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Math Help - Questions about procedure

  1. #1
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    Questions about procedure

    Hey all,

    I'm currently working on an assignment that has some "fairly" abstract ideas that need to be utilized, and we never went over these specific scenarios in class...

    I'm only looking for direction on how to go about solving...

    #1.) There is a line going through the origin that divides the region bounded by the line y = x - x^2 and the x-axis into two sections of equal area. Find the slope of this line.

    Now, I found the area of the region to be 1/6 and therefore, half of it should be equal to 1/12. I don't know how to determine where to cut the interval off into two pieces so that I can solve for the slope of the line...

    Any thoughts?

    #2.) How can I show that two regions have the same area without evaluating any integrals (the prof specifically told us that we needn't use any). The equations for the lines are:

    y = x / (1 + x^4) from 0 to 2
    y = 1 / (2(1 + x^2)) from 0 to 4

    I have no idea where to begin, other than using integrals to evaluate them...

    #3.) If R is the region bounded by y = sin (x^2) and the x-axis from x = 0 to x = pi^(1/2), let S be the solid of revolution if R is rotated about the y-axis. Find the Volume of S.

    I used Cylindrical shells to find the volume and ended up getting 0 for my final answer. Anyone else ever had that happen? (My calculations have been checked...)

    #4.) I need to find a function f that makes the following true:

    The integral from 0 to pi of x(f(sinx))dx = the integral from 0 to pi of x(sinx)

    I then need to evaluate the integral on the right side...any thoughts?

    Again, looking for direction, not mooching answers...

    Thanks,

    Ben
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Hey all,

    I'm currently working on an assignment that has some "fairly" abstract ideas that need to be utilized, and we never went over these specific scenarios in class...

    I'm only looking for direction on how to go about solving...

    #1.) There is a line going through the origin that divides the region bounded by the line y = x - x^2 and the x-axis into two sections of equal area. Find the slope of this line.

    Now, I found the area of the region to be 1/6 and therefore, half of it should be equal to 1/12. I don't know how to determine where to cut the interval off into two pieces so that I can solve for the slope of the line...

    Any thoughts?

    Cool question! This first thing I thought to do was find the equation of a tangent line to it and then find its antiderivative?? Just guessing. Trying to give you another option. I didn't try it out.

    #2.) How can I show that two regions have the same area without evaluating any integrals (the prof specifically told us that we needn't use any). The equations for the lines are:

    y = x / (1 + x^4) from 0 to 2
    y = 1 / (2(1 + x^2)) from 0 to 4

    I have no idea where to begin, other than using integrals to evaluate them...

    Maybe estimating rectangles and do a Reimann sum?

    #3.) If R is the region bounded by y = sin (x^2) and the x-axis from x = 0 to x = pi^(1/2), let S be the solid of revolution if R is rotated about the y-axis. Find the Volume of S.

    I used Cylindrical shells to find the volume and ended up getting 0 for my final answer. Anyone else ever had that happen? (My calculations have been checked...)

    \int\limits^{\sqrt\pi}_{0} 2\pi \cdot x sin(x^2) dx

    u = x^2

    du = 2xdx

    \frac{du}{2} = xdx

    2\pi \int\limits^{pi}_{0} sin(u) du

    \pi [-cos(u)]^{\pi}_{0}  =  \pi [-cos\pi - - cos0]

    =\pi[+1+1] = 2\pi

    Double check my calculations, but I think this is correct



    Sorry don't have time to finish, hope this helps!
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  3. #3
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    Thanks for the pointers, but I'm missing what you're proposing to do on the first question...

    Where should I take the tangent line and proceed from there?

    And on the second answer, using the rectangles and Reimann sums requires integration...

    -Ben
    Last edited by TyrsFromAbove37; February 12th 2009 at 08:49 PM.
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by TyrsFromAbove37 View Post
    Thanks for the pointers, but I'm missing what you're proposing to do on the first question...

    Where should I take the tangent line and proceed from there?

    -Ben
    You know when you are first learning about secant and tangent lines and how when one point on the secant line moves closer and closer along the curve towards the point of the tangent line? I was just thinking that using that method somehow you could find where it cut the area in half. I honestly have never had a question like this and I think it is probably one of those "make you think" kind of questions as opposed to getting the right answer. Know what I mean?

    I graphed it and also graphed y=x (which goes through the origin and it looks like y=x may be tangent to the curve at the origin. You also mentioned finding the area to be 1/6, so I also graphed \frac{1}{6}x and it looks promising, but I honestly don't know how to suggest you proceed. Good luck! Post the answer when you get it! I'm curious to see.
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  5. #5
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    I'll keep working on that one, but in the mean time, do you have any more thoughts on #'s 2 and 4?

    -Ben
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by TyrsFromAbove37 View Post
    I'll keep working on that one, but in the mean time, do you have any more thoughts on #'s 2 and 4?

    -Ben
    For #2, I'd say definitely use estimating rectangles and add up your areas that way. This is how we learned about integrals before we knew how to do integrals. It's the same concept, but the answer won't be as exact as if you used an integral. That is really the only way I would approach that one.

    This website has a good explanation:
    Pauls Online Notes : Calculus I - Area Problem

    Gotta go to bed, I'll try to look at #4 tomorrow. Good luck with these!
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  7. #7
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    Finally got the answer to the harder one...

    It involved setting the equation of the parabola equal to y=ax and solving for a. This was then used as the point where the line would intersect the graph on the opposite side...

    Thanks for the help though.

    -Ben
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