91Ó°ÊÓ

Parametric equations are a way to define a line or curve by using parameters. Instead of describing the curve in terms of only x and y, we use another variable, often called t, to determine the position of a point on the curve.
This is especially helpful when describing movements or paths that change with another factor—like time.
In our exercise, we're given two parametric equations:

• \( x = t \ln t \)
• \( y = \sin^2 t \)

These equations define a path as t varies.
To find points on the curve, simply substitute t values into these equations. For instance, when \( t = \frac{\pi}{4} \), you plug this into both equations to get specific x and y values.
This makes \( (x, y) \) descriptions in terms of \( t \), providing a dynamic view of the path traced on a graph.

Cartesian coordinates are a system that uses two values, usually labeled as x and y, to identify a specific point on a 2D plane.
They are very familiar because they allow us to describe locations in an easy-to-visualize format, where horizontal and vertical movements are tracked separately.
When dealing with parametric equations, converting to Cartesian coordinates helps place points on a regular graph.
In our exercise, when \( t = \frac{\pi}{4} \), we substitute this value into the parametric equations:

• The x-coordinate is \( \frac{\pi}{4} \ln\left(\frac{\pi}{4}\right) \)
• The y-coordinate is \( \sin^2\left(\frac{\pi}{4}\right) = \frac{1}{2} \)

Hence, we obtain the Cartesian coordinates \( \left( \frac{\pi}{4} \ln\left(\frac{\pi}{4}\right), \frac{1}{2} \right) \), which represents a fixed position on the 2D plane.

Derivative calculation is crucial for finding rates of change and slopes of curves at given points.
With parametric equations, you typically need to find the derivatives of x and y with respect to the parameter t.
For our exercise:

• The parametric function for x, \( x = t \ln t \), is differentiated using the product rule, resulting in \( \frac{dx}{dt} = \ln t + 1 \).
• For y, \( y = \sin^2 t \), you use the chain rule, giving \( \frac{dy}{dt} = 2 \sin t \cdot \cos t = \sin(2t) \).

Evaluating these derivatives at \( t = \frac{\pi}{4} \) provides the rates of change for each coordinate at that particular value of t.
These derivatives form the basis for calculating the slope of the tangent line to the curve.

Point-slope form is a method to represent the equation of a line if you know a point on the line and the line's slope.
Its formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) are the coordinates of the known point.
In our example, these are:

• \( m = \frac{1}{\ln\left(\frac{\pi}{4}\right) + 1} \)
• Point: \( \left( \frac{\pi}{4} \ln\left(\frac{\pi}{4}\right), \frac{1}{2} \right) \)

Substituting into the point-slope formula provides the tangent line's equation, helpful for understanding how the curve is oriented at that specific location.
Such lines are crucial in calculus, allowing us to approximate curves locally by straight lines.