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    So I'm being an idiot clearly but can't see what I'm doing wrong.

    Question 7 part 2. dx/dy is  ln(2y+3)+\frac{2y+8}{2y+3}.

    So then I said that at A dx/dy =0 but then i get stuck.
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    (Original post by Music99)
    So I'm being an idiot clearly but can't see what I'm doing wrong.

    Question 7 part 2. dx/dy is  ln(2y+3)+\frac{2y+8}{2y+3}.

    So then I said that at A dx/dy =0 but then i get stuck.
    I think you're thinking A is a turning point but the question doesn't say that.

    You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
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    (Original post by notnek)
    I think you're thinking A is a turning point but the question doesn't say that.

    You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
    Ohhhh! Facepalm!!
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    (Original post by notnek)
    I think you're thinking A is a turning point but the question doesn't say that.

    You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
    So I subbed y=0 into dx/dy and you get ln(3) +8/3 but that doesn't give you the correct gradient.
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    (Original post by Music99)
    So I subbed y=0 into dx/dy and you get ln(3) +8/3 but that doesn't give you the correct gradient.
    dy/dx gives you the gradient, not dx/dy..
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    (Original post by Music99)
    So I'm being an idiot clearly but can't see what I'm doing wrong.

    Question 7 part 2. dx/dy is  ln(2y+3)+\frac{2y+8}{2y+3}.

    So then I said that at A dx/dy =0 but then i get stuck.
    can you send me that paper please?
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    (Original post by notnek)
    dy/dx gives you the gradient, not dx/dy..
    I thought that because at A y=0 we would use dx/dy and at B use dy/dx ...
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    (Original post by Music99)
    I thought that because at A y=0 we would use dx/dy and at B use dy/dx ...
    The gradient is always dy/dx (if the y-axis is vertical). Substituting the x-coordinate into dy/dx (x) or the y-coordinate into dy/dx (y) will give you the same answer.
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    (Original post by notnek)
    The gradient is always dy/dx (if the y-axis is vertical). Substituting the x-coordinate into dy/dx (x) or the y-coordinate into dy/dx (y) will give you the same answer.
    Ah okay, so then other then finding the rate at which x changes w.r.t y is there a use for dx/dy
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    (Original post by Music99)
    Ah okay, so then other then finding the rate at which x changes w.r.t y is there a use for dx/dy
    That sounds like a good use to me. What use that dy/dx has would you like dx/dy to have?

    Maybe you want it to have a name like 'gradient function'? How about, 'reciprocal gradient function'?
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    (Original post by notnek)
    That sounds like a good use to me. What use that dy/dx has would you like dx/dy to have?

    Maybe you want it to have a name like 'gradient function'? How about, 'reciprocal gradient function'?
    Haha, :P, going to mark the paper soon, If I get things wrong mind If I quote you?
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    (Original post by Music99)
    Haha, :P, going to mark the paper soon, If I get things wrong mind If I quote you?
    Sure.
 
 
 
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