40-Calculus-of-PolarEqA

# 40-Calculus-of-PolarEqA - Calculus of Polar Equation...

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Calculus of Polar Equation Because polar equation is a parametric equation, the previous discussions on the calculus of parametric equations are directly applicable. However, because polar equation relies on a unique set of parameters and it is widely used, its characteristics are specifically studied and several convenient formulas can be derived. We will begin with finding the slope. For parametric equation, remember that the slope is defined as: dt dx dt dy dx dy = Polar equation is essentially a type of parametric equation where: θ θ sin cos r y r x = = So, for polar equation, the slope becomes: θ θ d dx d dy dx dy = However, the big difference is that θ , r are both varying. Therefore: You must use the product rule and implicit differentiation to find the derivative: θ θ θ θ θ θ θ θ θ θ θ θ d dr d d r d dy d dr d d r d dx sin cos cos sin + = + = Substituting θ d dx and θ d dy back into the dx dy equation and after some algebraic arrangement, the slope of a polar curve is defined as : θ θ θ θ θ θ d dr r d dr r dx dy cos sin sin cos + + = The above equation looks complicated but it is actually quite simple to use. Example: Find dx dy for 4cos r θ = 4cos 4sin dr r d θ θ θ = = The above formula gives: ( ) ( ) ( ) ( ) . 4cos cos sin 4sin 4cos sin cos 4sin Simplified as needed dy dx θ θ θ θ θ θ θ θ + = = + If the problem asks for a specific tangent point, say at 30 o θ = , just evaluate the above at 30 o θ =

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