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Math Help - Find Points if horizontal or vertical tanget line

  1. #1
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    Find Points if horizontal or vertical tanget line

    I was given an implicit function to solve for \frac{dy}{dx}

    Which I have done here f(x)=x^2+y^2-6x+10y=20

    2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20

    Which I have solved \frac{dy}{dx} for

    \frac{dy}{dx}=\frac{6-2x}{2y+10}

    I know the vertical tangent line point can be found when \frac{dy}{dx} numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

    However both the numerator and denom. have to satisfy the originial equation?

    for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

    and vice versa for the denom
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    I was given an implicit function to solve for \frac{dy}{dx}

    Which I have done here f(x)=x^2+y^2-6x+10y=20

    2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20

    Which I have solved \frac{dy}{dx} for

    \frac{dy}{dx}=\frac{6-2x}{2y+10}

    I know the vertical tangent line point can be found when \frac{dy}{dx} numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

    However both the numerator and denom. have to satisfy the originial equation?

    for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

    and vice versa for the denom
    Note that x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20 \implies (x-3)^2+(y+5)^2=54

    So the function you are dealing with is a circle of radius 3\sqrt{6} centered at (3,-5). This may be a helpful picture to have in your mind while doing this.



    Given that \frac{dy}{dx}=\frac{6-2x}{2y+10}, the horizontal tangent occurs when the numerator is zero (because then the derivative is 0), which is at x=3. Use the function to find that y=-5\pm 3\sqrt{6}.

    Similarly, the vertical tangent occurs when the denominator is zero (because then the derivative is "infinity"), which is at y=-5. Use the function to find that x=3\pm 3\sqrt{6}.

    So the horizontal tangent lines occur at (3,-5+3\sqrt{6}) and (3,-5-3\sqrt{6}) and the vertical tangent lines occur at (3+3\sqrt{6},-5) and (3-3\sqrt{6},-5).

    If you think about the picture of the circle, this makes a lot of sense.
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  3. #3
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    Quote Originally Posted by sk8erboyla2004 View Post
    I was given an implicit function to solve for \frac{dy}{dx}

    Which I have done here f(x)=x^2+y^2-6x+10y=20

    2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20

    Which I have solved \frac{dy}{dx} for

    \frac{dy}{dx}=\frac{6-2x}{2y+10}

    I know the vertical tangent line point can be found when \frac{dy}{dx} numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

    However both the numerator and denom. have to satisfy the originial equation?

    for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

    and vice versa for the denom
    You didnt differeniate Differentiate the right hand side
    d/dx(20) = 0
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  4. #4
    Super Member redsoxfan325's Avatar
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    At first I thought so too, but it seems like even though he didn't write it, he still got the correct derivative. 2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=0 \implies \frac{dy}{dx}(2y+10)=6-2x \implies \frac{dy}{dx}=\frac{6-2x}{2y+10}
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  5. #5
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    Quote Originally Posted by redsoxfan325 View Post
    Note that x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20 \implies (x-3)^2+(y+5)^2=54

    So the function you are dealing with is a circle of radius 3\sqrt{6} centered at (3,-5). This may be a helpful picture to have in your mind while doing this.



    Given that \frac{dy}{dx}=\frac{6-2x}{2y+10}, the horizontal tangent occurs when the numerator is zero (because then the derivative is 0), which is at x=3. Use the function to find that y=-5\pm 3\sqrt{6}.



    Similarly, the vertical tangent occurs when the denominator is zero (because then the derivative is "infinity"), which is at y=-5. Use the function to find that x=3\pm 3\sqrt{6}.

    So the horizontal tangent lines occur at (3,-5+3\sqrt{6}) and (3,-5-3\sqrt{6}) and the vertical tangent lines occur at (3+3\sqrt{6},-5) and (3-3\sqrt{6},-5).

    If you think about the picture of the circle, this makes a lot of sense.

    What if i plug x and y in the original equation unimplifed like you did my answer will not be the same such as when i plug in y i get

    \frac{-10+\sqrt{128}}{2}
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  6. #6
    Super Member redsoxfan325's Avatar
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    What are you using for x and what are you using for y when you plug them into the original equation?
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  7. #7
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    Well for the denom I solved y for -5 and for the num. x for 3 so i am trying to plug one in and then solve for the other unknown variable
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  8. #8
    Super Member redsoxfan325's Avatar
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    When I plug x=3 into the original equation, I get y=-5\pm3\sqrt{6}, and when I plug y=-5 into the original equation, I get x=3\pm3\sqrt{6}. Can you show your steps to how you ended up with \frac{-10+\sqrt{128}}{2}?
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  9. #9
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    Yeah sure thing I am sure it is something stupid I am doing

    Ok so here is the original function

    x^2+y^2-6x+10y=20

    Ok the num when equal to zero is x=3

    3^2+y^2-6(3)+10y=20

    So I get

    y^2+10y-9=20

    and then somewhere I think is where I make my mistake and plugged in incorrect numbers
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    Yeah sure thing I am sure it is something stupid I am doing

    Ok so here is the original function

    x^2+y^2-6x+10y=20

    Ok the num when equal to zero is x=3

    3^2+y^2-6(3)+10y=20

    So I get

    y^2+10y-9=20

    and then somewhere I think is where I make my mistake and plugged in incorrect numbers
    y^2+10y-29=0 \implies x=\frac{-10\pm\sqrt{100-(4\cdot-29)}}{2} =\frac{-10\pm\sqrt{216}}{2} = \frac{-10\pm6\sqrt{6}}{2} = -5\pm3\sqrt{6}
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  11. #11
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    Ah thanks Ill re-do my work now

    I forgot how to rewrite or do the sqrt of 216 the way u did :x mind showing that too
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