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Math Help - A question related to Derivative Rules.

  1. #1
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    A question related to Derivative Rules.

    Hey everyone.

    I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

    "Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

    Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

    How abouts do I solve this?

    Any insight would be helpful.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jeavus View Post
    Hey everyone.

    I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

    "Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

    Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

    How abouts do I solve this?

    Any insight would be helpful.
    Are you asking about two different problems? I don't see how \sqrt{x}+\sqrt{y}=1 relates to \frac{1}{x^2}+\frac{1}{y^2}=1.
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  3. #3
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    Oh, I'm sorry. I meant to write x^1/2, not 1/x^2.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jeavus View Post
    Hey everyone.

    I'm currently trying to solve a problem from my textbook that was assigned for homework. It reads;

    "Let P(a, b) be a point on the curve sqroot(x) + sqroot(y) = 1. Show that the slope of a tangent at P is -sqroot(b/a)."

    Using Graphmatica, (1/x^2)+(1/y^2)=1 shows four asymptotes, one in each quadrant, all approaching 1 or (-1) (depending on the positive/negative side of the y axis).

    How abouts do I solve this?

    Any insight would be helpful.
    x^{\frac{1}{2}}+y^{\frac{1}{2}}=1

    The derivative would be (using implicit differentiation):

    \frac{1}{2}x^{-\frac{1}{2}}dx+\frac{1}{2}y^{-\frac{1}{2}}dy=0

    \frac{1}{2\sqrt{x}}dx+\frac{1}{2\sqrt{y}}dy=0

    \frac{1}{2\sqrt{y}}dy=-\frac{1}{2\sqrt{x}}dx

    \frac{dy}{dx}=-\sqrt{\frac{x}{y}}

    Plugging in the point P(a,b), you'd get:

    \frac{dy}{dx}=-\sqrt{\frac{a}{b}}
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  5. #5
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    Thanks very much.

    I can see where I was thinking wrong.
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    \frac{dy}{dx}=-\sqrt{\frac{x}{y}}

    Plugging in the point P(a,b), you'd get:

    \frac{dy}{dx}=-\sqrt{\frac{a}{b}}

    I made a typo. This should have been:

    \frac{dy}{dx}=-\sqrt{\frac{y}{x}}

    Which, for P(a,b), would give us:

    \frac{dy}{dx}=-\sqrt{\frac{b}{a}}
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