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calculus

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-find the equation of the tangent line to the curve y=5xcosx at the point (pi,-5pi)
-the equation of this tangent line can be written in the form y=mx+b where
m=
and b=
-what is the answer to m and b?

  • calculus - ,

    f(x)=5xcos(x)
    f'(x)=5cos(x)-5xsin(x)
    =5(cos(x)-xsin(x))
    m=f'(π)
    and the equation of the tangent passing through the point (x0,y0)=(π, -5π) is:
    (y-y0)=m(x-x0)
    Substitute values and simplify to get the equation of the line.

  • calculus - ,

    m=pi
    and b= y-pi=pi(x+5pi)
    y-pi=pi(x)+5pi(pi)
    m=pie but what does b= though

  • calculus - ,

    Given
    (x0,y0)=(π, -5π)

    We have established:
    m
    =f'(π)
    =5(cos(π)-π*sin(π))
    =5(-1 - π*0)
    =-5

    The equation of a line of slope m passing through (x0,y0) is:
    (y-y0)=m(x-x0)
    (y-(-5π))=(-5)(x-π)
    Simplify to get the required equation.
    b=0 by coincidence.

    See graph:
    http://img10.imageshack.us/i/1298601541.png/

    Check my work.

  • calculus - ,

    you did not follow MathMate's suggestion

    f'(x) = 5(cosx - xsinx)
    f'(π) = 5(cosπ - πsinπ)
    = 5(-1 - π(0)) = -5 , so m = -5

    equation:
    y = -5x + b , but (π,-5π) lies on it, so
    -5π = -5(π) + b
    b = 0

    equation : y = -5x

    or using MathMate's suggestion:

    y + 5π = -5(x-π)
    y + 4π = -5x + 5π
    y = 5x

  • calculus - ,

    ok how about this one:
    f(x)=12x/sinx+cosx
    find f'(-pie)
    this is what i did:
    12(-pi)/(0+(-1))
    = 12pi
    is that right what am i doing wrong and what would be the answer

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