Mark Bly's Math 142 Webpage

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Math Tutorial Center

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Lecture Commentary

Jan 13
Jan 15
Snow Day
Jan 22
Jan 25
Jan 27
Jan 29
Review Day
Feb 3
Feb 5
Feb 8
Feb 10
Feb 15
Feb 17
Feb 19
Feb 22
Feb 26
Review Day
Mar 2
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Mar 28
Mar 30
Apr 1
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Apr 6
Apr 8
Apr 11
Series Test User Guide
Apr 15
Review Day
Apr 20
Apr 25

HW Help

Test 1
HW #1
Tue Jan 19
(5.3 #23) Since there is no product rule for integrating, you might want to multiply/foil and then take the antiderivative.
(5.3 #25) Since there is no division rule for integrating, you might want to divide each numerator term by the x^2 first.
(5.3 #41) Here, f is the derivative of the other functions graphed. So, if f takes a value of zero, the other functions would need to have a critical pt.
(5.3 #53) Similar to #23 above.
(5.3 #61) No simple way for your answer not to involve a power of e.
(5.3 #63) A two-stepper. You might try finding each missing constant at each step rather than involving two contants at once.
(5.3 #71) The first sentence essentially says that s(1)=0. Getting familiar with this kind of phrasing will help for these types of problems.
(5.1 #32b) This is just the sum a_1+...+a_10 minus the a_1 term. You can think about what the difference will be.
(5.1 #39) Similar to 5.3 #23.
(wksht #1) Similar to 5.3 #25, but with x^(1/2).
(wksht #3) Similar to 5.3 #63.
HW #2
Tue Jan 26
(5.1 #7b) The region is a trapezoid that can be broken into a rectangle and triangle.
(5.1 #43) 1/N^2 can be factored out of the summation since it does not involve the variable i.
(5.1 #58) The summation will distribute across the sum and factor out the constant to yield formula-friendly Simga(x^2) and Sigma(x) sums.
(5.1 #63) The constant multiple is the width of each rectangle. The terms of the sum are the rectangle heights/function values.
(5.1 #73) Squaring y=sqrt(1-x^2) and simplifying, you get x^2+y^2=1^2, the equation of a circle centered at the origin of radius 1. Since you are only given the positive sqrt, you are only conisdering the semicircle portion where y>=0.
(5.2 #7) Similar to 5.1 #73.
(5.2 #61) You have the area under f(x) on [2,9] and are subtracting the area under f on [2,5]. What amount of the x-axis remains considered?
(AP #2) Similar to 5.1 #63.
(AP #4) Similar to 5.1 #73.
(AP #6) Consider the min and max x-values given as the domain (ie. [1,8]). Then, break the area region into pieces as given.
Email Q&A
HW #3
Tue Feb 3
(5.4 #23) No quotient rule. Divide each term in numerator by t^(1/2) and then you have just a sum.
(5.4 #39) In general, the integral of 1/(mx+b) equals (1/m)*ln|mx+b|+C. The 1/m is to account for the chain rule.
(5.5 #29) d/dx of F(x^2) requires the chain rule. It computes to f'(x^2)*2x.
(5.5 #33) Similar to 5.5 #29, but you need to compute d/dx of F(x^2)-F(sqrt(x)). Since this expression is a sum, just take it term by term.
(5.6 #5) The integral from a to b of v(t) is V(b)-V(a), which is just s(b)-s(a)...or, change in position.
(5.7 #33) In general, when in the function you see a pair of polynomial expressions differing in degree by 1, set u equal to the higher power expression.
(5.7 #59) Hint: u=x^(1/2). When you compute du, you'll see why this works.
(5.7 #72) Hint: u equals either x, 1/x, lnx, or 1/lnx.
(5.7 #91) Similar to 5.7 #72.
(5.8 #11) In other words, x=3u.
(5.8 #25) Comparing the expressions in the numerator and denominator, one is the derivative of the other. Substitution can come to you aid.
(5.8 #44) Since you don't have an integration rule for tan(x), you might rewrite as sin(4x+1)/cos(4x+1) to get in terms of more familiar trig functions.
(5.8 #50) This one is not easy to see. Using u=sqrt(25x^2-1) and multiplying the top and bottom of the fraction by 25x, it is doable as an invtan integral (not invsec...which doesn't apply without absolute value bars).
(AP #13) You need x to be some multiple of u that when squared cancels the 9 and allows a sqrt(4) to be factored out of the denominator.
Email Q&A
5.5 #29
5.7 #54
5.8 #15
Test 2
HW #4
Tue Feb 9
(6.1 #29) The variable x is of degree 1 and y is not. Solve for x and graph in yx-plane.
(6.1 #38) You can drop the absolute value and find the point of intersection where x is positive. Then, using the fact thatthe parabola and aboslute value functions are symmetric about the y-axis, you have the other intersection point
(6.1 #41) Similar to 6.1 #29
(6.1 #47) Need to break the region between the three curves into a pair of regions each between two curves. See image.
(6.2 #9) For any given y, what is the width of the unit circle from left-to-right? This expression holds the key.
(AP #2) Similar to 6.1 #29
HW #5
Tue Feb 16
(6.2 #25) u=x+1?
(6.2 #27) Multiply out x(6-x) to be able to integrate.
(6.3 #15-19) On the region of interest, you need to determine which function is above and which is below.
(6.3 #27-31) Note region A is between y=x^2+2 and y=6, or (as a function of y) between x=2 and x=sqrt(y-2)
(6.3 #33-35) Note region B is between y=x^2+2 and y=0.
(AP #5) Even though this region is technically bounded by three curves, you should be able to complete the computation using only two of those functions in your integral.
HW #6
Tue Feb 23
(6.4 #43) Not rotating about the x- or y-axis. Must be careful accounting for the radius of rotation.
(6.4 #47) Function of y about y-axis. This is not shells, is it?!
(6.4 #51) Similar to 6.4 #47 above.
(6.5 #3-7) Note that k is given in N/m, but stretching/compressing measurements are given in cm.
(6.3 #17-29) Videos posted for Fri 2/19 may be of assistance. In #29, note the kg/m chain density can be computed directly from the information given.
(AP #2) Similar to 6.4 #43 above.
(AP #6) Similar to 6.3 #17 above.
Test 3
HW #7
Wed Mar 2
(7.1 #13, #21) Will need to iterate through integration by parts twice over.
(7.1 #22, #23) When you see ln(x) inside an integral, it is extraordinarily frequent that you set u=ln(x) and use integration by parts.
(7.1 #35) With du=(1/2)*x^(-1/2)*dx, you can rearrange to write dx in terms of just u and du to be able to integrate (by parts).
(7.1 #77) With u=x^4, you can write x^7*dx in terms of u and du to be able to integrate (by parts).
(7.2 #53) Recall that 1/cos(x) is just sec(x). To integrate sec(x), use the (sec(x)+tan(x))/(sec(x)+tan(x)) trick from class.
(7.2 #55) When you see an integral involving only tan(x), a path to consider always is rewriting as sin(x)/cos(x).
(AP #1, #4) Similar to 7.1 #22 and #23 above.
(AP #6) Will have to use the double-angle identity cos^2(x)=(1/2)+(1/2)*cos(2x) twice to reduce the cos^4(x) to a cos^2(2x). Then, you'll have to use the identity again to get everything down to single powers of cos.
(AP #9) Similar to 7.2 #55 above.
Email Q&A
HW #8
Tue Mar 8
(7.3 #9) When the integral is completely in terms of tan and sec, I found it easier to see how to integrate taking everything to sin and cos.
(7.3 #15) When you boil down to a sin^2 integral, you'll need to use the double-angle identity (like on last HW).
(7.3 #17) Converting from sec and tan into sin and cos, you can get down to integrals of csc(x). Similarly tricky to integrating sec(x), this time multiply top and bottom by (csc(x)-cot(x)).
(7.3 #21) Recall that cot has a derivative of -csc^2. This will help you integrate csc^2.
(7.3 #48) You'll need your double-angle identities to make sense of the sin^2*cos^2 integral.
(7.4 #9) Just like lone powers of tan, you might try dealing with this lone power of tanh by writing in terms of sinh and cosh.
(7.5 #31, #37) Expressions given are not proper rational functions. Need to divide down the numerator degree first.
(AP #1) Rewriting 1/cos^2 as sec^2 can help you integrate more straightforwardly.
(AP #2) Similar to AP #1, but with 1/sin^2 and csc^2.
(AP #4) When you get down to sin^2/cos^2, you might write as tan^2 and then swap for sec^2 using 1+tan^2=sec^2 to get something straightforward to integrate.
(AP #5, #6) The identity cosh^2(x)-sinh^2(x)=1 will be helpful on these. Analogous to odd powers of sin & cos problems.
HW #9
Tue Mar 22
(7.5 #13) Not a proper ratio of polynomials. Needs the division algorithm to be made proper.
(7.6 #41) If you FOIL, you can deal with the sin(x)cos(2x) term by using the double-angle identity in reverse...that cos(2x)=2cos^2(x)-1.
(7.6 #45) After using integration by parts, the 2x^2/(x^2+9) term can be dealt with via division algorithm and partial fraction decomposition.
(7.6 #55) The notation ln^2(x) means ln(ln(x)). And, letting u equal this gets you half way home via parts.
(7.7 #1) Does the integral have a bound of +/- infinity? Does the integral have a bound at a vertical asymptote?
(7.7 #11) u=4-x will get the function in the form of a familiar p-integral.
(7.7 #15) Similar to 7.7 #11.
(7.7 #45) Substitution with u=1+x^2 can evaluate the integral.
(7.7 #59) Since |1-sin(x)| is never more than 2, our function is sandwiched between 2/x^2 and -2/x^2. Sketching a picture of this might help.
(AP #2) Similar to 7.5 #13.
(AP #3, #4) Similar to 7.7 #11.
Email Q&A
Test 4
HW #10
Tue Apr 5
(7.9 #15, 17) Note N is even, as will always be the case with Simpson's rule. Reason described in Lecture Commentary Mar 23.
(7.9 #27) The data given contains your function values (ie. f(0 min)=550, f(5 min)=575, etc.).
(8.4 #33) To find our K, we need the maximum absolute value of the 6th derivative of f on [0,0.25]. This is the maximum of |-cos(x)| on [0,0.25], which can be determined definitively.
(8.4 #41) We just need to find a single n. So, we could use a K that we know bounds all absolute derivative values for cos(x) and then chase the n by substituting n=1,2,3,... into our error formula.
(8.4 #45) The expression on the right hand side is just |x|^4/4!. So, the K being used is 1. Why does K=1 work for f(x)=e^-x on [0,inf)?
(8.4 #46) Similar to 8.4 #45 above. Note that 5!=120. So, again, we must show that K=1 is a valid choice of K.
(10.1 #23) Multiplying top and bottom by 1/sqrt(x^2) might help.
(10.1 #25) Since ln() is a continuous function, it distributes in and out of a limit. That is to say, "the limit of ln(f)" is just equal to "ln(the limit of f)".
(AP #2) Similar to 7.9 #27.
(AP #3) Recall the derivative of tan^-1(x) is 1/(1+x^2). For subsequent derivaives, just use the chain rule.
(AP #4, #5) Similar to 8.4 #33.
(AP #6) Similar to 8.4 #41.
Email Q&A
HW #11
Tue Apr 12
(10.1 #41) If you separate 9 copies of 9 upstairs and 9! downstairs from the other terms in the fraction, it may be easier to see the pattern.
(10.1 #45) With trig function upstairs, this is a squeeze theorem candidate.
(10.1 #55) May be easier to see ratio when separated as a sum of two fractions, each written over the common denominator.
(10.2 #11) Note cancellation from telescoping terms.
(10.2 #12) Use partial fraction decomposition to see the telscoping terms of the sum.
(10.2 #25, 29) Note sum does not start at n=0. Must account for missing terms.
(10.2 #33-35) Can factor out common factor so the terms of the sum start with a 1 and can use the goemetric series formula.
(10.3 #5) u substitution?
(10.3 #17) Note that when n>=1, it is true that n>=sqrt(n).
(AP #1, #2) L'Hopital can be of service.
(AP #3) Similar to 10.2 #33-35.
(AP #6) Bounding the numerator and using comparison can make the numerator a lot easier on the eyes.
(AP #7) Recall your result from AP #1.
(AP #8) Note the terms of this sum form a decreasing sequence of positive numbers.
HW #12
Mon Apr 18
(10.3 #25) A series with a sin or cos upstairs can always have the numerator bounded in absolute value by 1.
(10.3 #51) Difference in powers between top and bottom is 1/2. Limit compare to 1/n^(1/2).
(10.4 #12-13) Using the fact that for an alternating series |S-S_n| (10.4 #24) The terms in the sum approach what value?
(10.4 #25) Might first separate into two fractions, each over the common denominator to get two geometric sums, one with r=3/5 and the other with r=-2/5.
(10.5 #7) Note that (n+1)^100 has leading term of n^100, and all other terms are of lower degree (and hence do not affect the limit as n->inf).
(10.5 #50) u=ln(x)?
(AP #3) Rather involved problem. Using ratio test, note the expression (n+1)^n/n^n can be rewritten as [(n+1)/n]^n...which can further be rewritten as [1+(1/n)]^n. Now use the result from AP #2 in HW 11 that says this expression has a limit of e^1.

Test Review

Test 4
Practice Problems
Solutions to Practice Problems

Internet Videos

Test 1
Wed 1/13
The antiderivative
Power rule for integrals
Integral of sin(x), cos(x) & e^x
Integral of x^-1
A more complicated integral expression
Initial value problem with pos/vel/acc
Fri 1/15
Summation notation
Approximating area under curve (4 rectangles)
Fri 1/22
(5.1, 5.2)
Approximating area under curve (N rectangles)
Intro to Riemann Sum
Mon 1/25
(5.2, 5.4)
The fundamental theorem
Evaluating a definite integral with FTC
Breaking up an integral inverval
Swapping the bounds on an integral
Area under curve using graph
Wed 1/27
(5.5, 5.6)
Derivative of an integral
Area under a rate function
Integral and pos/vel/acc
Water in tub problem
Fri 1/29
Integration by u-substitution
An example using u-substitution
Another example using u-substitution
Integral of tan(x) via u-substitution
Area under curve with u-substitution
Mon 2/1
Test 2
Wed 2/3
Area between two curves
Area between three curves
Fri 2/5
(See videos from Wed 2/3 above)
Mon 2/8
Volume by cross-sections
Another volume by cross-sections
Volume of a sphere by cross-sections
Wed 2/10
Average value of a function
Average value example
Mean value theorem for integrals
Fri 2/12
Volumes of revolution - Disks
Volumes of revolution - Between 2 disks
Mon 2/14
(6.3, 6.4)
Volume by shells
Shell Method Example
Wed 2/17
Shells with two functions of x
Shells with two functions of y
Fri 2/19
Work - Raising Weight
Work - Pumping Water (Triangular Prism)
Work - Pumping Water (Inverted Cone)
Work - Pumping Water (Cylinder)
Mon 2/22
(See videos from Fri 2/19 above)
Test 3
Wed 2/24
Integration by parts (one iteration)
Integration by parts (two iterations)
Integration by parts (manipulating equations)
Integral of ln(x) (by parts)
Fri 2/26
Odd power of sinx or cosx example
Another odd power of sinx or cosx example
Even power of sinx and cosx example
Integral of tan(x) via u-substitution
Mon 2/29
Wed 3/2
Integral using trig substitution
Another integral using trig substitution
Fri 3/4
(7.3, 7.4)
A trig substitution w/ an even power of cosx
Part two of the video above
Hyperbolic Trig Functions
Mon 3/7
Integration using partial fractions
Wed 3/9
See Mon 3/7
Fri 3/11
A convergent improper integral
A divergent improper integral
Mon 3/21
See Fri 3/11
Test 4
Wed 3/23
Trapezoids to approximate area under curve
An example using trapezoids
Explanation of Simpson's Rule formula
Complete justification for Simpson's Rule
Mon 3/28
Wed 3/30
Idea behind Taylor Polynomials
Taylor Expansion for e^x
Taylor Expansion for sinx
Taylor Expansion for cosx
Fri 4/1
Visualizing Taylor Approximation
Why Taylor error depends on f^(n+1)(x)
Proof of Taylor error bound formula
Mon 4/4
Intro to sequences
Limit of a sequence
Examples of convergent/divergent sequences
Wed 4/6
Partial sums
Infinite series are a limit of partial sums
Finite geometric series formula
Infinite geometric series formula
Infinite bouncing ball and geometric series
Fri 4/8
Integral test
Integral test example
Comparison test
Comparison test example
Mon 4/11
(10.3, 10.4)
Limit comparison test
Alternating series test
Absolute vs conditional convergence
Wed 4/13
Ratio test
Root test
After Test 4
Fri 4/15
Intro to Power Series
Mon 4/18
Wed 4/20
Using ratio test to find radius of convergence
Representing a function as a power series
Using integration to determine a power series
Fri 4/22
Taylor Approximations using series
Visualizing a Taylor Approximation
Finding a Taylor Series based on cos(x)
Mon 4/25
Arc length formula
Arc length example
Wed 4/27
Area between f(theta) and origin
Example with area between f(theta) and origin
Fri 4/29
Center of mass basics