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Using function of function rule

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data
    At any time t seconds the distance x metres of a particle moving in a straight line from a fixed point is given by x = 4t + ln(1-t). Using the function of a function rule, determine

    A.) the initial velocity and acceleration






    3. The attempt at a solution

    I had a go at the velocity section and got the following.

    x = 4t + ln(u)
    x = 12 - 4u + ln(u)

    du/dt = 0 - 1 = -1 (dx/dt = dx/du x du/ds)

    v = dx/dt = (-4 + 1/u) x (-1) - (4 - 1/u)

    = 4 - 1/u = 4 - 1/3-t

    t=0

    dx/dt = 4 - 1/3-0

    = 4 - 1/3

    3 1/3 (initial velocity)

    I think this is right, I know it looks complex but my question is how do I go about finding out the acceleration from this. Any help much appreciated
     
  2. jcsd
  3. Mar 21, 2007 #2
    What's u? If you've substituted u for 1-t, how did you arrive at the second equation?
     
  4. Mar 21, 2007 #3
    u is equal to 3-t. does that answer your question
     
  5. Mar 21, 2007 #4
    Why 3-t?

    [tex]x = 4t + \ln{(1-t)}[/tex]
    [tex]v = \frac{dx}{dt} = \frac{d(4t)}{dt} + \frac{d(\ln{(1-t)})}{dt}[/tex]

    Now, the first derivative on the right is straightforward. For the second, substitute u = 1-t and use the chain rule (the function of function rule, if you prefer), [itex]\frac{df}{du}\frac{du}{dt}[/itex].
     
  6. Mar 21, 2007 #5
    i asked my lecture the same thing and he told me what I told you. what about the acceleration
     
  7. Mar 21, 2007 #6
    Are you sure the question says 1-t, and not 3-t? Even then the answer wouldn't change much. Just apply the chain rule for the second term.

    Acceleration is the rate of change of velocity.
     
  8. Mar 21, 2007 #7
    absolutely sure. When I was sitting talking though this with him he used 3 - t. Now I'm confused. Using the function of the function rule change 1-t into 3-t.
     
  9. Mar 21, 2007 #8

    cristo

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    You've either copied the question down incorrectly, or misinterpreted his substitution. To do this question, one would use the substitution u=1-t. Try working it through with this substitution.
     
    Last edited: Mar 21, 2007
  10. Mar 22, 2007 #9
    show me the question as you would do it and I will put it to him.
     
  11. Mar 22, 2007 #10

    HallsofIvy

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    So all of this was to trick someone into doing the problem for you?
     
  12. Mar 22, 2007 #11
    I've shown you my working for the question like you asked, and you have told me something is wrong with it. Not asking you to do the whole question I'm asking you to tell me where I have gone wrong with the part on its initial velocity. Isn't that what this forum is for.
     
  13. Mar 22, 2007 #12

    cristo

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    The purpose of this section of the forum is to help with homework. I (and others!) have pointed out exactly where your problem is; namely that you used an incorrect substitution. Furthermore, if you look at my previous post, I have told you the correct substitution to use.

    Work through the question again, but using the substitution I gave you above. If you have any problems, then show your work and state where you are having difficulty.
     
    Last edited: Mar 22, 2007
  14. Mar 22, 2007 #13

    arildno

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    What you have done wrong, or your teacher has done wrong is to mix together 1-t and 3-t. That has already been pointed out to you.
     
  15. Mar 22, 2007 #14
    does it matter which you use. 1-t or 3-t. I think its a typo, can't get hold of him to ask him. But I started it in class and used 3-t. Let this be a lesson to finish what you start.
     
  16. Mar 22, 2007 #15

    cristo

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    Yes, it does matter! The point of the substitution is so that you can simplify ln(1-t) to ln(u) and then apply the chain rule: [tex]\frac{d}{dt}\ln(u)=\frac{d}{du}\ln(u)\frac{du}{dt}[/tex].

    If you make the substitution u=3-t, then you do not simplify the expression of the logarithmic term, and so you cannot apply the chain rule in this way.

    Trust me; try it with the substitution u=1-t!
     
  17. Mar 22, 2007 #16
    I'll do it your way. because I don't really understand this 3-t business, thanks for the help and patience.
     
  18. Mar 22, 2007 #17
    should the answer for velocity be dx/dt = 4 - 1/(1-t) & acceleration be d2x/dt2 = -1/(1-t)^2
     
  19. Mar 22, 2007 #18

    arildno

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    Yes&yes :smile:
     
  20. Mar 22, 2007 #19

    cristo

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    Yup, they are correct. Now you need to find the intial values don't you?

    [Sorry, I keep cross posting with arildno!]
     
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