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Math Help - differentiation help..

  1. #1
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    Post differentiation help..

    The question is:

    Find slopes of all lines tangent to the graph of the relation:

    y^3 + 15x^15y^2 - 50y = 150 - 150x^5

    at the points on its graph where x=1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by omibayne View Post
    The question is:

    Find slopes of all lines tangent to the graph of the relation:

    y^3 + 15x^15y^2 - 50y = 150 - 150x^5

    at the points on its graph where x=1
    there are three points for x = 1: (1,0),~ \left( 1, \frac {-15 + 5 \sqrt{17}}2\right),~\text{and } \left( 1, \frac {-15 - 5 \sqrt{17}}2 \right)

    differentiate implicitly with respect to x and plug each of these points in to find the slopes of each of the three tangent lines.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by omibayne View Post
    the correct answers for slopes are :- 15, -35 and -55 but i dont know how to get them :S:S
    i just told you how to get them

    you do know how to differentiate implicitly, right?
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  4. #4
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    I do know but it didnt work...:S
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by omibayne View Post
    I do know but it didnt work...:S
    y^3 + 15x^{15}y^2 - 50y = 150 - 150x^5

    \Rightarrow \frac {dy}{dx} = \frac {-(750x^4 + 225x^{14}y^2)}{3y^2 + 30x^{15}y - 50}

    now plug in the points given earlier
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  6. #6
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    Post help

    can smn help me differentiate this equation. i tried it several times but it didnt work..any help will be appreciated..thanks

    y^3 + 5x^15y^2 - 50y = 150 - 150x^5
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by omibayne View Post
    can smn help me differentiate this equation. i tried it several times but it didnt work..any help will be appreciated..thanks

    y^3 + 5x^15y^2 - 50y = 150 - 150x^5
    Can you show us what work you have so far?

    \frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}
    Last edited by 11rdc11; October 25th 2008 at 11:49 PM.
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  8. #8
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    double posting is against the rules. and it is disrespectful to do so when someone already tried to help you
    I agree with Jhevon but did you repost because you posted the problem wrong? If so just edit your original post and don't start a new thread.

    One post has

    y^3 + 15x^15y^2 - 50y = 150 - 150x^5

    and the other has

    y^3 + 5x^15y^2 - 50y = 150 - 150x^5
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  9. #9
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    [/tex]
    Quote Originally Posted by 11rdc11 View Post
    Can you show us what work you have so far?

    \frac{dy}{dx}=\frac{3y^2+5x^{15}2y-50}{-750x^4-75x^{14}y^2}
    i got \frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    double posting is against the rules. and it is disrespectful to do so when someone already tried to help you
    Quote Originally Posted by 11rdc11 View Post
    I agree with Jhevon but did you repost because you posted the problem wrong? If so just edit your original post and don't start a new thread.

    One post has

    y^3 + 15x^15y^2 - 50y = 150 - 150x^5

    and the other has

    y^3 + 5x^15y^2 - 50y = 150 - 150x^5
    yeah i posted the wrong question...i am new to this so sorry abt that...
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  11. #11
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by omibayne View Post
    [/tex]

    i got \frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}
    Your correct I made a mistake when typing it into latex. I edited my post
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  12. #12
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by omibayne View Post
    yeah i posted the wrong question...i am new to this so sorry abt that...
    It's kewl. We were all newbies at one point lol. Next time just make the correction to your original post ok
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  13. #13
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    Quote Originally Posted by 11rdc11 View Post
    It's kewl. We were all newbies at one point lol. Next time just make the correction to your original post ok

    ok thanx
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  14. #14
    Junior Member toraj58's Avatar
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    as a genral solution you can sovle all problems like yours with this approach:

    first sort it as:

    f(x,y) = 0

    then use this formula to differntiate:

    \frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}

    it means when you are differntiatig a respect to x then y assumed as constant and in reverse; be sure before differniation that all statements of the equation should be sored on one side and other side should be only Zero(0).
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  15. #15
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by toraj58 View Post
    as a genral solution you can sovle all problems like yours with this approach:

    first sort it as:

    f(x,y) = 0

    then use this formula to differntiate:

    \frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}

    it means when you are differntiatig a respect to x then y assumed as constant and in reverse; be sure before differniation that all statements of the equation should be sored on one side and other side should be only Zero(0).
    I never understood why we were taught that so late in the course. I find it easier than doing implicit diff
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