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Homework Help: Conceptual understanding of Derivates Question

  1. Sep 1, 2014 #1


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    Question:A bicyclist starts from home and rides back and forth along a straight East/West path. Her (instantaneous) velocity as a function of time is given by v(t), where time t is measured in minutes. Consider:

    [itex]d_{1} = \int^{30}_{0} v(t) dt[/itex] and [itex]d_{2} = \int^{30}_{0} |v(t)| dt[/itex]

    Choose what it represents from the following:
    (a) the total distance the bicyclist rode in 30 minutes
    (b) the bicyclist's average velocity over 30 minutes
    (c) the bicyclist's distance from the home after 30 minutes
    (d) none of the above.

    My thinking: Allright well, the two integrals (d1 and d2) are essentially the same thing other than the absolute value sign. This leads me to believe that d1 is (c) and d2 is (a). This is kind of a logical guess on my part. Can anyone explain what the equation is actually saying?

    To me, d1 is saying: The area under the curve from 0minutes to 30 minutes is represented by v'(t). This would mean to me that (c) should be the correct answer. While, d2 is saying the absolute value (negative and positive area combined) of v'(t) from 0mintues to 30 minutes. This would be the total distance traveled aka (a).

    If my conceptual understanding/writing is wrong, please do point it out to me.

    Is my thinking correct?

    Thank you all for your help.
  2. jcsd
  3. Sep 1, 2014 #2
    Do you mean, "Choose what each represents from the following"? What's v'(t)?

    General tip - draw (rough) graphs of v against t and |v| against t and remember the general definition of integration = "area under the curve".
  4. Sep 1, 2014 #3


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    Yes, I do mean that. Thank you.

    Hm. The only problem with graphing is, I don't have a function to graph.
    So this would be a purely conceptual question. I suppose.

    Maybe I can use a dummy function and try to graph that to help my conceptual understanding. I will try this.
  5. Sep 1, 2014 #4


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    Yes, you are correct- (d2) is the total distance ridden, (d1) is the distance from home. You can check that by taking a simple example. Suppose she rides directly away from home at constant velocity v (meters per minute)> 0 for 15 minutes, then rides directly back home with velocity -v for 15 minutes. After 15 minutes she will be 15v meters from home, then she turns around and after another 15 minutes is back home: 0 meters from home. But she rode a total of 30v meters.

    [tex]d1= \int_0^{30} v(t)dt= \int_0^{15} v dt+ \int_{15}^{30}(-v)dt= v(15- 0)+ (-v)(30-15)= 15v- 15v= 0[/tex]
    [tex]d2= \int_0^{30} |v(t)|dt= \int_0^{30} v dt= 30v[/tex]
    Last edited by a moderator: Sep 1, 2014
  6. Sep 1, 2014 #5


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    HallsofIvy, thank you for explaining the conceptual side AND the mathematical side of the problem to me. My hunch of logic has been verified.

    My kindest regards to you for helping me understand this.
  7. Sep 1, 2014 #6


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    One last Calculus question:

    So I understand that the integral of e^t^2 is not a normal integral and cannot be easily calculated.

    But in situation where

    the integral from 0 to sqrt(x) of e^t^2

    Is stated and then it asks me to Compute f'(x)

    Can I accomplish this?

    My thinking is that, Sure the Integral is not really possible for me to computer, by does f'(x) = the integral? I don't believe so.
  8. Sep 1, 2014 #7


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    Look into the Fundamental Theorem of Calculus, e.g.,http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus .

    WWGD :What Would Gauss Do?
  9. Sep 1, 2014 #8


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    Okay, so I am trying to work on this.

    I plug the integral into a calculator (for integrals) and receive the answer of:


    I'm just not quite sure what this is telling me. This seems to be telling me the area under the curve from 0 to sqrt(x) of the initial function.

    I am trying to find f'(x)

    So I've been supplied (post above) with the fundamental theorem of Calculus. This seems to help, but I'm not putting everything together.

    So, since f(x) = e^t^2 dt is f'(x) = 2t*e^t^2
    Is it that simple?
  10. Sep 1, 2014 #9

    Ray Vickson

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    No. I hope you realize that what you wrote is nonsense: you have x on one side and t on the other. On the other hand, maybe you do not actually mean what you wrote. Originally you had
    [tex] f(x) =\int_0^{\sqrt{x}} e^{t^2} \, dt \,[/tex]
    although you wrote something that could be interpreted as ##(e^t)^2##, which is very different. Use parentheses, like this: e^(t^2).

    Also: if you are just a beginner at this material, I would recommend that you avoid using the calculator, except for numerics; just do things directly, by hand, and reason it out step-by-step, taking as long as necessary and using as much paper as you need. With practice you will get better and faster---but likely only if you avoid the calculator.
  11. Sep 1, 2014 #10
    Suggestion on writing equations in physics forums - look up to the right of the editor and you will see an editing symbol that looks like X2. Click on that to write expressions like et2
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