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Help with Couple of Calculus Problems

  1. Apr 26, 2004 #1
    Hello, I am completely stumped by some of these problems, and have no where else to get help, can anyone help me with these problems? Any help would be greatly appreciated. Thank you very much.

    1. The function v(t)=12t(squared) - 16t is the velocity in m/sec of a particle moving along the x-axis, where t is measured in seconds. Use analytic methods to find the particles displacement for 0<(greater than or equal to)t<(less than or equal to)5. (round to nearest 10m). (Note: just t is squared in the equation, and it's 0 to 5 including both 0 and 5) (I got 120m but it doesn't seem right)

    2. A certain spring obeys Hooke's Law and requires a force of 10N to stretch it 8cm beyond its natural length. How much work would be done in stretching it to 12cm beyond its natural length? (Note: I got 90N*m but I'm not sure if it's right)

    3. Find the area of the region enclosed by y=sin2x and y=cosx for -(negative)PI/2 greater than or equal to X less than or equal to PI/6. (I got -4.5 which seems completely wrong :frown: )

    These next two I had no idea how to do

    4. The base of a solid is the region between the line y=4 and the parabola y=x(squared). THe cross sections perpendicular to the x-axis are semi-circles. Find the volume of the solid.

    5. A region is bounded by the lines: y=(square root of)X , y= X-2, and y=0. Find the volume of the solid generated by rotating this region about the x-axis.

    THanks again for any help, sorry if it is too much.
  2. jcsd
  3. Apr 26, 2004 #2
    Why don't you show your work so we can see where you're having trouble?

  4. Apr 27, 2004 #3
    For first question:
    by integrating your velocity function, you'll get

    4t^3 - 8t^2, with limits on t: 0 to 5 seconds

    applying the limits, you'll get 300m which is surely not what you were getting

    For your second question:
    since, F = Kx
    find K using F1 = 10 KN and x1 = 0.08m

    Now using this K and x2=0.12 m find F2. Your required work done is '(1/2) F2 x2'

    note: F2 or x2 doesn't mean a 'square'

  5. Apr 27, 2004 #4
    use 0<=t<=5 instead of what you've written, this will make your question more readible

  6. Apr 27, 2004 #5
    So, your range of 'X' for the third question is
    -pi/2 <= X <= pi/6

    Now draw the two curves on a graph so that you have a clear idea of the shape of the two and hence the limits (I hope you haven't missed this elementry step for your problem) OK!
    Next thing you must do is to calculate the areas in the positive and negative regions seperately.
    Now take a differential strip parallel to y with its upper and lower bounds defined by the two functions. the area of the strip is defined by (y2-y1)dx. Replacing y1 and y2 with the respective functions and knowing your limits on x, you can now evaluate the areas.

  7. Apr 27, 2004 #6
    Your statement of fourth question is not too clear, let me guess!

    You have a half paraboid with its axis parallel to y OK!, limits on y are 0 to 4 and limits on x are defined by the funciton y = x^2
    if this is what you mean then you can you can use the Pipus theorem to solve this problem. take half of this parabola then calculate its area and centroid, now multiply this area with the distance its centroid covers in order to complete the paraboloid by revolving around y axis. this distance should be 'pi(a)' where a is the centroid about y.

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