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Thread: 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 $\displaystyle \frac{dy}{dx}$

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

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

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

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

    I know the vertical tangent line point can be found when $\displaystyle \frac{dy}{dx}$ 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 $\displaystyle \frac{dy}{dx}$

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

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

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

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

    I know the vertical tangent line point can be found when $\displaystyle \frac{dy}{dx}$ 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 $\displaystyle x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20$ $\displaystyle \implies (x-3)^2+(y+5)^2=54$

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



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

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

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

    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 $\displaystyle \frac{dy}{dx}$

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

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

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

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

    I know the vertical tangent line point can be found when $\displaystyle \frac{dy}{dx}$ 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. $\displaystyle 2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=0 \implies \frac{dy}{dx}(2y+10)=6-2x \implies \frac{dy}{dx}=\frac{6-2x}{2y+10}$
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  5. #5
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    Quote Originally Posted by redsoxfan325 View Post
    Note that $\displaystyle x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20$ $\displaystyle \implies (x-3)^2+(y+5)^2=54$

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



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



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

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

    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

    $\displaystyle \frac{-10+\sqrt{128}}{2}$
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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 $\displaystyle x=3$ into the original equation, I get $\displaystyle y=-5\pm3\sqrt{6}$, and when I plug $\displaystyle y=-5$ into the original equation, I get $\displaystyle x=3\pm3\sqrt{6}$. Can you show your steps to how you ended up with $\displaystyle \frac{-10+\sqrt{128}}{2}$?
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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

    $\displaystyle x^2+y^2-6x+10y=20$

    Ok the num when equal to zero is x=3

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

    So I get

    $\displaystyle y^2+10y-9=20$

    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

    $\displaystyle x^2+y^2-6x+10y=20$

    Ok the num when equal to zero is x=3

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

    So I get

    $\displaystyle y^2+10y-9=20$

    and then somewhere I think is where I make my mistake and plugged in incorrect numbers
    $\displaystyle y^2+10y-29=0 \implies x=\frac{-10\pm\sqrt{100-(4\cdot-29)}}{2} =\frac{-10\pm\sqrt{216}}{2}$ $\displaystyle = \frac{-10\pm6\sqrt{6}}{2} = -5\pm3\sqrt{6}$
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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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