Find parametric equations for the semicircle x^2+y^2=a^2, y > 0, using as parameter the slope

Find parametric equations for the semicircle x^2+y^2=a^2, ...

Key Concepts

Implicit Differentiation

Implicit differentiation is a method used to find the derivative of a function when it is given in an implicit form, rather than explicitly solved for one variable. In this context, differentiating the circle's equation with respect to x allows us to relate the derivatives of x and y directly, which is essential for determining the slope of the tangent at any point on the curve.

Parametric Equations

Parametric equations represent a curve by defining both x and y as functions of a third variable, known as a parameter. This representation is particularly useful for describing curves where the relationship between x and y is more naturally expressed in terms of another varying quantity, such as the slope of the tangent in this problem.

Tangent Line Slope

The slope of the tangent line to a curve at a given point provides key information about the instantaneous rate of change of the function at that point. In our problem, using the slope as a parameter links the rate at which y changes with respect to x to the geometric coordinates of the circle, thereby facilitating a unique parametric form.

Circle Geometry

The geometry of a circle, described by the equation x² + y² = a², defines the set of points that are at a constant distance from the center. Understanding the inherent symmetry and properties of the circle is crucial when deriving relationships between its coordinates and characteristics such as the tangent slope.

Transcript

0:00 Hi.

00:01 Today we're going to talk about parameters and we want to write the parametric equations.

00:11 We are given x squared plus y squared equals a squared.

00:24 And we're told that the parameter is the derivative.

00:27 So the first thing i want to do is find the derivative of our equation.

00:43 I'm going to do that implicitly using implicit differentiation.

00:47 So i'm going to take the derivative with respect to x and get 2x, take the derivative with respect to y, and then follow that up with d, y, d, d, x.

01:01 And the derivative of our constant, a squared, is zero.

01:11 So over here, because of our parameter, we can replace that with t.

01:27 Now, we can solve each of these, or solve this equation for each variable x and y.

01:36 So if we solve for x, we're going to subtract 2y from both sides.

01:46 And divide by 2, and we get x equals negative y t.

01:58 And if we solve for y, we're going to subtract negative 2x and divide by 2t, and that is going to give us y equals negative x over t we're almost complete we now have to take this and substitute it in for x and that will give us one of our parametric equations and then we will take y and substitute it in for y and that'll give us our other parametric equations.

02:47 So let's get started with that.

03:00 So i have x squared plus y squared equals a squared.

03:08 I'm replacing x with negative y t which gives me y squared t squared plus y squared equals a squared.

03:35 We solve an equation that has two variables like that by factoring.

03:41 We're going to factor the y squared out and we'll have t squared plus one remaining and then we divide both sides by t squared plus one and then we simply square root both sides so that we have y...

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