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Homework Help: Equation of a curve homework

  1. Apr 24, 2006 #1
    i really need some help with this question, my assignment is due in tommorow, i don't really understand the question - please help!

    A curve has a gradient (2x-3)(3x+2) at the point (x,y) and passes through the point (2,-3). Find the equation of the curve. illustrate the answer with a sketch.
     
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  3. Apr 24, 2006 #2

    Hootenanny

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    Well you know that;

    [tex]\frac{dy}{dx} = (2x-3)(3x+2)[/tex]

    How do you suppose you can find [itex]y=...[/itex]?

    ~H
     
  4. Apr 24, 2006 #3
    i think i have to integrate the gradient but not sure how to.

    how does this help me find the equation of the curve?
     
  5. Apr 24, 2006 #4

    Hootenanny

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    When you differentiate a curve, you obtain the gradient. Therefore, if you integrate the gradient you will obtain the curve.

    Here's a simple example.
    If the gradient of a curve is given as [itex]2x[/tex] then;

    [tex]\frac{dy}{dx} = 2x[/tex]

    [tex]y = \int \frac{dy}{dx} dx = \int 2x dx[/tex]

    [tex]y = x^2 + C[/tex]

    Therefore, the equation of the curve is [itex]y=x^2 + C[/itex]. In your case you are given a point, which will allow you to find the constant of integration.

    Do you follow?
    ~H
     
  6. Apr 24, 2006 #5
    so if i differentiate i get,

    6x^2-9x-6

    and then integrate,

    2x^3-9x^2/2-6x ?
     
  7. Apr 24, 2006 #6

    Hootenanny

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    You don't need to differentiate, you are given the gradient of the curve ([itex]\frac{dy}{dx}[/itex]), so all you need to do is integrate.

    ~H
     
  8. Apr 24, 2006 #7
    so i get,

    (x^2 - 3x)(3x^2/2 +2x) ?
     
  9. Apr 24, 2006 #8
    If you distribute before you integrte you get 6X^2-5x-6
    if you integrate that you get 2x^3-(5/2)x^2-6x+c
    Plug in you x value to find c.
     
  10. Apr 25, 2006 #9

    Hootenanny

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    You cannot integrate the brackets seperately as you have done. I would recommend expanding the brackets as tangents suggests, then integrating. As I said before, once you have integrated, you can use the given point to find the constant of integration.

    ~H
     
  11. May 1, 2006 #10
    if i expand the brackets from (2x-3)(3x+2) i get,

    6x^2 + 4x - 9x - 6

    and then integrate,

    y = 2x^3 + 2x^2 - 9x^2/2 - 6x +c

    do i then just plug in the x and y values to find c by rearranging,

    y - c = 2x^3 + 2x^2 - 9x^2/2 - 6x ?
     
  12. May 1, 2006 #11

    Hootenanny

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    Yes, thats correct. It may have been simpler to collect the terms before integrating thus;

    [tex]6x^2 + 4x - 9x - 6 \equiv 6x^2 - 5x -6[/tex]

    But your working is correct, as long as you collect you terms before presenting you final answer that is fine.

    ~H
     
    Last edited: May 1, 2006
  13. May 1, 2006 #12
    do you mean 6x^2 - 5x -6

    the result of this using x = 2 gives 8
    does this mean,

    -3 - c = 8

    adding 3,

    -c = 11,

    c = -11 ????

    sorry to keep on but i need to get this correct!
     
  14. May 1, 2006 #13

    Hootenanny

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    Yes, I've corrected my post above.

    No, you have to plug you numbers (x = 2, y = -3) into your intergrated equation. In my previous post I was simply pointing out that it would have been easier to integrate the equation if you had simplified first.

    ~H
     
  15. May 1, 2006 #14

    VietDao29

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    Okay, I think I'll make it a bit clearer for you.
    A curve has a gradient (2x-3)(3x+2) at the point (x,y) simply means that the slope of the tangent line to the function at the point (x, y) is (2x-3)(3x+2). Or in other words, the derivative of that function is (2x-3)(3x+2). Can you get this?
    So what you should do is to find the function by knowing its derivative, and 1 point it passes through.
    You should first find the functons whose derivative is (2x-3)(3x+2), you can do this by integrating the derivative. And only 1 of those functions will pass the point given, with this information, you'll be able to obtain the desired function.
    ---------------
    Example:
    Find the curve whose derivative is 4x3 + 5, and passes through the point (0, 1).
    ---------------
    First, you integrate the derivative:
    [tex]\int (4x ^ 3 + 5) dx = x ^ 4 + 5x + C[/tex]
    For every value of C, you'll have a function, and those functions that have the form x4 + 5x + C will have the derivative of 4x3 + 5 (they are parallel to each other, pick 2 distinct C's, graph it, and see what I mean, e.g x4 + 5x, and x4 + 5x + 4).
    And 1 and only 1 of them will pass through (0, 1)
    So:
    1 = 04 + 5.0 + C
    <=> C = 1.
    So the curve is y = x4 + 5x + 1.
    Can you get this? :)
     
  16. May 1, 2006 #15
    would my final equation be,

    y = 2x^2 - 5x^2/2 - 6x + 3 ??

    i also need to sketch the graph, to find the points where the line intercects the axis do i use,

    y = 0 for x axis
    x = 0 for y axis
     
  17. May 1, 2006 #16

    Hootenanny

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    Almost, I think it is just a typo, but the first term should be 2x3.

    Yes, you are correct.

    ~H

    P.s. I edit my last post because the 2x2through me a bit.
     
  18. May 1, 2006 #17
    sorry it was supposed to be

    2x^3-5x^2/2-6x+c

    so that gives me,

    c = -3-16+10+12 = 3
     
  19. May 1, 2006 #18
    sorry that was a bit of a marathon, thank you for your help
     
  20. May 1, 2006 #19

    Hootenanny

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    Yeah, you've got it. No problem.

    ~H
     
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