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Math Help - Gradient problems URGENT

  1. #1
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    Exclamation Gradient problems URGENT

    Find the coordinates of the points on the curve y=x^3 where the gradient is 12

    Please help URGENT
    P.S Can people explain this in detail as I am horrible at maths
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  2. #2
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    Re: Gradient problems URGENT

    Gradient means inclination and inclination points to something like 'Slope'
    Now, what is slope ?
    It is simply  \frac{dy}{dx} for any function y=f(x)
    so here you have f(x)=x3
     \frac{dy}{dx} = 3x^2
     x^2 = 4
     x=+4, -4
    Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
    f(4)=64
    f(-4)=-64
    Thanks from terone71
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    Re: Gradient problems URGENT

    Quote Originally Posted by pratique21 View Post
    Gradient means inclination and inclination points to something like 'Slope'
    Now, what is slope ?
    It is simply  \frac{dy}{dx} for any function y=f(x)
    so here you have f(x)=x3
     \frac{dy}{dx} = 3x^2
     x^2 = 4
     x=+4, -4
    Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
    f(4)=64
    f(-4)=-64
    Thanks. This some has been bothering me for some time!!!!
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    Question Re: Gradient problems URGENT

    Quote Originally Posted by pratique21 View Post
    Gradient means inclination and inclination points to something like 'Slope'
    Now, what is slope ?
    It is simply  \frac{dy}{dx} for any function y=f(x)
    so here you have f(x)=x3
     \frac{dy}{dx} = 3x^2
     x^2 = 4
     x=+4, -4
    Now you have the value of x. To find the complete set (x,y) just plug the values of x in to f(x)
    f(4)=64
    f(-4)=-64
    One more question! How did you get x^2 = 4 in the first place? After getting dy/dx = 3x^2 . You then lost me!
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    Re: Gradient problems URGENT

    i put the value of dy/dx that is given as 12
    so cancelling 3 gives me x^2 = 4
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  6. #6
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    Re: Gradient problems URGENT

    But then after that how did you still find x? If you cancel 3 then its x^2=4
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    Re: Gradient problems URGENT

    okay, now take the entire thing to one side (i.e subtract 4 from both sides)
    so x^2 = 4
    become: x^2 - 4 =0
    now use a^2 - b^2 = (a+b) \cross (a-b)
    so you have:- (x-2) \cross (x+2)=0
    this is true when either  (x-2)= 0 or  (x+2)=0 or both are zero.
    So x=  \pm 2
    Last edited by pratique21; July 24th 2012 at 09:16 PM.
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  8. #8
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    Re: Gradient problems URGENT

    Thanks dood
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