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Math Help - Implicit differentiation - not defined?

  1. #1
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    Implicit differentiation - not defined?

    I have the problem:

    Find the linear approximation about (-1,1) to the implicit function given by the relation: 2xy-x8+y2=-2

    My problem is that trying implicit differentiation, I always get a slope which is not defined, because the denumerator is always (x+y) and therefore 0.

    Did I do an error doing my calculation? If my calculation was right, can I answer that the tangent line is not defined at (-1,1)?
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  2. #2
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    Re: Implicit differentiation - not defined?

    No, there is no error. You can, however, say that the "linear approximation" is the vertical line x= -1.

    Another way of looking at it is to treat this as an implicitly defining x as a function of y. Then, differentiating both sides with respect to y you would get x'= 0. That gives x= 0(y-1)- 1 or, again, x= -1.
    Last edited by HallsofIvy; June 4th 2012 at 01:45 PM.
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    Re: Implicit differentiation - not defined?

    Hm, I always used the formula:

    x-x0= x'/y' (x-x0)

    In that case for this example I get:
    y-1=0*(x+1)
    so
    y=1

    what was my mistake? Did I use the wrong formula in order to get the tangent line?
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  4. #4
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    Re: Implicit differentiation - not defined?

    the tangent line is x = -1
    Attached Thumbnails Attached Thumbnails Implicit differentiation - not defined?-implicit-graph.jpg  
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    Re: Implicit differentiation - not defined?

    Quote Originally Posted by infernalmich View Post
    Hm, I always used the formula:

    x-x0= x'/y' (x-x0)
    What do x' and y' mean here? (And surely, you don't mean x- x_0 on both sides?)

    In that case for this example I get:
    y-1=0*(x+1)
    so
    y=1

    what was my mistake? Did I use the wrong formula in order to get the tangent line?
    I can't answer that until I know what you mean by x' and y'. Normally, I would interpret them as derivatives, but with respect to what variables? If x and y were given as functions of some other variable, and x' and y' were the derivatives with respect to that third variable, the tangent line would be given by y- y_0= (y'/x')(x- x_0) where the slope is the reciprocal of what you give.
    Last edited by HallsofIvy; June 5th 2012 at 05:21 PM.
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    Re: Implicit differentiation - not defined?

    The function in my example is given implicitly:
    2xy-x8+y2=-2
    find the tangent line at the point: (-1,1) - I refer to this points as x0 and y0

    hence y'(x)=- (af/ax) / (af/ay)
    so y=y0-y'(x) (x-x0)

    In this case y=1-0
    so y=1

    ____

    Trying to do the same with:
    x=x0-x'(y)(y-y0)
    x=-1- 0*(y-1)

    now my stupid question:
    Why is x=-1 the tangent line and not y=1?
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  7. #7
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    Re: Implicit differentiation - not defined?

    Quote Originally Posted by infernalmich View Post
    The function in my example is given implicitly:
    2xy-x8+y2=-2
    find the tangent line at the point: (-1,1) - I refer to this points as x0 and y0

    now my stupid question:
    Why is x=-1 the tangent line and not y=1?
    first, 2xy-x^8+y^2=-2 is not a function ... why?

    second, all points on a vertical line have the same x-value, hence, the equation of the vertical line passing thru the point (-1,1) is x = -1

    fyi, all points on a horizontal line have the same y-value , hence the equation of a horizontal line would be y = a constant.
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    Re: Implicit differentiation - not defined?

    Now I try with a second one:

    Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:
    2x3y3-3x2y3=5

    af/ax=-5
    af/ay=10

    than I can calculate:
    y=y0-(-2)(x-x0)
    y=-1+2*(x-1)
    y=2x+2
    0,5y=x

    and for x:
    x=1-(-0,5)(y--1))
    x=0,5y

    am I right?
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    Re: Implicit differentiation - not defined?

    Quote Originally Posted by infernalmich View Post
    Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:
    2x3y3-3x2y3=5
    the point (1,-1) is not on the curve 2x^3y^3 - 3x^2y^3 = 5

    did you copy the problem correctly?
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    Re: Implicit differentiation - not defined?

    no I didn't, sorry.

    The correct problem is:
    2x3y2-3x2y3=5
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    Re: Implicit differentiation - not defined?

    Quote Originally Posted by infernalmich View Post
    no I didn't, sorry.

    The correct problem is:
    2x3y2-3x2y3=5
    I get ...

    \color{red}{y + 1 = \frac{12}{13}(x - 1)}
    Attached Thumbnails Attached Thumbnails Implicit differentiation - not defined?-implicit2.jpg  
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  12. #12
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    Re: Implicit differentiation - not defined?

    Oh I understood my error. I messed up one of the - signs :-)

    Thank you!
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    Re: Implicit differentiation - not defined?

    I have another question. Could someone check this for me please:

    find the tangent line to the equation:

    y2+e2x+xln(y)=4 through (0,2)

    First I take the derivatives:

    fy(x,y)=2e2+y2+ln(y)
    fx[SUB](x,y)[/SUB=2y*e2x+x/y

    through the point (0,2) I have:
    fy=ln(2)+2
    fx=4

    so my tangent line is:
    y-2=-(ln(2)+2)/4*x
    y=x/2-ln(2)+2

    Is this true? I really hope so, because I am trying this examples since 2 hours :-)
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  14. #14
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    Re: Implicit differentiation - not defined?

    Quote Originally Posted by infernalmich View Post
    I have another question. Could someone check this for me please:

    find the tangent line to the equation:

    y2+e2x+xln(y)=4 through (0,2)

    First I take the derivatives:

    fy(x,y)=2e2+y2+ln(y)
    This is wrong to start with. fy= 2y+ x/y.

    fx(x,y)=2y*e2x+x/y
    No, fx= 2e2x+ ln(y).

    through the point (0,2) I have:
    fy=ln(2)+2
    fx=4
    No fy(0, 2)= 4 and fx(0, 2)= 2+ ln(2). You have "x" and "y" reversed.

    so my tangent line is:
    y-2=-(ln(2)+2)/4*x
    Oddly enough, you have again mixed up "x" and "y" so that this is correct!

    y=x/2-ln(2)+2
    This, however, is not correct. You should have -((ln(2)+ 2)/4)x+ 2. The x is not just multiplied by the "2/4".

    Is this true? I really hope so, because I am trying this examples since 2 hours :-)
    You really need to review basic differentiation rules and algebra.
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    Re: Implicit differentiation - not defined?

    I have another question:

    Find all values for which the tangent line of the following equation is horizontal:
    2x^2+xy+y^2-4=-2

    first I get the derivatives with respect to x and y:
    af/ax=4x+y
    af/ay=x+2y

    Than I use implicit differentiation: - (4x+y)/(x+2y)

    As I remember, if the numerator is 0 the tangent line is horizontal, so the tangent line is horiziontal for y=4x or x=y/4 - if I would be asked for the vertical tangent line than the denumerator would have to be 0.

    Is this true?
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