journal 16.docx - Journal 16 Problem 3 What is the normal...

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Journal 16 Problem: What is the normal line to the polar curve r = 3 + cosθ at θ = 3 π 2 ? Then convert your Cartesian equation (of the normal line) into polar form. Solution: First, we must parameterize our polar equation. Recall that x = rcosθ y = rsinθ . We can then find dy dx and evaluate it at our angle θ = 3 π 2 . r = 3 + cosθ { x = ( 3 + cosθ ) cosθ y = ( 3 + cosθ ) sinθ Recall dy dx = y ' x ' : y ' = ( 3 + cosθ ) ( cosθ ) + ( sinθ ) ( sinθ ) x' = ( 3 + cosθ ) ( sinθ ) +(− sinθ )( cosθ )
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dy dx = ( 3 + cosθ ) ( cosθ ) + ( sinθ ) ( sinθ ) ( 3 + cosθ ) ( sinθ ) +(− sinθ )( cosθ ) We now evaluate dy dx at θ = 3 π 2 : dy dx = ( 3 + cosθ ) ( cosθ ) + ( sinθ ) ( sinθ ) ( 3 + cosθ ) ( sinθ ) +(− sinθ )( cosθ ) θ = 3 π 2 dy dx = ( 3 + cos 3 π 2 )( cos 3 π 2 ) + ( sin 3 π 2 )( sin 3 π 2 ) ( 3 + cos 3 π 2 )( sin 3 π 2 ) +(− sin 3 π 2 )( cos 3 π 2 ) ¿ ( 3 + 0 ) ( 0 ) + ( 1 ) ( 1 ) ( 3 + 0 ) ( 1 ) + ( 1 ) ( 0 ) ¿ 1 3 The slope of the tangent line is 1 3 . Recall that the normal line is perpendicular to the tangent line. To find the slope of the normal line m ( ¿¿ ) ¿ , we simply take the opposite reciprocal of the slope of the tangent line.
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