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Math Help - Intersection of the Slopes of Curves

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Intersection of the Slopes of Curves

    If m is the slope of the curve xy=3 and m2 is the slope of the curve x^2-y^2=4, then at a point of intersection of the two curves:

    1. (A) m1 = m2
    2. (B) m1= -m2
    3. (C) m1*m2 = 1
    4. (D) m1*m2 = -1
    5. (E) m1*m2 = -2


    I think I might be doing it wrong... >_<

    First, I found the derivative of the first one explicitly.
    <br />
\frac{dy}{dx}[3x^-1]<br />

    which is:

    <br />
 \frac{-3}{x^2}<br />

    And then, the derivative of the second one implicitly:

    <br />
\frac{dy}{dx}[x^2-y^2]=\frac{dy}{dx} [4]

    which is:

    <br />
\frac{dy}{dx}= \frac{x}{y}

    But I can't see a relationship between the two slopes...?
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  2. #2
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    Quote Originally Posted by StarlitxSunshine View Post
    If m is the slope of the curve xy=3 and m2 is the slope of the curve x^2-y^2=4, then at a point of intersection of the two curves:

    1. (A) m1 = m2
    2. (B) m1= -m2
    3. (C) m1*m2 = 1
    4. (D) m1*m2 = -1
    5. (E) m1*m2 = -2

    I think I might be doing it wrong... >_<

    First, I found the derivative of the first one explicitly.
    <br />
\frac{dy}{dx}[3x^-1]<br />

    which is:

    <br />
\frac{-3}{x^2}<br />

    And then, the derivative of the second one implicitly:

    <br />
\frac{dy}{dx}[x^2-y^2]=\frac{dy}{dx} [4]

    which is:

    <br />
\frac{dy}{dx}= \frac{x}{y}

    But I can't see a relationship between the two slopes...?

    Multiply both slopes...

    Tonio
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  3. #3
    Junior Member StarlitxSunshine's Avatar
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    O.o I tried that...

    <br />
\frac{-3}{x^2} * \frac{x}{y} = \frac{-3}{xy}<br />

    That doesn't help... ? -3 isn't a choice in the answers...
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  4. #4
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    You need to consider the slopes "at the point of intersection". So you need to find that point first, evaluate both gradients (at that point) then look for the answer.
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  5. #5
    Junior Member StarlitxSunshine's Avatar
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    Quote Originally Posted by Debsta View Post
    You need to consider the slopes "at the point of intersection". So you need to find that point first, evaluate both gradients (at that point) then look for the answer.
    Ohh...Umm.. :33

    To find the point of intersection, should I make the table of values for each equation ? Or is there another way ?
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  6. #6
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    Sub y = 3/x from first eqtn into the second one and do it algebraically.
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    Quote Originally Posted by StarlitxSunshine View Post
    O.o I tried that...

    <br />
\frac{-3}{x^2} * \frac{x}{y} = \frac{-3}{xy}<br />

    That doesn't help... ? -3 isn't a choice in the answers...

    But you're given what xy is, girl!

    Tonio
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    Quote Originally Posted by StarlitxSunshine View Post
    Ohh...Umm.. :33

    To find the point of intersection, should I make the table of values for each equation ? Or is there another way ?

    You don't need to find the intersection point at all: it is a very nasty thing (square roots inside square roots...you don't want to meet such a thing in an obscure alley at night!)!

    Tonio
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  9. #9
    Junior Member StarlitxSunshine's Avatar
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    Quote Originally Posted by tonio View Post
    But you're given what xy is, girl!

    Tonio
    O_O I am!!

    I'm so sorry !! X3

    Then it would be -3/3 = -1 which is a choice @_@

    Sorry, sorry !
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  10. #10
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    That's true. Tonio's way is much easier. You're nearly there.
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