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Homework Help: Several Derivative Problems

  1. Sep 30, 2008 #1
    Hello there.

    I actually need help with several problems

    1) If f(x)=4x^2-4 x+3, then f'(5) = 36
    Use this to find the equation of the tangent line to the parabola y=4x^2-4x+3 at the point (5,83). The equation of this tangent line can be written in the form y = mx+b where
    m= ?
    b= ?

    2)For what values of x does the graph of f(x)=6x^3-9x^2-216x+18 have a horizontal tangent?

    3)f(x) = x^{8}h(x)
    h(-1) = 5
    h'(-1) = 8

    I need to calculate f'(-1)

    4) A particle moves along a straight line with equation of motion s=t^{3}-3t^{2} Find the value of t (other than 0 ) at which the acceleration is equal to zero.

    5) A particle moves along a straight line and its position at time t is given by s(t)=2t^3-15t^2+24t where s is measured in feet and t in seconds.
    Find the velocity (in ft/sec) of the particle at time t=0

    The particle stops moving (i.e. is in a rest) twice, once when t=A and again when t=B where A < B
    What is the position of the particle at time 10?
    What is the TOTAL distance the particle travels between time 0 and time 10?

    Thank you,
  2. jcsd
  3. Sep 30, 2008 #2


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    Science Advisor
    Homework Helper

    Hi Neil! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Sep 30, 2008 #3
    I did problems 1-4. I was able to get help from my roommate and we worked it out. For problem 5 i was able to get the first 4 questions. The 5th part of the 5th question is: what is the TOTAL distance the particle travels between time 0 and time 10? i tried adding all the values of t(0) to t(10) and its not correct. I have no idea what to do.
  5. Sep 30, 2008 #4


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    Science Advisor

    Well, if the particle were always moving to the right, you would just subtract the starting position from the final position, right?

    And, if the particle were always moving to the left, you would subtract the final position from the starting position (because "total distance" is always positive).

    So the problem really is to separate "moving to the right" (positive velocity) from "moving to the left (negative velocity). Do you see that the change MUST occur where the velocity is 0? For what values of t is v= s'(t) equal to 0?
  6. Sep 30, 2008 #5
    1 and 4
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