Find the slope of the tangent line to the given polar curve at the point specified by the value

Find the slope of the tangent line to the given polar curve...

Key Concepts

Differentiation in Polar Coordinates

Differentiating a polar function with respect to the angle requires using the chain rule and product rule. This differentiation is necessary to compute the derivatives dx/d? and dy/d?, which in turn are used to find the slope of the tangent line in the Cartesian coordinate system.

Slope of a Tangent Line to a Polar Curve

The slope of the tangent line to a polar curve at a given point can be determined by converting the polar equation to its parametric Cartesian form and then applying differentiation. The formula for the slope is (dy/d?)/(dx/d?), which involves differentiating both x = r cos(?) and y = r sin(?) with respect to ?.

Conversion to Cartesian Coordinates

Converting from polar to Cartesian coordinates is essential for analyzing properties like slopes of tangent lines because many calculus tools, including differentiation, are naturally formulated in the Cartesian system. The conversion is done using the formulas x = r cos(?) and y = r sin(?).

Polar Coordinates

Polar coordinates provide a system for representing points in the plane using a distance from a fixed point (the pole) and an angle from a fixed direction. This system is especially useful for curves that have a natural circular symmetry or are easier to describe in terms of angles and radii.

Transcript

00:04 Since we're given a polar equation, r equals 2 times sine of theta, and we need to find the slope of the tangent line.

00:11 We want to first rewrite the pull equation using parametric equations.

00:17 So let's think about the fact that x equals r times cosine of theta and y equals r times sine of theta.

00:27 We're in this case, our r is given by 2 times sine of theta.

00:33 So therefore, x equals 2 times sine theta times cosine of theta.

00:46 And likewise, we have y equals.

00:50 Again, since our r in this equation is 2 times sine of theta, replace the r with 2 times sine of theta, multiplied by sine of theta, to obtain that y equals 2 times sine squared theta.

01:08 So now we have our x and y equations in terms of theta, where theta is our parameter, and we now have a parametric form.

01:20 So now to find the derivative of y with respect to x, we're going to find the derivative of y with respect to theta, divided by the derivative of x with respect to theta.

01:38 So the derivative of a y with respect to theta, we use a power rule, which really is a chain rule.

01:48 We're going to take the power of sine squared theta and multiply it by the two in the front to give us four times sine of theta.

01:58 But next times the derivative of what was being squared, the derivative of sine theta is cosine of theta.

02:10 So the derivative of y with respect to theta is four times sine of theta times cosine of theta.

02:19 We're next going to divide that by the derivative of x with respect to theta.

02:25 Since x is 2 sine theta times cosine theta, define the derivative of x with respect to theta, we need to use a product rule.

02:35 In this case, i'm going to do it in the following form.

02:38 The first function times the derivative of the second function, plus the second function times the derivative of the first function.

02:50 Our first function will take to be 2 times sine of theta.

02:53 Our second function was cosine theta.

02:57 The derivative of cosine theta is negative sine of theta, plus our second function, which is cosine theta, times the derivative of our first function, which will use a constant times a function rule.

03:12 The two will remain, and the derivative of sine theta is cosine of theta.

03:19 Or in other words, dxd theta is negative 2 times sine squared theta, plus 2 times cosine squared of theta.

03:31 Again, that's our derivative of x with respect to theta.

03:35 So back in our formula, we're going to replace the dx over d theta with negative 2 times sine squared theta plus 2 times cosine squared of theta...

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