91Ó°ÊÓ

Parametric equations express a set of related quantities as explicit functions of an independent parameter. Typically, in a two-dimensional space, parametric equations for a curve are given in the form of two equations, where x and y are each defined as functions of a third variable, often denoted as t. For example, the parametric equations x = f(t) and y = g(t) together define the set of points (x, y) that form the curve as t varies. This is a powerful way to describe curves that are not functions, such as circles and ellipses, as it allows us to compute points on the curve and to find the direction of movement along the curve at any point in time.

In the exercise provided, the parametric equations are \(x = 2 \cos t\) and \(y = 2 \sin t\), which describe a circle of radius 2. The parameter \(t\) represents the angle in radians, measured from the positive x-axis. As \(t\) changes, the value of \(x\) and \(y\) also change, tracing out the circle in the Cartesian plane.

When dealing with parametric functions, the derivative is critical for understanding how the function changes with respect to the parameter, which is often time. The derivative of parametric functions involves finding the rate of change of both the x- and y-coordinates with respect to the parameter t.

The derivatives \(dx/dt\) and \(dy/dt\) represent the slopes of the functions with respect to t. To find the slope of the tangent line to the parametric curve at a particular point, one must divide \(dy/dt\) by \(dx/dt\) to obtain \(dy/dx\). This quotient is the slope of the tangent line to the curve at the given point. In the given exercise, after differentiating the functions for x and y with respect to t, and evaluating at \(t = \pi/4\), we get the derivative that corresponds to the slope of the tangent line at that specific point.

The point-slope form of a line is essential for writing the equation of a line when you know a single point on the line and its slope. This form is expressed as \(y - y_1 = m(x - x_1)\), where (x₁, y₁) is the point the line passes through, and m is the slope of the line.

After determining the slope of the tangent line from the parametric derivative and the coordinates of the point of tangency by substituting the given parameter value into the parametric equations, you can apply these values to the point-slope form to write the equation of the tangent line. In the solution to the exercise, the slope of the tangent line is found to be -1, and the point of tangency is \((\sqrt{2},\sqrt{2})\). Using the point-slope form, the equation of the tangent line at \(t = \pi/4\) is then computed as \(y = -x + 2\sqrt{2}\).