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Drawing a Graph using Integral Equation

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Ok guys, its been a while since I have done this one so I don't even know where to begin. Here is the problem I have to complete:

    The power (in watts) from an engine is represented by the equation below, where t is the time in seconds.

    P = 30t^2.2 + 3t

    1) Draw a table showing values of P for the range t=0 to t=9 seconds in intervals of 1 second. (Stuck on how to do this with the above equation)
    2) Plot the graph that represents power against time from 0 to 9 seconds for this engine. Ensure the graph is correctly and fully labelled and titled. (Easy part)
    3) Show the area on the graph which represents the energy converted between t = 5s and 9s (Easy part)
    4) Using the trapezium rule, estimate the energy converted between t=5 and t=9 for intervals of 1 sec (never done this before)
    6) Check your answer by integrating the expression between the limits t=5 and t=9. Account for the differences obtained.

    2. Relevant equations
    1) How do I solve my table of co-ordinates to plot the graph?
    2) How do I apply the trapezium rule once I have plotted the graph?
    The rest I can do
    3. The attempt at a solution
    I have integrated the expression to 9.375t^3.2 +3t^2/2

    Please be nice, I am still relatively new to all of this so where you may be fully experience in dealing with calculus, trapezium rules and definite integration i'm still learning and I can't see how to solve this task.
     
  2. jcsd
  3. Nov 29, 2014 #2

    PeterDonis

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    Your integral looks correct. However, you need to provide more of an attempt at a solution. Since you say items #2 and #3 are "easy", please post your solutions to those items.

    Also, does your textbook describe how to use the trapezoidal rule (or trapezium rule)? The Wikipedia page seems fairly informative:

    http://en.wikipedia.org/wiki/Trapezoidal_rule
     
  4. Nov 29, 2014 #3

    Hi, I didn't post anything up for 2 and 3 because I don't know how to solve the table to be able to draw the graph and the area between 5 and 9 seconds. Once I get the table I can draw the graph and area on easily. We have to leave our textbooks at work so I don't have it with me.
     
  5. Nov 29, 2014 #4

    PeterDonis

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    What have you tried? Have you tried plugging various values for ##t## into the formula?
     
  6. Nov 29, 2014 #5
    Hi, I have tried plugging in the time constants I.e. 0,1,2,3 but that then makes them 1^2.2, 2^2.2, 3^2.2 which produces high values, it didn't seem right?
     
  7. Nov 29, 2014 #6

    Ray Vickson

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    Just TELL us what values you finally obtained. How can we possibly be of help if you won't give us the information?
     
  8. Nov 30, 2014 #7
    Ok, firstly, please could you atleast be civil. I don't appreciate the tone in your message, I would have happily given you the answers had you asked me nicely. I'm here for help with my homework because I don't know how to solve it to get the right answers, I would appreciate it if you reconsidered your approach to me from here forward.

    The t is for the time constants 0-9. And has a power of 2.2 in the equation. So 0 to the power of 2.2 is 0
    1 to the power of 2.2 is 158.49 (2dp)
    2 to the power of 2.2 is 316.98 (2dp)
    3 to the power of 2.2 is 475.46 (2dp)
    4 to the power of 2.2 is 633.96 (2dp)
    5 to the power of 2.2 is 792.45 (2dp)
    6 to the power of 2.2 is 950.94 (2dp)
    7 to the power of 2.2 is 1109.43 (2dp)
    8 to the power of 2.2 is 1267.91 (2dp)
    9 to the power of 2.2 is 1426.40 (2dp)
     
  9. Nov 30, 2014 #8

    PeterDonis

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    Just as a note, if you read the instructions given on how to post homework questions, they will tell you to show everything you've already done, even if you don't understand how to go any further. If you had already made the calculations you gave in post #7, according to the instructions, they should have been part of the "attempt at solution" section of your original post.

    Please bear in mind that we are trying to help you, but we can't just give you the answers or tell you how to get them; that's not the purpose of the homework help forum. The purpose is for you to show us what you've done, and for us to try to help you see for yourself how to get the rest of the way to the solution.

    And a coefficient of 30, and a second term with 3t; so the complete equation, as you posted it in the original post, is ##P = 30 t^{2.2} + 3 t##.

    Given the above, I don't understand how you got this. 1 to any power is 1; 30 times 1 to the power 2.2 is therefore 30; and 3 times 1 is 3, so plugging ##t = 1## into the above equation for ##P## should give ##P = 33##, shouldn't it?
     
  10. Nov 30, 2014 #9

    Ray Vickson

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    Nothing in my response was uncivil. Besides, what you really need to show are the values of ##P##, not of ##t^{2.2}##, because ##P## is what you have been asked to tabulate and plot.
     
  11. Dec 1, 2014 #10
    Ok sorry, it was a misunderstanding of context. I have found out that my powers have calculated incorrectly because I was using a poor and unbranded calculator. This is why I was having trouble solving the formula. I will re attempt it in the next 48 hours and let you know if I have any problems.
     
  12. Dec 1, 2014 #11

    Ray Vickson

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    There are several free on-line calculators available. Alternatively, if your computer is loaded with a spreadsheet, you can use that.
     
  13. Dec 1, 2014 #12

    Mark44

    Staff: Mentor

    IMO, Ray WAS being civil, if a bit curt, which is understandable given the number of posts in this thread asking you for more information.
     
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