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Math Help - Integral and a slope problem

  1. #1
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    Integral and a slope problem

    find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*



    also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 333arizona View Post
    find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*



    also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated.
    \int\tan^2(x)+\sin(x)dx

    \tan^2(x)+1=\sec^2(x)\Rightarrow\tan^2(x)=\sec^2(x  )-1

    So \int\tan^2(x)+\sin(x)dx=\int\sec^2(x)-1+\sin(x)dx=\tan(x)-x-\cos(x)+C
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 333arizona View Post
    find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*



    also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated.
    3x^2+2xy+6y^2-3x-8y=0

    differntiating implicity we get

    6x+2[xy'+y]+12y\cdot{y'}-3-8y'=0

    Substituting x=1 and y=1

    we get 6(1)+2[y'+1]+12\cdot{y'}-3-8y'=0\Rightarrow{5+6y'=0}\Rightarrow{y'=\frac{-5}{6}}
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