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Parametric equations offer a way to describe a set of related quantities as functions of an independent parameter, often denoted as \( t \). This is particularly useful in physics and engineering where the position of a moving object can depend on time. Instead of expressing \( y \) directly in terms of \( x \), parametric equations describe both \( x \) and \( y \) separately as:

• \( x = f(t) \)
• \( y = g(t) \)

Each different value of \( t \) gives a point \((x, y)\) on the curve. For instance, in the problem, \( x = 1 + \ln t \) and \( y = t^2 + 2 \). These equations define a curve in the \( xy \)-plane as \( t \) varies.
This form is powerful since it can express more complex relationships than simple \( y = f(x) \), particularly when dealing with curves that loop back on themselves or have other complications.

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It describes the curve's direction at that precise spot, acting like a local linear approximation of the curve. To find a tangent line, you need:

• A point on the curve, \((x_1, y_1)\)
• The slope of the curve at that point

The slope is essentially the rate of change of \( y \) with respect to \( x \) at that point. Using the point-slope form, the equation of a tangent line is: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope. For example, here we found the tangent line at the point \((1, 3)\). Once you find the slope from the derivative, you can use the point-slope form to write the equation of the tangent.

The derivative in calculus gives the slope of a function at any point, crucial for drawing tangent lines to curves. When you have parametric equations, the derivative \( \frac{dy}{dx} \) captures how \( y \) changes with \( x \) and can be found using:

• Calculate \( \frac{dy}{dt} \), the rate of change of \( y \) with respect to \( t \).
• Calculate \( \frac{dx}{dt} \), the rate of change of \( x \) with respect to \( t \).
• Combine them using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

In this exercise, we computed these derivatives to obtain \( 2t^2 \) at the given \( t \), finding that the slope was 2 at \( t = 1 \). Therefore, the derivative reveals the steepness and direction of the tangent line at a specific point on the curve.

Sometimes, it can be helpful to eliminate the parameter \( t \) and find a relationship directly between \( x \) and \( y \). This process makes it possible to express the curve as a standard function \( y = f(x) \). By eliminating \( t \), we simplify our analysis when focusing just on the \( x \)-\( y \) plane.
Here's how to eliminate a parameter:

• Solve one of the parametric equations for \( t \) in terms of \( x \) or \( y \).
• Substitute this expression into the other equation.

For our problem, we found \( t = e^{x-1} \) from \( x = 1 + \ln t \). Then plugging this into \( y = t^2 + 2 \) gives \( y \) as a function of \( x \). Eliminating parameters is a valuable tool in calculus as it lets you work more easily with the mathematical relationships in a straightforward \( y = f(x) \) form.