91Ó°ÊÓ

When we talk about parametric equations, we're diving into a way of using parameters to describe curves, which is pretty cool if you think about it! Instead of using the usual format, like the standard y = f(x), parametric equations let us express both x and y as functions of a third variable, usually denoted by t. This is particularly handy for describing complex curves that can't be easily represented with just x and y.

In the problem given, we have the parametric equations:

• \( x = t^3 - t \)
• \( y = t^2 \)

Each of these equations tells us how x and y change as the parameter t changes. By manipulating t, we can trace out the path of a curve on a graph, allowing us to understand and analyze the curve's behavior at various points.

Calculating derivatives is a big part of finding tangent lines, especially with parametric equations. Derivatives tell us how a function changes at any given point, which is crucial when determining the slope of a tangent line.

Here are the steps we followed in the exercise:

• For \(x = t^3 - t\), the derivative \(\frac{dx}{dt}\) is calculated as \(3t^2 - 1\). This derivative shows how the x-coordinate changes as t varies.
• For \(y = t^2\), the derivative \(\frac{dy}{dt}\) is calculated as \(2t\). This represents how the y-coordinate changes with t.

These derivatives are essential for the next step: finding the slope of the curve using these rates of change. Remember, the process of differentiation uncovers the rate at which one quantity changes compared to another.

The slope of a curve at a particular point tells us how steep the curve is at that point. For parametric equations, the slope of the tangent line is calculated using a ratio of the derivatives from our previous section.

To find the slope (m) of the curve, we divided \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\). So, the formula becomes:

• \( m = \frac{dy/dt}{dx/dt} = \frac{2t}{3t^2 - 1} \)

By substituting \(t = 2\), we find the specific slope to be \(\frac{4}{11}\). This value tells us how the y-value changes compared to the x-value at the point where \(t = 2\). It's like finding out the angle at which the curve climbs or descends at any particular point.

Finally, to write the equation of the tangent line, we use the point-slope form. This is a super useful formula for writing equations of lines when you know the slope and a specific point on the line.

The general point-slope form is given by:

• \( y - y_1 = m(x - x_1) \)

Where \((x_1, y_1)\) is the point on the line, and m is the slope. For our problem, we knew the point at \( t = 2 \) was \((6, 4)\) and the slope was \(\frac{4}{11}\).

By plugging these values into the point-slope form, the equation became \( y - 4 = \frac{4}{11}(x - 6) \). After simplifying, we got the neat final equation of the tangent line: \( y = \frac{4}{11}x + \frac{26}{11} \). Using the point-slope form is like having a shorthand recipe to quickly figure out the formula of the line.