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Math Help - FTC & Motion along a line.

  1. #1
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    FTC & Motion along a line.

    I was doing some homework and came across a couple of problems I was unsure of. The first set is FTC and the latter is motion along a line. These are quite a few problems and most of them are already solved. They don't have to be checked; I just need help on the blank ones.

    FTC:
    Suppose h(x)= integral of f(t) [0,x], where f(t) is shown in the graph below.
    The image is attached.
    1. Find the critical points of h(x).
    -At first, I had 5 and 11 because those are the points that f(t) =0. I changed the critical point to 0, though, because the integral with the bounds [0,0] would equal 0.

    2. Identify the values of x when h(x) has a relative minimum and/or maximum.
    -This is where I'm having the most trouble. I THOUGHT I had an idea of what to do, and found where the values of f(x) increased and decreased....yeah, I'm very confused on this problem.

    3. Find the absolute minimum and maximum values of h(x) on [0,12].
    -The absolute min I got was h(12)=-1 and the absolute max I got was h(0)=-4.

    Motion along a line:
    From its initial position of s(0)=-3, a particle moves along a linear path with a known velocity of v(t)=t^2-3t-4, where v(t) is measured in meters per second and t is greater than or equal to 0.

    1. Find the time when the particle has an acceleration of 9 meters per squared second.
    -I solved this and got s(6)=147

    2. Find a function to represent the position of the particle at any time t.
    -s(t)=s(0) + integral of u^2-3u-4 du [0,t]

    3. Find the position of the particle when its velocity is zero.
    -I was unsure so I just solved v(0)=-4.

    4. Determine when the velocity of the particle is positive and when it is negative.
    -This is the problem I had no clue on.

    5. Find the distance traveled by the particle on the interval 0 less than/equal to t less than/equal to 5.
    -I solved it and got 21.50.

    Thanks!
    Attached Thumbnails Attached Thumbnails FTC & Motion along a line.-d.jpg  
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  2. #2
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    I changed the critical point to 0, though, because the integral with the bounds [0,0] would equal 0.
    What does the fact that h(0)= 0 have to do with being a critical point?

    The derivative of h(x) is f(x). The critical points of h are where f= 0, not where h= 0. Note that for x< 5, f(x)< 0 so h(x) is decreasing while for x> 5, f(x)> 0 so h(x) is increasing. There is a relative minimum of h(x) at x= 5. The opposite happens at x= 11. f(x) goes from positive to negative, so h(x) goes from increasing to decreasing. There is a relative maximum of h(x) at x= 11.

    h(12) is NOT -1 and h(0) is NOT -4. Those are the values of f(x), not h(x)!

    It's not that difficult to find h(x) explicitely. For x= 0 to 3, f(x)= -4 so h(x)= -4x. For x= 3 to 7, f(x)= 2x- 10 so [tex]h(x)= x^2- 10x+ 21 (the constant, 21, is so that -4(3)= -12= (3)^2- 10(3)+ 21). For x= 7 to 12, f(x)= -x+ 11 so -(1/2)x^2+ 11x- 52.5 (the constant, -52.5, is so that 7^2- 10(7)+ 21= 0= -(1/2)(7^2)+ 11(7)- 52.5). Calculate the values at the critical points, x= 5 and x= 11, and at the endpoints x=0 and x= 12 to see what the absolute maximum and minimum are.


    1. Find the time when the particle has an acceleration of 9 meters per squared second.
    -I solved this and got s(6)=147
    That doesn't even make sense. The problem asked for a time, not a value of s. Do you mean that your answer is 6 seconds?

    2. Find a function to represent the position of the particle at any time t.
    -s(t)=s(0) + integral of u^2-3u-4 du [0,t]
    Okay, what is that integral?

    3. Find the position of the particle when its velocity is zero.
    -I was unsure so I just solved v(0)=-4.
    That says that when t= 0, v= -4 which has nothing to do with "when its velocity is zero." You want to solve v(t)= 0 for t, then put that value of t into the position function. Since the velocity function is quadratic, there will probably be two solutions.

    4. Determine when the velocity of the particle is positive and when it is negative.
    -This is the problem I had no clue on.
    Surely you know what "positive" and "negative" mean! Solve the inequalities v(t)> 0 and v(t)< 0. Hint: v(t) can change from positive to negative only where v(t)= 0- look at (3) again.
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