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33 - Lecture 4 Reference Miller Analytic Geometry and...

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Lecture 4 January 10, 2012 Reference: Miller Analytic Geometry and Calculus 1 Calculus in Polar coordinates Example 1.1. Find an expression for the angle of the radius vector to the tangent vector of a curve in polar coordinates. ADD PICTURE The slope OP is given by y x and the slope of PT is given by d y d x . Thus tan θ = y x and tan φ = d y d x . Using geometric properties of triangles we deduce that 180 - φ + θ + ψ = 180 and hence that φ = θ + ψ and so tan ψ = tan( φ - θ ) = sin( φ - θ cos( φ - θ ) = sin φ cos θ - sin θ cos φ cos φ cos θ - sin φ sin θ = tan φ - tan θ 1 - tan φ tan θ hence tan ψ = d y d x - y x 1 + y x d y d x = x d y - y d x x d x + y d y 1

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Now x = r cos θ , y = r sin θ so d x = - r sin θ d θ + cos θ d r and d y = r cos θ d θ + sin θ d r so tan φ = r cos θ ( r cos θ d θ + sin θ d r ) - r sin θ ( - r sin θ d θ + cos θ d r ) r cos θ ( - r sin θ d θ + cos θ d r ) + r sin θ ( r cos θ d θ + sin θ d r ) = r 2 d θ r d r = r d θ d r Theorem 1.1. In polar coordinates r d θ d r gives the angle between the tangent to a curve and the radius vector. Corollary 1.1. Suppose that two curves intersect at a point P . Let ψ 1 and ψ 2 be the angles from the radii to the tangent vectors. ADD PICTURE Then the angle α between the curves can be found by means of the formula tan α = tan( ψ 2 - ψ 1 ) = tan ψ 2 - tan ψ 1
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33 - Lecture 4 Reference Miller Analytic Geometry and...

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