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calculus help please

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Find the points where the tangent line is horizontal or vertical. Find the area under one period of the curve.

x= r(deta - cos deta)
Y= r(1- sin deta)

deta= pie/3

  • calculus help please - ,

    %pi;/3 is not correct. There is a value of θ for the horizontal tangent, and another for the vertical tangent.

    x= r(θ - cos θ)
    Y= r(1- sin θ)

    For the given parametric equations, r is a magnification factor, so for all intents and purposes, we can assume it to be 1.0. If it assumes any other value, the graph will be magnified or reduced accordingly.

    To find dy/dx or dx/dy, we can apply the chain rule
    dy/dx = (dy/dθ) / (dx/dθ)
    and similarly
    dx/dy = (dx/dθ) / (dy/dθ)

    where dx/dθ = r(1+sinθ), and
    dy/dθ = r(-cosθ)

    A vertical tangent occurs at a point where dy/dx=0, and a horizontal tangent occurs where dx/dy=0.

    Note that at θ=-π/2, both dx/dθ and dy/dθ are zero. Use l'hôpital's rule to evaluate.

    You should get a complete cycle if you work with θ from -2π to 2π.

    To help you visualize the function, I include two graphics plots:

    Post your results for verifications if you wish.

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