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Math Help - Tangent to the curve.

  1. #1
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    Tangent to the curve.

    Ok, I have a question here, however the problem i have is more simple than the actual question itself, it's a basic math lack of understanding on my part.

    Consider the curve defined by:

    x^2 + 4xy + y^2 = 6

    Find an expression for \frac{dy}{dx} in terms of x and y, and hence give the equation of the tangent to the curve at the point (x,y) = (1,1)

    So, to begin, I make:

    x^2 + 4xy + y^2 = 0

    then differentiate with respect to X:

    giving \frac{dy}{dx} = 2x+4y

    Next, differentiate with respect to Y:

    giving \frac{dx}{dy} = 4x + 2y

    Now, I know that the next step is \frac{dy}{dx} = -\frac{2x+4y}{4x+2y}

    But I don't understand where the - comes from to make the fraction negative. I know that \frac{dx}{dy} needs to be inverted to make it into \frac{dy}{dx} and so moving to the bottom ofthe fraction, but where does the minus sign come from?

    Could someone please explain that before we move onto the next step? Thanks
    Last edited by Alias_NeO; August 27th 2007 at 10:31 AM. Reason: Math code error.
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  2. #2
    Super Member Rebesques's Avatar
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    Why bother so much? Just implicitly write

    <br />
x^2 + 4xy(x) + y^2(x) = 6<br />

    and differentiate for x:

    <br />
2x+4y(x)+4xy'(x) + 2y(x)y'(x) =0<br />

    so at points where y(x) is defined,

    <br />
y'(x) =(-2x-4y(x))/(4x+2y(x))<br />

    and use (x,y(x))=(1,1) to find y'(1)=\ldots...
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  3. #3
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    Sorry

    I do it the other way because it is the way i learned and I can't understand the way you gave, can you help fcomplete my method? I have a lot to learn in the next two days and learning a new method from scratch isn't the best way to get that done.
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  4. #4
    Super Member Rebesques's Avatar
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    But I don't understand where the - comes from to make the fraction negative.

    Call the given function F(x,y)=0. Assuming y=y(x), differentiate using the chain rule:


    <br />
\frac{dF}{dx}\frac{dx}{dx} +\frac{dF}{dy}\frac{dy}{dx}=0<br />

    and since \frac{dx}{dx}=1, solve the last one for \frac{dy}{dx} to obtain:

    <br />
\frac{dy}{dx}=\frac{-\frac{dF}{dx}}{\frac{dF}{dy}}<br />
.

    This is your method, isn't it?
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  5. #5
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    Argh

    I'm sorry, it's this general sort of maths notation that causes me problems. Bearwith my while i try write all you have just said out again using the full numerical functions.

    I can't do general equiations/maths. Need to see the figures and values.
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  6. #6
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    Nope.

    Sorry to be a nusiance, could you please tell me what each of the terms refers to, i can't quite work it out, also what does "y =y(x)" mean?

    I gathered that:

    F(x,y) = x^2+4xy+y^2

    I get lost after that

    Not looking good for me.
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  7. #7
    Super Member Rebesques's Avatar
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    y=y(x) means that the y in the equation F(x,y)=0 is a function of x.



    Anyhow, you just need the formula <br />
\frac{dy}{dx}=\frac{-\frac{dF}{dx}}{\frac{dF}{dy}}<br />
.
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  8. #8
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    Ok.

    Ok, so when i have a fraction of a negative number over a positive one, and i take the negatiev sign out, do i invert all signs on the top?

    so +\frac{-1}{+2} to -\frac{+1}{+2}

    Would doing the following be the same and correct?

    +\frac{-x-1}{+2} to -\frac{+x+1}{+2}

    Shown the signs to explain what I mean, is this the correct manipulation?
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  9. #9
    Super Member Rebesques's Avatar
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    Yes it is.

    Work hard to overcome those algebra shortcomings!
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  10. #10
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    Next step.

    Quote Originally Posted by Rebesques View Post
    Yes it is.

    Work hard to overcome those algebra shortcomings!
    Tell me about it, but right now i just need to able to do the papers, starting wednesday i have 3 maths papers, if i don't pass them all i fail my year, i resit the yearagain losing another year of my life, i end up being a year behind all my friends and it will cost me an extra 3000 so on the whole, i really don't want to fail these.

    That being said, I don't have any more maths exams after this year.

    Ok, so now I got that part, next is the next step.

    "Give the equation of the tangent to the curve at the point (x,y) = (1,1)

    How do i sole this last step.

    The solution I am given is:

    y = -(x-1) + 1 = 2-x

    How do i find this? Thanks
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  11. #11
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    Quote Originally Posted by Alias_NeO View Post
    ...
    Ok, so now I got that part, next is the next step.
    "Give the equation of the tangent to the curve at the point (x,y) = (1,1)

    How do i sole this last step.

    The solution I am given is:

    y = -(x-1) + 1 = 2-x
    Hello,

    you already know the gradient of the curve at T(1, 1):

    \frac{dy}{dx} = -\frac{2x+4y}{4x+2y} . Plug in the given values:

    y'(1,1) = -\frac{2\cdot 1+4\cdot 1}{4\cdot 1+2\cdot 1} = -\frac{6}{6} = -1

    You are looking for the equation of a straight line passing through T (the tangent point) with a slope m = -1. Use the point slope-formula of a straight line:

    If a line passes through the point P_1(x_1, y_1) with the slope m then the equation of the line is:
    \frac{y-y_1}{x-x_1}=m . Plug in all values you know (coordinates of T, slope m):

    \frac{y-1}{x-1}=-1~\Longrightarrow~y-1=-1 \cdot (x-1) ~\Longrightarrow~ y=-(x-1)+1~\Longrightarrow~ y=-x+2

    I've attached a diagram of the curve and the tangent.
    Attached Thumbnails Attached Thumbnails Tangent to the curve.-hyperb_tangent.gif  
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  12. #12
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    Thanks

    Thank you for the excellent explanation.
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