Math 110.109, Calculus II (Physical Science and Engineering)

Spring 2011 Course Lecture Synopses

 

Week 2:  February 7  11, 2011

 

http://www.mathematics.jhu.edu/brown/courses/s11/109.htm

 

 

 

 

·       Monday, February 7:  Today I introduced another integrand form whose structure appears quite difficult to work with, but has an interesting pattern that we can exploit.  The expression  shows up in many applications due to the integrand’s relationship to functions whose graphs are circles:  Think  as the equation for a circle centered at the origin of radius .  Solving for , we get a function of  if we restrict to the positive part, .  It is difficult to integrate directly using the techniques you already know, however.  So we seek to exploit the structure with a new technique.  Since the graph os this expression is a circle, why not use trigonometric functions to study it?  In this case, let   (the reason why will be clear in a moment).  Then .  Hence for a substitution like  in , where then , the new integrand would simply be a rather straightforward trig function and arguably much easier to solve, no?  Using this as a motivation, I introduced the notion of a trigonometric substitution for integrands that include a form similar to what we just studied.  There are three such forms, and the choice of a substitution really just depends on setting up what is under the radical to be something where a quick trig identity can make the radical easier to play with.  Above, the substitution  works since it takes the expression under the radical which is a differences of squares, and through the identity , or , makes it a product of squares, which is easy to simplify, eliminating the radical.  One important thing to worry about is the fact that, when switching variables like this  and  takes  you must keep the relationship between  and  one-to-one.  The reason is that one integrates with respect to a rising .  If the new variable  were not one-to-one, then as  increases, one would sometimes be integrating with a rising  (from left to right) and sometimes with a falling  (from right to left).  I will construct an example for you soon.  So we look for domains of substitutions where the trigonometric function is one-to-one.  Notice that the substitution makes the integral more complex.  But the idea is that now it is doable.  I did the example  to illustrate, using the substitution  to rewrite the integral .  And since on the interval , the function , this integral equals .  Do this calculation!  But we already knew that, since the graph of  on the interval  is one quarter of the circle centered at the origin of radius 2.  So its area is .  There are two other forms, with easy to implement substitutions.  One is , with a substitution , and  with its substitution .  I talked about hese in detail, about why they work. I finished the lecture with the example , using the substitution  and  to get that this integral equals  (Note that the absolute value signs go away since if we choose one of the two intervals where the substitution  is one-to-one (in this case also positive), then  will be positive also.  And the absolute values are not necessary.  Finally, I showed that the last integral equals .  We are not done yet, cine we need to return to the variable  in our answer.  We will do this next time.

·       Wednesday, February 9:  asdf.  To complete the previous calculation, we need to return to the  variable from the  variable, via the substitution , or .  To this end, set up a right triangle with one of the acute angle  Then, by the substitution, the hypotenuse has length , and the adjacent side has length .  Then the opposite side has length  by Pythagoras’ Theorem.  All of the other trigonometric functions are now available to us.  For example,   Our final answer is .  I talked much about how to set up these right triangles for the three core types of integral forms in this section.  I then did another example:   This one I will highlight in an extra problem writeup.  But the thing to watch here is the choice of substitution:  You want to make the term  into a single term (and not a sum) so that you can deal with the square root in the denominator (do you see the square root in the denominator?).  Here the appropriate substitution is .  Do you see why?  Here then .  The other thing to think about is the limits.  See the Extra Problem set for Section 7.3.  I then talked about Expressions that fit the form but do not look like it.  General quadratic expressions like  can be made to look like a sum or difference of two squares by the process of completing the square.  I used the example of  to illustrate:   Do you see how this is useful?  For the integral , we get (via the substitution ,  ),    which can now be solved with the trigonometric substitution .  I also detailed a portion of the similar problem .  Perhaps I will place this one also in the Extra problem set.  I finished the lecture by beginning the next section, on partial fraction decomposition, as way to write a general rational function as a sum of proper rational function, all of whose denominators have degree 2 or less.  Since we pretty much know how to always solve for the anti-derivative of such rational function, this is a nice procedure in general.  The only tricky part is in knowing how to factor the denominator of the original rational function, and how to make sure the original rational function is proper (degree of the numerator is less than the degree of the denominator).  I talked about what a proper rational function is, and how via long division, one can always reduce the problem to working with proper rational function.  I talked about factoring the denominator and then talked about the general procedure. 

·       Friday, February 11:  Continuing with the discussion on the algebraic technique called Partial Fraction Decomposition, I re-showed the basic steps of the method used to rewrite a proper rational function as a sum of proper rational functions whose denominators are all of degree 1 or 2.  Since we already know how to integrate rational functions whose denominators are linear or quadratic polynomials (see previous sections or Calculus I), this is a great way to integrate rational functions in general.  After writing down the method, I looked at the example I mentioned last class:  .  Here the denominator factors as , so we know by PFD that we can write .  Remember that the summands all have the factors of the denominator of the original rational function, and since the fractions on the right-hand side are proper, there can only be constants on the tops.  Since we do not know what they are yet, we leave them as unknown constants.  It is only algebra to find values for them;  recombine the fractions on the right, to get .  Hence   Now anytime two polynomials are equal, it must be the case that ALL respective coefficients are equal.  This means that the  coefficient on each side must be equal, and also the constants on each side.  This leads to two  equations (one for the monomial  and the other for the constants):   and .  This is solved by  and .  Hence   I then mentioned two extra complications that are also quite straightforward.  1) when I factor of the denominator of a rational function is quadratic, then the proper rational function corresponding to this factor may have a linear numerator instead of simply a constant.  This is not really a complication, but rather an acknowledgement that there is another constant to find.  I used the example .  In this case, the decomposition equation would look like .  Since the right hand side fractions must be proper, we must allow for the possibility that the first one may be of degree 1 on the top.  Thus we need two unknowns to represent a general form of a linear polynomial.  Nothing else changes here.  If you perform the same procedure, you will get three equations in three unknowns (the coefficients of , , and the constant coefficient), and the result will be ,  and .  Thus   Now how would you finish this problem?  Finally, I talked about how to handle the situation when the denominator has repeated factors.  One cannot simply add a new term or each factor here, as in , since the last three are really only one fraction with the numerator .  Instead, the process is to acknowledge that the factor to each intermediate power needs its own term in the sum.  After talking about this a bit, I did the example:  .  Here one plays the same game using the PFD .  Here one can solve the resulting four equations in four unknowns to get  and , and the final result is .  The final answer is .  I finished the class by starting directly into Section 9.1 with an introduction to differential equations and mathematical modeling.