Page 1 of 2 12 LastLast
Results 1 to 10 of 13

Thread: Find the line that halves the area of a cubic function: y=0.00111111x^3-0.1x^2+120

  1. #1

    Find the line that halves the area of a cubic function: y=0.00111111x^3-0.1x^2+120

    I have to find a line that splits the area under the graph y=0.00111111x^3-0.1x^2+120 into two equal halves. The line also must pass through x=-30. The graph is shown by the attachment below, in which the line runs from A (30, 0) to B (unknown).

    Cubic Graph.png

    I have integrated the function to find that the area under the graph is equal to 6075. (sorry I am unsure of how to insert all of the symbols and cannot show working). I understand that the line AB separates the graph into two halves, both with an area of 3037.5. However, I am unsure of how to do this step. I have looked at other resources but they only explain how to find a horizontal or vertical line without intersecting a specific point. If someone could assist in explaining this it would be greatly appreciated. Thanks.

  2. #2
    Elite Member stapel's Avatar
    Join Date
    Feb 2004
    Posts
    15,850

    Cool

    Quote Originally Posted by JacDevo View Post
    I have to find a line that splits the area under the graph y=0.00111111x^3-0.1x^2+120 into two equal halves.
    Shame on whoever wrote this exercise, expecting you to use "0.00111111" rather than some nice neat exact fraction. Oh, well.

    Quote Originally Posted by JacDevo View Post
    The line also must pass through x=-30. The graph is shown by the attachment below, in which the line runs from A (30, 0) to B (unknown).

    Cubic Graph.png
    Does this mean that the point "B" is on the graph of y? (Is the blue-ish line the graph of y?) And that the segment AB is meant not to be vertical or horizontal, but diagonal-ish?

    Also, on the above graph, where are the x- and y-axes? Where is the origin?

    Quote Originally Posted by JacDevo View Post
    I have integrated the function to find that the area under the graph is equal to 6075.
    "The area under the graph" will be infinite, unless some lower bound (such as, say, the x-axis) is specified. Please provide the rest of the information for this exercise.

    Quote Originally Posted by JacDevo View Post
    (sorry I am unsure of how to insert all of the symbols and cannot show working).
    To learn how to type math as text, please try here. Then please show things such as your bounds of integration, and how you determined them.

    Please be complete. Thank you!

  3. #3
    Quote Originally Posted by stapel View Post
    Shame on whoever wrote this exercise, expecting you to use "0.00111111" rather than some nice neat exact fraction.
    The graph that I attached was all that was given, so I had to find the equation from its information, sorry.

    Quote Originally Posted by stapel View Post
    Does this mean that the point "B" is on the graph of y? (Is the blue-ish line the graph of y?) And that the segment AB is meant not to be vertical or horizontal, but diagonal-ish?
    Yes, the line AB passes through the graph of y at point A and B. Also AB is meant to be diagonal if I am interpreting the question correctly.

    Quote Originally Posted by stapel View Post
    Also, on the above graph, where are the x- and y-axes? Where is the origin?
    I am assuming that the y-axis is the middle vertical line (120 metres) and that the x-axis is the horizontal line labelled 'road', since these are not specified in the question. The origin is therefore the point at which these two lines meet.

    Quote Originally Posted by stapel View Post
    "The area under the graph" will be infinite, unless some lower bound (such as, say, the x-axis) is specified. Please provide the rest of the information for this exercise.
    I should have said this but yes, it is bound by the x-axis.

    Quote Originally Posted by stapel View Post
    To learn how to type math as text, please try here. Then please show things such as your bounds of integration, and how you determined them.
    I will attach the question so that my working makes more sense. "A farmer owns a paddock that is bounded by part of a river and a straight road running east-west as in the diagram below. The part of the river forming the paddock can be regarded as cubic (ax3+bx2+cx+d) in shape. Sam wishes to divide the paddock into two paddocks of equal area by running a straight fence from point A (the western junction of the river and the road) to point B on the river. The diagram is not to scale. All measurements are in metres." The diagram is the same as I attached in my initial post.

    From this information, it is evident that the bounds of integration are from x=-30 to x=60.

    From this, I found the antiderivative of the initial cubic equation, which = x^4/3600-x^3/30+120x

    I then substituted x=60 into the antiderivative and subtracted it by x=-30. This gave the area under the graph bound by the x-axis, which equals 3037.5. This is as far as I have reached, as I do not know where to start in finding the line AB that halves the area of the cubic. Thank you for your help .
    Last edited by stapel; 04-04-2017 at 02:41 PM. Reason: Fixing quotes.

  4. #4
    Elite Member stapel's Avatar
    Join Date
    Feb 2004
    Posts
    15,850
    Quote Originally Posted by JacDevo View Post
    The graph that I attached was all that was given...I will attach the question...:



    A farmer owns a paddock that is bounded by part of a river and a straight road running east-west as in the diagram below. The part of the river forming the paddock can be regarded as cubic (ax3+bx2+cx+d) in shape. Sam wishes to divide the paddock into two paddocks of equal area by running a straight fence from point A (the western junction of the river and the road) to point B on the river. The diagram is not to scale. All measurements are in metres.
    Ah, so they gave you lots of additional information. Excellent!

    Quote Originally Posted by JacDevo View Post
    From this information, it is evident that the bounds of integration are from x=-30 to x=60.
    Assuming that the horizontal line is the x-axis, the vertical line is the y-axis, and that the intersection displayed is the origin, then, yes, "it is evident". But you'll want to specify all of these assumptions in your hand-in work.

    Quote Originally Posted by JacDevo View Post
    From this, I found the antiderivative of the initial cubic equation, which = x^4/3600-x^3/30+120x
    How did you obtain this equation, given that you have been provided with only two data points in the exercise, which is at least two points short of what is necessary for finding a cubic model? Please show all of your work.

    Thank you!

  5. #5
    Quote Originally Posted by stapel View Post
    How did you obtain this equation, given that you have been provided with only two data points in the exercise, which is at least two points short of what is necessary for finding a cubic model? Please show all of your work.
    As I am concerned with finding the line for the moment, I plugged the numbers into a cubic function calculator to find a quick result. That function came out after using the coordinates (-30,0), (0,120) and (60,0), which were found assuming that the x and y-axis did meet at the origin. They have provided little information in the question so it would be great if you could help in the next stage by assisting in finding the line. Thankyou .
    Last edited by stapel; 04-06-2017 at 01:40 PM. Reason: Fixing quotes.

  6. #6
    Elite Member stapel's Avatar
    Join Date
    Feb 2004
    Posts
    15,850

    Cool

    Quote Originally Posted by JacDevo View Post
    As I am concerned with finding the line for the moment, I plugged the numbers into a cubic function calculator to find a quick result. That function came out after using the coordinates (-30,0), (0,120) and (60,0)...
    Unfortunately, even assuming that the river does touch the x-axis at x = 60, three points is sufficient only to determine a quadratic. We can have very little confidence in any cubic regression. And all further computations, being dependent upon that cubic, will be suspect.

    Under that proviso, however, and since you feel you have an accurate determination of the entire area, one need only create an expression for the diagonal-ish line AB to proceed. The line will be straight, with one point being (-30, 0) and the other point being B = (xB, (1/900)xB^3 - (1/10)xB^2 + 120). Then you'll integrate between the straight line and the x-axis from x = -30 to x = xB, and integrate between the cubic and the x-axis from x = xB to x = 60. Set the result equal to half of the total area, and solve for the value of xB.

  7. #7
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,626
    I have a different interpretation of how to proceed. We are still lacking information, which I must guess from the original diagram, namely that there is a local maximum 30 meters from point A. On that assumption, we can proceed as follows.

    [tex]f(x) = ax^3 + bx^2 + cx + d \implies f'(x) = 3ax^2 + 2bx + c.[/tex]

    Instead of messing around with minus signs, I put the origin at point A.

    This gives four equations to find the cubic exactly.

    [tex]f(0) = 0 \implies a(0^3) + b(0^2) + c(0) + d = 0.[/tex]

    [tex]f(30) = 120 \implies a(30^3) + b(30^2) + c(30) + d = 120.[/tex]

    [tex]f'(30) = 0 \implies 3a(30^2) + 2b(30) + c = 0.[/tex]

    [tex]f(90) = 0 \implies a(90^3) + b(90^2) + 90c + d = 0.[/tex]

    Solve that system of four linear equations with four unknowns to get the cubic exactly.
    Last edited by JeffM; 04-06-2017 at 08:38 PM.

  8. #8
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,626
    As for your line, having chosen the origin at point A and letting the unknown co-ordinates of B be p and f(p)

    the equation of the line is just:

    [tex]g(x) = x * \dfrac{f(p)}{p}.[/tex]

    So the only remaining question is how do you find p?

  9. #9
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,626
    I solved the system of equations. The cubic is not what is given in the OP. In fact, it is not even close according to my computations.

  10. #10
    Quote Originally Posted by JeffM View Post
    I solved the system of equations. The cubic is not what is given in the OP. In fact, it is not even close according to my computations.
    What did you find?

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •