the volume of a sphere. In doing one of these, we caught the fact that at one point in our calculation we had lost a factor of two; this should serve to emphasize the importance of looking over your answers to make certain that they seem reasonable. The problem of the cored sphere was one that we could compute equally well by the methods of washers and cylindrical shells. We closed with a problem that seems to be impossible to do by washers. We wanted to compute the volume obtained by taking the closest region in the first quadrant that is bounded by the x-axis and the curve y=((sin(x))^2)/x, and rotating this about the y-axis. We had to use L'Hospital's Rule to compute that lim_(x-->0+) ((sin(x))^2)/x = 0, in order to see what this region looks like. Then we noticed two problems. If we were to set this up as a "washers" problem, then we would need to integrate from y=0 to y=b, where b is the maximal value of the function y=((sin(x))^2)/x on the interval [0,Pi]. Although determining b is in principle just a first semester calculus problem, it doesn't look particularly pleasant, and it turns out that there is not a nice closed form expression for b. Also, as the function ((sin(x))^2)/x is not invertible on [0,Pi], we cannot write explicit formulas for the inner and outer radii of the washer. Thus the method of washers is doomed because we don't have explicit ways of writing either the upper limit of the integration or the integrand. On the other hand, if we use shells, the picture is not so bleak. We write V=Int(SA(x),x) where SA(x) is the surface area of the cylinder that cuts the x-axis at x. Then SA(x) = base*height = circumference*height = 2*Pi*x*((sin(x))^2)/x = 2*Pi*(sin(x))^2. Our only problem now is that we don't know how to integrate (sin(x))^2; we'll take care of that on Wednesday.
Wednesday, October 8. Today we discussed how to integrate (nonnegative) powers of sines and cosines. The basic idea was to take the addition formula for cosine (it is not necessary to memorize this): cos(alpha+beta) = cos(alpha)*cos(beta) - sin(alpha)*sin(beta), and specialize it to the case where alpha=beta=x. We then get the double angle formula for cosine (which you should also not bother memorizing): cos(2x) = (cos(x))^2 - (sin(x))^2. We then combined this in two different ways with the fundamental trigonometric identity (you should definitely know this): (cos(x))^2 + (sin(x))^2 = 1. Eliminating the (cos(x))^2, we obtained the half-angle formula for sine (this is a very useful formula that you need to remember): (sin(x))^2 = (1-cos(x))/2. Also, eliminating the (sin(x))^2, we obtained the half-angle formula for cosine (this is an equally useful formula that you also need to remember): (cos(x))^2 = (1+cos(x))/2. We then set about systematically evaluating the integrals Int((sin(x))^m*(cos(x))^n,x) for several small (nonnegative, integer) values of m and n.
m=n=0 ==> Int(1,x) = x + c
m=1, n=0 ==> Int(sin(x),x) = -cos(x) + c
m=0, n=1 ==> Int(cos(x),x) = sin(x) + c
m=2, n=0 ==> Int((sin(x))^2,x) = (1/2)*[x-(1/2)*sin(2x)] + c
m=0, n=2 ==> Int((cos(x))^2,x) = (1/2)*[x+(1/2)*sin(2x)] + c
In the last two integrals we made use of the half-angle formulas. We checked the answer in the case m=2, n=0 in two ways. If we just differentiate (1/2)*[x-(1/2)*sin(2x)] + c, we get (1/2)*[1-cos(2x)], which is just (sin(x))^2 by the half-angle formula for sine. Alternatively, we can try applying an appropriate trigonometric identity before the differentiation. This would be the double angle formula for sine (this is a moderately important formula that you would be well served to remember): sin(2x) = 2*sin(x)*cos(x) --- which is obtained by setting alpha=beta=x in the addition formula for sine it is not necessary to memorize this): sin(alpha+beta) = cos(alpha)*sin(beta) - sin(alpha)*cos(beta). Then (1/2)*[x-(1/2)*sin(2x)] + c = (1/2)*[x-sin(x)*cos(x)] + c, which has the derivative (1/2)*[1 - cos(x)*cos(x) + sin(x)*sin(x)] = (1/2)*[(sin(x))^2 + (sin(x))^2] = (sin(x))^2. We ended by looking at the case m=n=1: Int(sin(x)*cos(x),x). We evaluated this integral in five different ways. There were two ways to do this by u-substitutions: (a) u=sin(x) and (b) u=cos(x). There were two ways to do this through integration by parts -- this is a cyclic integral: (c) let u=sin(x) and dv=cos(x)dx, and (d) let u=cos(x) and dv=sin(x)dx. These methods gave either (1/2)*(sin(x))^2 + c or (1/2)*(cos(x))^2 + c [these answers can be reconciled by observing that the two constants of integration are not the same]. Finally, the double angle formula for sine can be used since Int(sin(x)*cos(x),x) = Int((1/2)*sin(2x),x) = (-1/4)*cos(2x) + c [this c is different from both of the ones found in the other methods]. It was strongly advised that students acquaint themselves with the procedure found in the boxes on pages 443 and 444 in the text for evaluating integrals of powers of sines and cosines.
Friday, October 10. Today we began to discuss the general strategy behind integrating trigonometric integrals. We saw -- our example was, I think (we used student generated numbers that I might not be recalling exactly), Int((sin(x))^6*(cos(x))^13,x) -- that it was often possible to evaluate integrals of powers of sines and cosines by making either the u-substitution u=cos(x) or u=sin(x). Note that if one tries u=cos(x), then du=-sin(x)dx, then one needs to do is save one of the powers of sine for the du and convert the others to cosines via the fundamental trigonometric identity: (sin(x))^2+(cos(x))^2=1. The problem here is that if we save a single sine for the du, then we have 5 remaining powers of sine which becomes (sin(x))^5 = ((sin(x))^2)^(5/2) = (1-(cos(x))^2)^(5/2) = (1-u^2)^(5/2), and (because of the square root) it's not clear that we will be able to evaluate this integral. On the other hand if we instead save a single cosine in preparation for the substitution u=sin(x) [where du=cos(x)dx], then we have 12 remaining powers of cosine, and these can be swapped for powers of sines (two at a time) through the fundamental trigonometric identity without introducing any square roots. One detail in this problem was that we needed to compute (1-u^2)^6; we did this by using the Binomial Theorem (see the first endcover in the course text); we simplified this computation by using Pascal's triangle to generate the various coefficients in the Binomial expansion. We realized that there was a general principle at work here which allows us to integrate products of nonnegative integer powers of sine and cosines as long as one of these powers is odd (see the table on page 443 of the text). We next discussed the use of half-angle formulas for sines and cosines in the situation where both powers are even; our example was Int((sin(x))^2*(cos(x))^2,x). We also saw that since the powers of the sines and cosines were the same, we could also use the double angle formula for sine. (See the box on page 444 in the text.) We turned to the case where some of the powers can be negative by evaluating Int((sin(x))/(cos(x)),x) -- by making the u-substitution u=cos(x) we found that Int(tan(x),x) = ln|sec(x)| + c. Finally, a very preliminary draft of "A Note on Trigonometric Integrals" was distributed to the class.
Monday, October 13. We began by integrating sec(x) in two different ways. First we rewrote sec(x) as 1/cos(x) and then multiplied both the denominator and the numerator by cos(x). We replaced the (cos(x))^2 in the denominator with 1-(sin(x))^2, and then made the substitution u=sin(x). This gave us the integral Int(1/(1-u^2),u), which we could evaluate by using partial fractions [note that 1-u^2 = (1+u)(1-u)]. We did this and found that Int(sec(x),x) = (1/2)*ln|(1+sin(x)/1-sin(x)| + c. This method took a little while to work out (and we used all three of the whiteboards), but the ideas all followed naturally one after another. The second method was much faster (and it only used one whiteboard), but it was much less intuitive ... after reviewing the differentiation formulas for sec(x) and tan(x), we decided to multiply sec(x) by (sec(x)+tan(x))/(sec(x)+tan(x)), and then we noticed that the numerator was sec(x)*(sec(x)+tan(x)) = (sec(x))^2+sec(x)*tan(x) = D(sec(x)+tan(x),x). From this we saw that Int(sec(x),x) = ln|sec(x)+tan(x)| + c. The two answers can be reconciled by noting that (1/2)*ln|(1+sin(x)/1-sin(x)| = ln|sec(x)+tan(x)| -- see the handout from last class. Having now integrated both of the products of powers of secants and tangents where the total powers added to only one, we investigated those integrals where the powers summed to 2: Int((sec(x))^2,x), Int((sec(x))*tan(x),x) and Int((tan(x))^2,x). The first two followed immediately by rewriting the differentiation formulas for tan(x) and sec(x) as antidifferentiation formulas. The key idea in the third was using a version of the fundamental trigonometric identity, (tan(x))^2 + 1 = (sec(x))^2 [obtained by dividing both sides of (sin(x))^2 + (cos(x))^2 = 1 by (cos(x))^2], to rewrite (tan(x))^2 as (sec(x))^2-1. We discussed the boxed note, "How to evaluate Int((tan(x))^m*(sec(x))^n,x)," on page 445 in the text, as well as the comment on page 446, "If n is odd and m is even, we express the integrand completely in terms of sec(x). Powers of sec(x) may require integration by parts." This last comment was amply demonstrated by working through Int((sec(x))^3,x). We made the choice u=sec(x) and dv=(sec(x))^2dx (guided by our wanting to make the calculation of v as simple as), then used the fundamental identity to replace a (tan(x))^2 in the resulting integral with (sec(x))^2-1, and finally recognized that this was a "cyclic" application of integration by parts. We then changed topics and began to discuss work. When an object moves along a straight line under the influence of a constant force, we define the work done by that force to be the product of the force applied and the distance moved: W=F*d. To consider situations where the force is not constant, we need to use calculus. We invoke our basic philosophy of integration. We break the distance moved up into a large number of subintervals. If the force is a continuous function of the spatial coordinate, x, then we can argue that as the object is moved through each subinterval, the force is nearly constant, so the work that is done is nearly the force at one of the endpoints of the interval times the length of the interval. Adding up the works done to cross each subinterval and then taking the limit as the number of subintervals diverges and the lengths of all subintervals tend to 0 we get that W=Int(F(x),x). We illustrated this in a common context, that of a spring. Here the force is given to us by Hooke's Law which tells us that the restoring force of a spring is proportional to the distance that the spring has been stretched from its natural length. We worked through Exercise 6 from section 5.4 in Stewart's text. Finally we ended with a brief discussion of how particles in equilibrium settings can often be modeled as being inside a quadratic potential energy well, which means that the particle is subject to a linear restoring force which tends to keep it in its natural position, in exactly the same way that a spring's linear restoring force acts to try to keep a spring at its natural length. At the end of class, a review sheet to help students prepare for the second exam was distributed.
Wednesday, October 15. We worked one more work problem (pun intended). We computed the work required to roll up a vertically hanging non-uniform cable. The problem at first appear quite daunting, different pieces of the cable are lifted different distances (the higher pieces don't need to be lifted as far as the lower pieces), and the nonuniform density also means different forces have to be applied to lift different segments of the cable. We fell back on our general philosophy of integration, broke the cable into a lot of little pieces and approximated the work required to lift each of these. We then added these works together and took the limit as the subinterval lengths all went to 0 to obtain an integral formula for the work. The non-uniform density in this problem was chosen so that we essentially ended up needing to compute Int(sqrt(1-u^2),u=0..1). We were able to evaluate this by recognizing that this integral gave the area of one quarter of the unit circle, so it must be Pi/4. We then wondered how we might compute this integral if either the upper or lower limits of integration had been different. This served as our entrance to the topic of trigonometric substitutions. We want to learn how to collapse sums (or differences) of squares into a single square -- this will be particularly useful when the sum or difference of squares appears in a denominator or (as in this case) inside a square root. The key idea is to look at the sum or difference of squares -- in this case, 1-u^2 -- and compare it to the fundamental trigonometric identities: (sin(x))^2+(cos(x))^2=1 and (tan(x))^2+1=(sec(x))^2. Here such a formula is 1-(sin(x))^2=(cos(x))^2. This suggests that we should make a substitution that allows us to replace u with sin(x), because then sqrt(1-u^2) = sqrt(1-(sin(x))^2) =
sqrt((cos(x))^2), and if cos(x) is positive, this simplifies to cos(x). We'll talk more about this, and related substitutions after the second exam. A practice version of the second exam (and a solution set) was distributed at the end of class.
Thursday, October 16. We held a review session from 7:00-9:00 in ACD 304.
Friday, October 17. The second exam was given today.
Monday, October 20. . The graded exams (together with a solution key) were returned today. We returned to the integration problem discussed at the end of class last Wednesday, except that now we generalized it to trying to evaluate Int(sqrt(R^2-x^2),x), for some R>0. We outlined a basic strategy for making trigonometric substitutions.
Step 0. First make certain that there is not an easier way to evaluate the integral.
Step 1. Find the part of the integral where you want to collapse a sum or difference of two squares into a single square; in this case it is the R^2-x^2.
Step 2. To simplify matters at first, replace any of the constants in this part on which you are focusing with 1's; in this case, temporarily replace the R^2-x^2 with 1-x^2.
Step 3. Next ask yourself if you know of a version of the fundamental trigonometric formula that collapses a and the square of a trigonometric function into the square of another trigonometric function in a way that has something to do with the 1 and the x^2 that you got in Step 3. There are three possibilities at this point. If you have 1+x^2, then you should use the identity 1+(tan(theta))^2 = (sec(theta))^2; if you have x^2-1, then you should use the identity (sec(theta))^2-1 = (tan(theta))^2; and if you have 1-x^2, then you should use the identity 1-(sin(theta))^2 = (cos(theta))^2. In our case we have 1-x^2, so we think of the identity 1-(sin(theta))^2 = (cos(theta))^2.
Step 4. Now rewrite this identity in a way that brings back the constant(s) that you had ignored in Step 2; for our example, this means writing R^2-(R*sin(theta))^2 = (R*cos(theta))^2.
Step 5. Compare the original expression from Step 1 with the expression that you have in step 4 to see what you want to substitute for x; in this case we see that we want to let x^2=r^2*(sin(theta))^2, which we can do by letting x=R*sin(theta). Note that we are really making a theta-substitution here: theta=arcsin(x/R).
Step 6. Relate the differential dx to the differential d(theta) and make the trigonometric substitution from Step 5. In our example we have dx=R*cos(theta)*d(theta), and so Int(sqrt(R^2-x^2),x) = Int(sqrt(R^2-R^2*(sin(theta))^2)*R*cos(theta),theta) = Int(R^2*sqrt((cos(theta))^2)*cos(theta),theta). We then discussed how thinking of the trigonometric substitution as the substitution theta=arcsin(x/R) told us that theta was in the interval [-Pi/2,Pi/2]. Thus cos(theta) is nonnegative, which in turn implies that sqrt((cos(theta))^2) is not just |cos(theta)|, but it actually is equal to cos(theta).
Step 7. Evaluate the trigonometric integral form Step 6 using the methods we have already developed. In this example we had to use the half angle formula for cosine to get that Int(R^2*((cos(theta))^2),theta) = (R^2/2)*Int(1+cos(2*theta),theta) = (R^2/2)*(theta + sin(2*theta)/2) + c.
Step 8. Convert your answer from being expressed in terms of theta to an expression involving the original variable, x. This usually involves constructing a right-triangle diagram of the type that we used when we were deriving the differentiation formulas for the inverse trigonometric functions. If, as was the case in this example, we have used a half-angle formula in Step 7, then we will need to use the double angle for sine at this stage. In our example we looked back at our original substitution to see that theta=arcsin(x/R). Then we used the double angle formula for sine to rewrite sin(2*theta)/2=sin(theta)*cos(theta). Of course, sin(theta)=x/R, and after we draw a right triangle diagram we see that cos(theta)= sqrt(R^2-x^2)/R. Therefore Int(sqrt(R^2-x^2),x) = (R^2/2)*( arcsin(x/R) + x*sqrt(R^2-x^2)/R^2) + c.
We did two more examples Int(1/sqrt(3+x^2),x) and Int(1/(x*sqrt(x^2-1)),x) that respectively demonstrated how to make a tangent and a secant substitution. In working the second example we saw the reason that some mathematicians (like Stewart) prefer to create the inverse secant function by restricting the domain of the secant function to the union of the intervals [0,Pi/2) and [Pi,3*Pi/2), rather than the union of [0,Pi/2) and (Pi/2,Pi] -- with the choice of domain that Stewart makes, when we make a secant substitution like x=sec(theta), we find that tan(theta) is nonnegative, and so sqrt((tan(theta)^)=tan(theta); if we had used the other domain for the restricted function then we would instead have that sqrt((tan(theta)^)=|tan(theta)|. Finally, at the end of class we began to discuss differential equations. We talked about how the general solution was actually infinitely many solutions, what mathematicians sometimes call a "family of solutions." In applications we don't usually care about the entire family of solutions; we just want to be able to use the one special solution corresponding to the exact situation we are studying to be able to understand some phenomenon or to make a prediction. We get this by adding an initial condition (IC) [a specification of the value of the solution for some value of the independent variable] to our differential equation (DE). The combination of a DE and an IC is called an Initial Value Problem (IVP).
Wednesday, October 22. We resumed our discussion of differential equations, but focused our attention on a very special class of differential equations: separable differential equations, which can be written in either of the equivalent forms dy/dx = g(x)*f(y) or dy/dx = g(x)/h(y). [To go back and forth between these two versions, just let f(y)=1/h(y).] We began by looking at the special differential equation y'=ky (where '=d/dt), which we solved three times.
(i) First we used a method from first semester calculus (see Section 6.5 in the course text) in which after we divided both sides by y, we recognized y'/y as being (ln|y|)'. We then had (ln|y|)' = k; integrating both sides gave us ln|y| = k*t+c, which is an implicit solution to the differential equation. At this point, the calculus is done, and if we want to get an explicit solution to the differential equation (i.e., if we want to be able to write y as a certain function of t) we just need to do some algebra. Note that it is not always to obtain an explicit solution from an implicit one, but we can in this case. First exponentiate both sides (base e) to get that |y| = exp(k*t+c) = e^c*e^(k*t). Then y = (ħe^c)*e^(k*t) = C*e^(k*t). An interesting comment is that by allowing C to also equal zero we regain a solution to the original DE that we had lost when we divided both sides of the original equation by y.
(ii) Next we repeated this method except that instead of immediately recognizing that y'/y was the derivative of ln|y|, we just integrated both sides of the equation obtained after dividing the original DE by y. We had that Int(y'(t)/y(t),t) = Int(k,t) = k*t+c. In the integral on the left-hand side we made the substitution u=y(t) (so du=y'(t)dt). This gave us Int(1/u) = k*t+c, from which we immediately get ln|u| = k*t+c, which then yields the same implicit solution as we got the first time: ln|y(t)|=k*t+c.
(iii) Finally we looked at a streamlined version of the second method where we were able to avoid having to make the substitution u=y(t) by taking advantage of the Leibniz notation for derivatives and integrals. Rewrite the differential equation in differential notation as dy/dt = k*y, and then rearrange the terms so that only y-terms appear on the left-hand side of the equals sign, only t-terms appear on the right-hand side of the equals sign, and every term contains a factor of either dy (if it is on the left-hand side) or dt (if it is on the right-hand side). Here we got dy/y = k*dt. Now we integrate both sides, each with respect to the variable indicated by the differential: Int(1/y,y) = Int(k,t) ==> ln|y| = k*t+c. Comparing this technique with that of method (ii) shows that this method "build in" the substitution that we need to make to evaluate the integral on the left-hand side.
In the remainder of the class time we solved the Initial Value Problem, Exercise 14 in Section 8.1 of the course text. We first solved the differential equation to obtain the general implicit solution. Once we had done this, we had finished the calculus part of this problem and only had two more algebraic steps to perform: trying to solve for y so as to obtain an explicit solution, and imposing the initial condition to determine the value of the constant of integration in our answer. These last steps can be done in any order, but it is usually simplest to impose the initial condition first, and that's what we did!
Friday, October 24. Dr. Barsky was away at a conference in San Francisco, but he left a lecture which was given by Dr. Chien in his place. The class began with a separable differential equation: y' = (y^2+1)*y. In solving this we were led to the integral, Int(1/((y^2+1)*y),y). In order to evaluate this we needed to extend our work on partial fractions to cover the case of irreducible quadratic factors: the term y^2+1 is called irreducible because it is impossible to factor a linear term out of it (to convince yourself of this impossibility, ask yourself, can y^2+1 ever equal zero?). It was decided that we should look for a decomposition in the form 1/((y^2+1)*y) = (A*y+B)/(y^2+1) + C/y. Putting the right-hand side over a common denominator, and then equating the coefficients in the numerator of y^2 (A+C=0), y (B=0) and the constant terms (C=1) gave us the decomposition form 1/((y^2+1)*y) = -y/(y^2+1) + 1/y, from which it was not difficult to evaluate the integral. At this point , Dr. Chien led a discussion on how to set up a partial fraction decomposition when the denominator contains a repeated quadratic factor. [See Case IV in Section 7.4 in the course text.] The class ended by starting work on evaluating Int(1/(x^3+1),x). Notice that the denominator is equal to 0 when x=-1. This indicates that x-(-1) or x+1 is a factor of x^3+1. Doing a long division we find that x^3+1 = (x+1)*(x^2-x+1). Using the quadratic formula we see that the quadratic equation x^2-x+1=0 does not have real solutions which tells us that the quadratic polynomial x^2-x+1 is irreducible. We got as far today as setting up the partial fraction decomposition 1/(x^3+1) = A/(x+1) + (B*x+C)/(x^2-x+1), and getting the system of equations A+B=0, -A+B+C=0 and A+C=1. (To be continued on Monday.)
Monday, October 27. We completed the evaluation of Int(1/(x^3+1),x). Picking up where we left off on Friday, we found that A=1/3, B=-1/3 and C=2/3. This told us that Int(1/(x^3+1),x) = (1/3)*(Int(1/(x+1),x) - Int((x-2)/(x^2-x+1),x)). The first integral is just ln|x+1|. To evaluate the second integral, we fell back on an old algebraic trick that we could give a new use in calculus: completing the square. Comparing the denominator x^2-x-1 with (x-a)^2=x^2-2*a*x+a^2, we realized that we could write x^2-x-1 = x-2*(1/2)*x+(1/2)^2+3/4 = (x-(1/2))^2+3/4, so the remaining integral is Int((x-2)/(x^2-x+1),x) = Int((x-2)/((x-(1/2))^2+3/4),x). We then made the substitution u=x-(1/2) [with du=dx] to get that this integral was just Int(((u+(1/2))-2)/(u^2+3/4),u) = Int(u/(u^2+3/4),u) - Int((3/2)/(u^2+3/4),u). The first integral could be computed by making the substitution v=u^2+(3/4) [this gave a term which involved the logarithm of v (the absolute value bars not being necessary since v is positive), while the second integral could be evaluated by recognizing that it looked like an arctangent and then making the substitution w=2*u/sqrt(3). In a certain sense, this integral evaluation is a sort of high-water mark in the course; we used almost every integration technique we have learned except for integration by parts: partial fractions, completing the square, substitutions, and differentiation formulas for inverse trigonometric functions (and if we hadn't noticed that the last integral would lead to an inverse tangent, we would have made a trigonometric substitution). We then began to wonder how useful Computer Algebra Systems might be in evaluating similar integrals, and we then went to the computers where we looked at a Maple Worksheet "CAS Integration." Together the class worked through the part of the worksheet that showed how Maple evaluates integrals of the form Int(1/(1+x^n),x). One thing that we saw was that it could do the integration of Int(1/(1+x^3),x) that we had spent almost an hour (between Friday and today) doing in a few seconds. On the other hand, we noted that although Maple could integrate Int(1/(1+x^4),x) it was incapable of factoring it into a product of irreducible quadratics even though that is a calculation that we can perform without too much difficulty. The class was given some time to explore other sections of this worksheet: "An integral for Arclength", "Let's see how Maple does on Int(x/(1+exp(-x^2)),x)", "Let's see how Maple does on Int(sec(x)^3,x)", and "Integrals of sines of powers." Finally, we began to discuss how to compute arclengths. This is one of our basic integration applications, where we have to figure out how to compute a quantity that (at first) it seems we don't know how to calculate. The trick is to break the curve up into many small segments. We then approximate each of these segments of the curve by its secant line segment. The length of this secant line segment can be computed by using the Pythagorean Theorem; it's just sqrt((Delta x)^2+(Delta y)^2) where Delta x (respectively, Delta y) is the change in the x-coordinate (resp., the y-coordinate) between the two endpoints of the line segment. So an approximation to the arclength is just the sum of many terms each of which is in the form sqrt((Delta x)^2+(Delta y)^2). In order to pass to an integral in the limit as the number of segments tends to infinity and the lengths of all of the little segments tend to zero, we need to extra a factor of Delta x from the square root. We do this by writing sqrt((Delta x)^2+(Delta y)^2) = sqrt(1+((Delta y)/(Delta x))^2)*Delta x. Note that when we take the Delta x --> 0 limit, (Delta y)/(Delta x) --> dy/dx; this is just the slope of the secant lines converging to the slope of the tangent line, i.e., to the derivative. In this way we obtained the formula for arclength L = Int(sqrt(1+(dy/x)^2),x=a..b). We ended by noting that it was possible to obtain a formula giving the arclength as an integral with respect to y by factoring Delta y instead of Delta x out of the square root before passing to the limit that transformed the sum into an integral.
Wednesday, October 29. We began by reviewing the derivation of the integral formula for the (arc)length of a curve, and then we did two arc-length calculations. First we calculated the length of a hanging cable [recall that we had seen on September 12] that this curve is the graph of the hyperbolic cosine. Our second example seemed straight-forward enough (at first), but it turned out to be quite complicated if we didn't approach it the right way. We tried to compute the length along the curve y^2=x^3 between the points (1,1) and (8,4). If we tried to calculate the arclength as an integral with respect to x we'd get L = Int(sqrt(1+(4/9)*x^(-2/3)),x=1..8) -- note that although Maple can numerically evaluate this integral, it cannot compute the indefinite integral Int(sqrt(1+(4/9)*x^(-2/3)),x) -- this was the arclength integral treated in the Maple worksheet that we were using on Monday. With some effort we can make the substitution u=sqrt(1+(4/9)x^(-2/3)) [to find the relation between dx and du, it is simplest to first reverse this substitution to write x=((9/4)*(u^2-1))^(-3/2)]. Doing so we obtain L = (-8/9)*Int((u^2-1)^(-5/2)*u^2,u=sqrt(13)/3..sqrt(10)/3). Maple can compute the corresponding indefinite integral now, and with much more work (try a secant-substitution) so can we. At this point we went back to see if there was a simpler way to compute the desired arclength, and it turned out that there was! If we wrote the length as an integral with respect to y then we'd just get L = Int(sqrt(1+(9/4)*y),y=1..4) = ... = (8/27)*(10^(3/2) - (13/4)^(3/2)) ~= 7.634. Perhaps the most important point to take away from this lecture is this: Just as there are two ways to compute volumes of revolution, there are also two ways to compute arclengths. In a particular problem, if one method leads to an integral that seems to be difficult to evaluate, investigate setting up a different integral to calculate the arclength. Here are a pair of interesting comments on today's lesson. (1) The fact that Maple cannot integrate sqrt(1+(4/9)*x^(-2/3)), but that it can evaluate the integral after we have made an appropriate u-substitution is a compelling argument for learning integration methods even if you have access to a powerful Computer Algebra System such as Maple. (2) It is also interesting to note that the straight-line distance between (1,1) and (8,4) is sqrt((8-1)^2+(4-1)^2) = sqrt(58) ~= 7.616. The fact that the distance along the curve is so close to the straight-line distance between the endpoints tells us that the curve y^2=x^3 is nearly linear between x=1 and x=8.
Friday, October 31. We began with one last arclength calculation, one where we've known the answer for a long, long time but where we've never seen before why that answer is right: we tried to calculate the circumference of a circle. Now the equation for a circle of radius r is x^2+y^2=r^2. The equation of the upper semicircle is then y=sqrt(r^2-x^2), so we could calculate the arclength of the semicircle, and then just double our answer. We decided to take an additional advantage of the symmetry of the circle and just calculate the arc-length of the quarter-circle in the first quadrant; then we would obtain the circumference by multiplying this arclength by 4. We then set-up the integral in a straightforward way: y' = -x*(r^2-x^2)^(-1/2), so (after some algebra) sqrt(1+(y')^2) = 1/sqrt(1-(x/r)^2). Therefore the arclength is given by Int(1/sqrt(1-(x/r)^2),x=0..r), which after the substitution u=x/R becomes r*Int(1/sqrt(1-u^2),u=0..1). We wanted to say that this was just r*arcsin(u) evaluated between 0 and 1, but there's a problem with this last step. The justification for writing that Int(f(u),u=a..b) = F(b)-F(a) is the Fundamental Theorem of Calculus, which requires that the integrand be continuous on the entire (bounded) interval [a,b]. That's not what happens here; the integrand is undefined (hence discontinuous) at the upper limit of integration u=1. We first proposed a way of getting around this problem that was particular to the problem of the circle and then went back and proposed a more systematic way to deal with such problematic definite integrals. We decide that we could compute the circumference of the circle by calculating the arclength of the eighth of the circle that is in the first quadrant and above the line y=x. Following our work from above, this arclength is L = Int(1/sqrt(1-(x/r)^2),x=0..sqrt(2)*r/2) = r*Int(1/sqrt(1-u^2),u=0..sqrt(2)/2) = r*(arcsin(sqrt(2)/2) - arcsin(0)) = r*Pi/4, and so the circumference is C = 8*L = 2*Pi*r -- as we had expected. There was no problem with using the Fundamental Theorem of Calculus with this last approach because we stayed away from the "bad point" u=1 where the integrand was discontinuous. We then talked more generally about definite integrals. We said that a definite integral was proper if it was an integral to which we could apply the Fundamental Theorem of Calculus. Specifically, for a definite integral to be proper the interval of integration needed to be a (closed and) bounded interval [a,b], and the integrand needed to be continuous on the entire interval of integration. An definite integral that is not proper is called an improper integral. Following Stewart we use the following terminology: an integral that is improper because the interval of integration is unbounded (either (-Infinity,b] or [a,Infinity) or (-Infinity,Infinity)) is called an improper integral of Type 1, while an integral with a bounded interval of integration but a discontinuous integrand is called an improper integral of Type 2. We went on to discuss the Type 2 improper integrals, deferring the study of the Type 1 integrals to next week. We saw that we could cut out the bad point u=1 in Int(1/sqrt(1-u^2),u=0..1) by only integrating to some point u=t (with t<1) and then taking the limit as t tends to 1 from below. (Think of this as the mathematical way of scraping off the mold at u=1 from the cheese that is the interval [0,1].) Indeed, we DEFINED Int(1/sqrt(1-u^2),u=0..1) to equal lim_(t-->1-) (Int(1/sqrt(1-u^2),u=0..t) = lim_(t-->1-) arcsin(t) = arcsin(1) = Pi/4, which when we multiply it by 4*r also leads to the formula C=2*Pi*r for the circumference of the circle. We ended by discussing how to define the improper integral Int(f(x),x=a..b) where f is discontinuous at some point c in [a,b]. There are three cases to be considered: c = b, c = a, and a < c < b. The treatment in class can also be found in box (3) on page 491 of the course text. One last important point that was made in class is that improper integrals are LIMITS of proper definite integrals, and as such, there are generally two possibilities: either that limit fails to converge (in which case we say that the improper integral is divergent) or the limit converges to some number (in which we say that the improper integral converges to that number, although we sometimes shorten this to saying that the improper integral equals that number).
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