91Ó°ÊÓ

Parametric equations are a way to express a set of related quantities as explicit functions of a parameter.
In this exercise, we're given the equations \(x = t^2\) and \(y = t^3\). These equations allow each variable, \(x\) and \(y\), to depend on a third variable, \(t\).
This method provides an effective way to describe curves that might be difficult to define with a standard function like \(y = f(x)\). For example, the route an object travels through a wind path can be described more naturally in parametric form.

To find the position of the object at different moments in time, simply substitute specific values of \(t\) into the equations. Here, when \(t = 2\), substituting in these equations gives us the point \((4, 8)\) on the plane.

Determining the slope of a tangent line to a curve at a specific point is crucial for understanding how the curve behaves at that point.
For parametric curves, this involves finding the rate of change of \(y\) with respect to \(x\), which is defined by the ratio \(\frac{dy}{dx}\).

In this exercise, the slope is obtained using the derivatives of the given parametric equations.

• Differentiate \(x = t^2\) to get \(\frac{dx}{dt} = 2t\).
• Differentiate \(y = t^3\) to get \(\frac{dy}{dt} = 3t^2\).

These derivatives reflect how \(x\) and \(y\) move with respect to \(t\). The actual slope of the tangent line becomes \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3t}{2}\).
Evaluating this at \(t = 2\) results in a slope of 3, which gives us the steepness of the line.

Understanding how to calculate derivatives is essential for finding the slope of tangent lines for parametric equations.
Let's take a closer look at how derivatives were computed in this example.

For \(x = t^2\), apply the basic power rule for differentiation. The derivative \(\frac{dx}{dt} = 2t\) arises because the power rule simplifies differentiation to multiplying by the power and reducing it by one.
For \(y = t^3\), using the same power rule results in \(\frac{dy}{dt} = 3t^2\).

These steps reveal how variables change in relation to the parameter \(t\), enabling us to measure how one quantity changes relative to another through the use of parametric equations. Calculating these derivatives correctly is pivotal as they directly impact the accurate calculation of the slope of the tangent line.

Finding the equation of a tangent line involves applying the point-slope form of a line, which is a straightforward way to illustrate the tangent at a particular point on the curve.
In this exercise, with the known slope of 3 and the point \((4, 8)\) from the curve, the equation is constructed as follows:

The point-slope formula for a line is: \(y - y_0 = m(x - x_0)\), where \((x_0, y_0)\) is the known point and \(m\) is the slope.
Substituting the values from our problem gives \(y - 8 = 3(x - 4)\).
Simplifying and solving for \(y\) yields \(y = 3x - 4\).

This equation is essential for accurately representing the direction and rate of change at the point where the tangent meets the curve, providing valuable insight into the curve's behavior at that specific location.