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For the following exercises, find the slope of a tangent line to a polar curve $r=f(\theta) .$ Let $x=r \cos \theta=f(\theta) \cos \theta$ and $y=r \sin \theta=f(\theta) \sin \theta, \quad$ so the polar equation$r=f(\theta)$ is now written in parametric form.Use the definition of the derivative $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$ and the product rule to derive the derivative of a polar equation.

we use product rule to derive the derivative of a polar equation.

Calculus 2 / BC

Chapter 7

Parametric Equations and Polar Coordinates

Section 4

Area and Arc Length in Polar Coordinates

Parametric Equations

Polar Coordinates

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Harvey Mudd College

University of Michigan - Ann Arbor

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So in this problem were asked to find D Y d x at a certain data for a polar function. So what is the slope of the tangent line for a polar function at some data. And we know that if we have some polar function, R equals F data than X equals. If time stayed effort data times co sign data and y equals FF data times signed data. So what we can do is find d x. Oh, I mean, find d Y d theta and divide that by d x dif data and then the d fate is in the denominators will cancel. So we get d Y d X, which is what we were looking for. And it will be in terms of theta because we had dif ada in both of our in our numerator and our denominator. So first we'll find de y d theta by taking the derivative of why with respect to data. So we know by the product rule that's going to give us af of state, uh, times and the derivative of sign is co sign data and then plus f derivative of they'd, uh, times signed data. So just the product rule being applied here and now we'll find DX dy theta, which is going to be again product rule. But on X of data, which is f if data times negative sign data is the derivative of co sign data is negative. Sign data. You said a little glitch there. Times negative sign data. Plus And now we have f derivative of data Times Co sign data. So now we just need to divide the two by each other. So d y eat, they'd ah, over de X de fate. Ah, equals we say, actually, I'll just bring these using the power of computer magic. I'll just bring this where we need it now. So we're taking our d y d theta and we're dividing it by r d x d theta And we know that d y d faded, divided by D X teeth Data is also equal to d just d Y d axe. So that's how you that is our definition of the derivative for a polar with respect to X and y

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