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How do u find gradients again?

  1. May 14, 2005 #1
    i need to find the gradient of the line y-3x=2 how do i do it again?
     
  2. jcsd
  3. May 14, 2005 #2
    Put it in the form y = mx + c and m is the gradient.
     
  4. May 14, 2005 #3

    dextercioby

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    What's the connection between the slope of the line & its gradient...?

    Daniel.
     
  5. May 15, 2005 #4
    hey,

    {y-displacement}/{x-displacement} = gradient displacement is how much y changes and x changes from two points.

    The gradient IS the slope of the line


    Regards,

    Ben
     
    Last edited by a moderator: May 15, 2005
  6. May 15, 2005 #5
    Rise over Run - is how I remember it.
     
  7. May 16, 2005 #6
    Wait, is this gradient the same gradient as in gradient of a vector field or does it mean gradient like slope of the line?
     
  8. May 16, 2005 #7
    Slope of line - y=mx +c

    Regards,

    Ben
     
  9. May 16, 2005 #8

    HallsofIvy

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    The original question was about y- 3x= 2, a real valued linear function of a single real variable. I'm afraid all that talk about "vectors" and even "derivatives" will just confuse the original poster.

    Slessi: solve for y. In this example, solving for y gives y= 3x+ 2 so the gradient (slope) is 3. (Almost) any linear equation can be solved for y in the form y= mx+ b and the gradient is the number m.

    (I say "(Almost)" because a vertical straight line, like x= 1, cannot be put in that form: it has NO gradient.
     
  10. May 16, 2005 #9
    I was just staring at this statement whilst drinking tea and began to think, there is a gradient because gradient is [y-displacement]/[x-displacement] and in this case, 0/change in x. Change in x is always a non-zero integer value and therefore the gradient is 0/non-zero integer = 0. Looking at the graph, this fits, the gradient looks to be 0.

    However, if the graph is of y=1 ,then the equation becomes: [change in y]/0. Change in y is always a non-zero integer value and therefore the gradient is non-zero integer/0 which is mathematically undefined. However, if one looks at the graph in a topolgical sense (Pemrose is my source on the idea of topolgy), then the gradient looks to be -infinity or +infinity. And this leads one to consider what result dividing by 0 gives.

    Anyway, those are just my thoughts,


    Regards,

    Ben
     
  11. May 17, 2005 #10

    HallsofIvy

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    Yes, that's exactly right.
    No, that's exactly wrong. The equation in question is x= 1. x is always 1: x doesn't change, y can be anything: the equation is change in y/0.
    Where in the world did you get that idea? If x= 1 the change is 0! x= 1 means exactly that: x is always 1, not the "change" in x!
    Same error: if y= 1 then y does not chage: the slope is 0/change in x= 0.
     
  12. May 18, 2005 #11
    Hi, I commented on x=1 instead of y=1 and visa versa - please reread the post in its now corrected form (below) and give your throughts.

    I was just staring at this statement whilst drinking tea and began to think, there is a gradient because gradient is [y-displacement]/[x-displacement] and in the case of y=1, 0/change in x. Change in x is always a non-zero integer value and therefore the gradient is 0/non-zero integer = 0. Looking at the graph, this fits, the gradient looks to be 0.

    However, if the graph is of x=1 ,then the equation becomes: [change in y]/0. Change in y is always a non-zero integer value and therefore the gradient is non-zero integer/0 which is mathematically undefined. However, if one looks at the graph in a topolgical sense (Pemrose is my source on the idea of topolgy), then the gradient looks to be -infinity or +infinity. And this leads one to consider what result dividing by 0 gives.

    -Ben
     
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