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Homework Help: Derivative problems

  1. Oct 10, 2005 #1
    1) i'm having trouble with this problem, it looks like their isn't enough information. can someone give me some hints please
    http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg [Broken]

    2)
    http://i2.photobucket.com/albums/y15/seiferseph/untitled2.jpg [Broken]
    i solved parts a-c fine, but i'm having some trouble with d and e
    i solved v(t) 6t^2 - 30t + 2 and a(t) = 12t - 30 (taking the derivative). for d and e it is looking for a range, right? heres what i got

    d) it is moving left when v(t) < 0, so between the zeros, or 0.676 < t < 4.93
    and it is moving right when v(t) > 0, or t < 0.676 and t>2.5

    e) p is speeding up when both v(t) and a(t) are in the same direction, and slowing down when they are opposite directions (signs)
    i got a(t) > 0 when t > 5/2, and a(t) < 0 when t < 5/2. so its speeding up when t > 4.93 and slowing when 0.676 < t < 2.5

    thanks in advance!
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 10, 2005 #2

    Fermat

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    For 1),

    y = x²
    y' = 2x

    From this you can get the slope of the tangent at A(1,1) , hence also the slope of the normal. You know the coordinates of a point and the slope of the line passing through that point, A, hence you can find the eqn of that line.

    Will that get you started ?
     
  4. Oct 10, 2005 #3

    Fermat

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    I agree with your answers for parts d) and e).
     
  5. Oct 10, 2005 #4
    yes that helped, thanks! i guess i just saw it and got confused :surprised

    so heres what i did, i sovled the slope of the normal line to be -1/2, which gave it the equation y = -1/2x + 3/2; i equated this to y=x^2 to solve for the intersection points, which i got to be (1, 1) and (-3/2, 9/4). i solved for the tangent line for x = -3/2 to get the slope to be -3, and i get the line equation to be y = -3x - 9/4. is this correct?

    btw, thanks for confirming the other one too! i just worked my way through it, i haven't seen a problem like that before
     
  6. Oct 10, 2005 #5

    Fermat

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    I get the same as yourself for all the results you gave :smile:
     
  7. Oct 10, 2005 #6
    thanks again, but i was thinking about part e of that second question, and when is it actually slowing down? would it be when they have the opposite sign? how does that work?
     
  8. Oct 10, 2005 #7

    Fermat

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    It is slowing down when velocity and acceleration have opposite signs.

    Consider the eqn of motion,

    vf = vi + at

    where vf is the final velocity after time t, and vi is some inital velocity.

    Let vf be +ve (positive) and let a be -ve (negative).
    If a is -ve, then vi +at will be continually getting smaller as t increases, hence vf (= vi + at) will be decreasing. So the speed is getting (numerically) smaller.

    Now, let vf be -ve and let a be +ve.
    If a is +ve, then vi +at will be continually getting larger as t increases, hence vf (= vi + at) will be increasing.
    But vf is -ve. And when -ve numbers increase they get numerically smaller. i.e the speed is getting smaller.
     
  9. Oct 10, 2005 #8
    so for this problem would it be

    speeding up when t>4.93 and when 0.676 < t < 2.5 ? and slowing down when they are opposite signs, at 2.5 < t < 4.93 and t < 0.676? i don't completely understand when it would be speeding or slowing in this problem...
     
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