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Math Help - Tangent at point...

  1. #1
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    Tangent at point...

    Ok so i was assigned some math problems over the summer but lost my math binder from last year and i dont remember how to solve this:

    x^2 -xy + y^2 = 9
    Find the point at which a vertical tangent exists

    I remember i used to be able to solve this so its really frustrating i forgot how... Just need a little help remembering. Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ijleyton View Post
    Ok so i was assigned some math problems over the summer but lost my math binder from last year and i dont remember how to solve this:

    x^2 -xy + y^2 = 9
    Find the point at which a vertical tangent exists

    I remember i used to be able to solve this so its really frustrating i forgot how... Just need a little help remembering. Thanks
    steps:

    - differentiate implicitly
    - solve for dy/dx
    - vertical tangents occur when dy/dx is undefined. solve for this
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  3. #3
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    Ok so i did:

    x^2 -xy + y^2 = 9
    2x - x(dy/dx)*y + 2y(dy/dx) = 0
    Now if i continue solving for this i would get an 'x' and a 'y' in my answer.. How do i fix that? :s
    -This used to be really easy.. the summer made me rusty
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ijleyton View Post
    Ok so i did:

    x^2 -xy + y^2 = 9
    2x - x(dy/dx)*y + 2y(dy/dx) = 0
    Now if i continue solving for this i would get an 'x' and a 'y' in my answer.. How do i fix that? :s
    -This used to be really easy.. the summer made me rusty
    rusty indeed, you differentiated incorrectly (you need the product rule to differentiate -xy)

    the actual answer is 2x - y - x \frac {dy}{dx} - 2y \frac {dy}{dx} = 0

    now solve for dy/dx

    you will get a rational function. these are undefined where the denominator is zero (provided the numerator isn't zero at the same time). so set it equal to zero and solve for y. once you have y, plug it into the original equation and solve for x. this will give you the vertical lines you seek. they are of the form x = something of course
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  5. #5
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    haha i feel really stupid for not being able to figure this out
    Ok so i have:
    2x-y-x(dy/dx) -2y(dy/dx)=0
    i tried to isolate (dy/dx) and was left with:
    (dy/dx)= (2x+y)/(x+2y)
    Now if i understood you correctly you wanted me to set (x+2y)= 0
    That would mean 2y=-x ... now what?
    -The answer choices have radicals.. i dont see how i'm gonna get that
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ijleyton View Post
    haha i feel really stupid for not being able to figure this out
    Ok so i have:
    2x-y-x(dy/dx) -2y(dy/dx)=0
    i tried to isolate (dy/dx) and was left with:
    (dy/dx)= (2x+y)/(x+2y)
    Now if i understood you correctly you wanted me to set (x+2y)= 0
    That would mean 2y=-x ... now what?
    -The answer choices have radicals.. i dont see how i'm gonna get that
    i told you exactly how to finish the problem

    i got \frac {dy}{dx} = \frac {2x - y}{x + 2y}

    set x + 2y = 0 \implies y = - \frac 12x, plug this in for y in the original, we get:

    x^2 - x \bigg( - \frac 12x \bigg) - \bigg( - \frac 12x\bigg)^2 = 9

    \Rightarrow x^2 + \frac 12x^2 - \frac 14x^2 = 9

    solve that to get your (radical) solutions
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