Math 122 — Fall 2013

Instructor: Dr. William DeMeo

Office: LeConte 314C

Office Hours: Wed, Thu, Fri 11–12

Course Syllabus

Homework and Textbook Information

Tentative Schedule

Exam Information

Supplemental Instruction Schedule (You may attend any of the sessions listed for Math 122, regardless of what section you are registered for.)

Other Notes and Resources

  • Free tutors are available in the Math Department.
  • Related rates. This topic is not covered in our textbook.  If you are interested in learning about it, a good resource is this webpage.

49 thoughts on “Math 122 — Fall 2013

  1. Professor DeMeo,
    I see that when f”(x)>0, f'(x) is increasing … can I assume that f(x) is concave up considering that f'(x) is increasing? Does increasing always mean concave up?? (I am referring to problem 6 on the chapter 2 review)

    Grace Powlas

    • Dear Grace,

      Good question. Assuming the derivative f'(x) exists at the point x, then we have f'(x) > 0 if and only if f(x) is increasing.

      Therefore (since the second derivative is the derivative of the first derivative), we have f”(x) > 0 if and only if f'(x) is increasing.

      We also know that f”(x) > 0 if and only if f(x) is concave up. So, the answer to your question is yes, all of the following statements are equivalent (assuming f”(x) exists at x):

      1. f(x) is concave up
      2. f'(x) is increasing
      3. f”(x) > 0

      See also: http://en.wikipedia.org/wiki/Second_derivative#Concavity

  2. Hi. I’m Olivia from your calc class on MWF. At 1:00 and I’m stuck on problem 32 from section 7.3 on the homework. I’ve tried it several ways and can’t figure it out. I’m down to my last attempt. Can you please explain to how to do this?

    • Dear Olivia,
      Good question. You are asked to “Use the Fundamental Theorem to determine the value of b if the area under the graph of f(x) = 6x between x=1 and x=b is 144. Assume b > 1.”

      Since the area under the graph between 1 and b is 144, we know that

      \int_1^b 6x \, dx = 144

      We need to compute the definite integral on the left, set it equal to 144, and then solve for b.

      \int_1^b 6x \, dx = \left. \frac{6}{2}x^2 \right|_1^b = 3 b^2 - 3.

      So we have the equation 3b^2 - 3 = 144 and we must solve for b. We will get two answers, one positive and one negative, but we know that b must be greater than 1.

  3. I had just one more quick question for you… I understand how to do both u and v, and just w substitution, I’m just having a hard time figuring out when to do which one. What attributes in a problem will tell me which to use? Also, on our tests, will we have to know which form of substitution to us or will you tell us? Thank you again so much!

  4. I’m having trouble understanding a problem that is on both the homework and the review for Exam 3. I think we may have gone over this in class but I still don’t really understand it. In the homework it is question 30 from section 7.3. It asks for the exact region bounded by (x^3)x. Do I just find where the graph intersects the x-axis and use those as my a and values and then integrate the equation and solve it using the first fundamental theorem of calculus? Also am I supposed to include where the original equation crosses the x-axis on the left side of the y-axis?

  5. Hello! I’m working on 7.4 question 3, and I am stuck. My problem is identical to the one in the book except it has e^4z instead. I unfortunately used up all my tries on my homework, but as I was using the option to see the solution, I was very confused on the last line. I understand all of the substitution and integral parts, but I am confused how the second to last line of the solution turns into the final solution. Can you please help me see how it simplifies into that answer? Thank you!

    • Dear Courtney,

      The last line of the solution they give is just a simplified version of the expression in the second-to-last line. If you had submitted the expression in the second-to-last line as your answer, you would have gotten it right. Here are the expressions in question:

      \frac{1}{4}(z+1)e^{4z} - \frac{1}{16}e^{4z} + C = \frac{1}{16}(4z + 3)e^{4z} + C

      To see that these are equal, use common denominators, and then factor out e^{4z}, as follows:

      \frac{1}{4}(z+1)e^{4z} - \frac{1}{16}e^{4z} + C = \frac{4}{16}(z+1)e^{4z} - \frac{1}{16}e^{4z} + C = \frac{4(z+1)-1}{16}e^{4z} + C

  6. Hi,
    For homework 7.3/7.4 question 24, I chose use -t^2 as my w value. My tutor told me that I would have to flip the bounds to go from -4 to 0 to account for the negative. This produced an answer of e^0 -e^-4, which was incorrect. He was very confident that it should be right. What are we doing wrong?

    • Sorry that this signed me in automatically with something else, this is Alex Voight.

    • We solved this one in lecture, but I’ll briefly review what we did. The integral is

      \int_0^4 2t e^{-t^2} dt.

      As you correctly answered on WileyPLUS, an appropriate substitution is w = -t^2, which yields dw = -2tdt and the integral becomes

      \int e^w (-1) dw

      Notice that I did not include limits of integration at this point. Probably the simplest way to proceed is to leave off the limits of integration until after you have computed the antiderivative of -e^w and expressed it as a function of t. After expressing the antiderivative in terms of t using the substitution w = -t^2, then you can use the original limits of integration, 0 and 4, to compute the definite integral.

      Please let me know if anything I’ve written is unclear.

    • I have worked out 7.3 question 16 about five different ways and continue to get 15.465 as my answer. What am I doing wrong

      • Dear Alex,

        It’s impossible for me to know what you are doing wrong if you don’t tell me what you are doing. Please explain what you have tried, so I can try to help you.

  7. Never mind I figured it out. Figured out you had to use the absolute value of the function. Unfortunately after the 11:00 due date…

  8. I have a question about two of the questions on the homework. I used the U substitution for question 14 I got ln(y+4) and for question 39 i got ln(e^t+6t). I was just wondering what I did wrong or if these are right. Thanks

  9. This isn’t specifically a mathematical question, but on WileyPlus, I’ve gotten a few questions wrong on my first try and when I try to correct them, it won’t let me retype the equation into the box, It says I have only used 1 of 5 attempts, but the box has no cursor for me to type the answer in. I thought it may have just been a glitch in the site, but this first happened a few days ago and it’s still acting weird. It’s really strange and a bit frustrating…not really sure what’s going on. Is anyone else having problems?

  10. I have a question about some of the 7.2 problems from the homework. For example, for question 14 from section 7.2 I’m asked to find the integral of dy/(y+6) and I got the answer to be log(y+6) and even checked my answer with wolfram alpha and found that it gave me the same answer. I’m not sure if its an issue with how I am entering my answer or if it’s something else? I have tried lg(y+6) and ln(y+6) and gotten it wrong both times. Just wondering if I’m making a mistake somewhere that I’m missing! Thanks!!

    • Dear Meghan,

      We discussed this in lecture. Recall, the natural logarithm of x, which our book denotes by ln(x), is only defined for x>0. For such positive values of x the derivative of ln(x) is 1/x. When x<0, on the other hand, ln(x) is not defined so the derivative of ln(x) doesn't exist for such x values. However, if x<0, then we saw that the derivative of ln(-x) is 1/x.

      I would try your best to refrain from using internet resources other than our textbook and the lecture notes for two reasons: 1. The notation for things like natural logarithm may differ, and will cause confusion when you enter alternative notation into WileyPLUS. 2. There are *many* false statements on the internet. In particular, for the problem you mentioned, the answer provided by Wolfram’s so called "knowledge engine" is simply wrong.

      If you're stuck and having trouble making sense of what we did in lecture, probably the best thing to do is to consult the textbook or come see me in office hours. For this problem, for example, the textbook tells us an antiderivative of 1/x. In general, it's not ln(x) (unless you already know x>0), and ln(y+6) is not a general antiderivative of 1/(y+6) (unless you already know that y>-6).

  11. I can’t seem to get question 26 correct on the homework. My version has the equation as y=(x^2)-2 and I am looking for the area between 0 and 4. My calculator and various web programs both bring up 13.333 (repeating). I know that I need to find the two areas and subtract the area below the x-axis, but I am still getting this same answer regardless.

    • Dear Alex, I believe that for this problem WileyPLUS actually asks for the area between the curve and the x-axis. If you take the integral of the function x^2 – 2 as x goes from 0 to 4, the answer will be the area above the x-axis minus the area below the x-axis. This is not what you want. You must compute two separate integrals, one for the area below the x-axis and one for the area above the x-axis, and then add the absolute value of these these integrals. If what I’ve written isn’t clear, please post another comment.

      • I am having complications on this problem as well. I have used my 4 out of 5 attempts and am having a hard time distinguishing between the lower and upper areas especially when putting it into my calculater

        • Dear Natalie, Please tell me what function appears in your version of the problem, what are your limits of integration, and what expression you came up with for the answer. Also, what did you enter in your calculator or Sage to solve this problem. For example, you might have tried in Sage: integral(x^2-2, x, 0, 4) (which would be incorrect).

  12. For question 19 in section 4.2 you have to find the critical points and inflection point of f(x)=x^5-25x^4+25. I found the critical points to be a max at x=0 and a min at x=20 but I cannot find the inflection point. Thank you.

    • Dear Stephen, What have you tried? Using the second derivative, you will find two x values where there could possibly be inflection points. Then, when you graph it, you will find that only one of these gives an inflection point. I graphed it at sagemath.org with this command: plot(x^5-25*x^4+25, (x,-20,30))

  13. For the supplemental questions in the 4.1-4.3, how do you solve for the critical points after finding the first derivative? I’m ending up with 2 variables and i’m not sure how to eliminate the upper-case R

  14. When you get a chance, take a look at 3.3.1 Question 23 of Chapter 6’s homework. I got the right answer, but it isn’t registering the lack of multiplication symbol between the number and the square root function as implied.

    • Dear Kyle, The answers you entered are incorrect. The problem is not related to a lack of multiplication sign, but rather your use of parentheses. For 3.1.23, the first term of the answer you submitted is (5t)4, which is equal to 54 t4, and is not correct. Similarly, on 3.2.15, you have (-2 e)-0.1t, which is -2-0.1t e-0.1t. Do you see why this is wrong?

  15. I’m having trouble solving question 3.1 number 22 on the homework. I can’t figure out what i’m doing wrong.

    • Dear Olivia, The version of Problem 3.1.22 given in the book asks for the derivative of \sqrt{1/x^3}. Can you rewrite this in a simpler form, e.g., as xa for some a? If so, can you tell me what a would be in this case?

  16. The graphs for 1.9 question 19 and 2.1 question 19 are not showing up, what should I do?

    • Dear David, I’m sorry you are having this problem. If you look on our Homework and Textbook Info page, you will see that another student had the same problem. Please read my reply to this student’s message. Also, you might try this link: http://help.wileyplus.com/browsercheck/index.html which will confirm that your browser meets some of the necessary requirements for using WileyPLUS. If changing or upgrading your browser doesn’t help, I suggest you call WileyPLUS tech support. If they cannot help you, I’d like to know about it. In that case, please send me an email explaining what happened.

  17. Dear Dr. Demeo, How do you do question 3 from section 2.2 chapter 2 in the homework? I know how the graph would look but I am unsure how to place the slope correctly. Thank you.

    • Dear Stephen, Were you in class today when we went over this problem? I will summarize our discussion as follows: if you look at the graph, it’s clear that there is one point where the derivative is zero, so you know where the graph of the derivative function crosses the x-axis. Let’s assume that the graph of the derivative is a straight line. We already have one point on the line, so the line will be determined once we have just one more point on it. For example, suppose the graph above is the graph of f(x), and consider x=0. Can you estimate the slope of the line tangent to f(x) at x=0? If so, then you have another point on the line, namely the point (0,m), where m=f'(0) is the slope of the line tangent to f(x) at x=0. (If you’re having trouble estimating the slope of the tangent line, then I would suggest printing out the graph and using a ruler and a pencil.)

      • Dear Dr. Demeo, I was in class I was just having some trouble piecing all of the information together into application for the problem. I think I understand it now. Thank you very much for your help and time

  18. Is our homework due tomorrow night or Sunday night since we haven’t learned 2.2 yet?

  19. I have a question about number 22 in section 1.9 on the homework. How many decimals do you want us to make x when writing the formula. I figured out the number but don’t know how many decimal places to make it. I know its the right answer because when I used it for the second part, I got the right answer.

    • Dear Andrew, You’ve asked how many decimals to make x, but I’m not sure what x is. The question asks for the formula for blood mass, M, as a function of body mass B. It also says that M is proportional to B, so the general form of the answer would be M = k*B. Your job is to find k. (Judging from the most recent WileyPLUS answer you submitted, it appears you are squaring some number to find k. Squaring is not necessary for this problem. Also, I believe that for most versions of this problem, yours in particular, no rounding is necessary since k will have only a few nonzero decimal places.)

      • Alright thank you! I figured it out but forgot to take the square out of the equation. So thats why I still missed that question, but I did figure it out! Thank you.

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