Question

Calculate the derivative \(d y / d x\) for the plane curve defined by the equations $$ x(t)=t^{2}-4 t, \quad y(t)=2 t^{3}-6 t, \quad-2 \leq t \leq 3 $$ and locate any critical points on its graph.

Short answer

The derivative is \( \frac{3(t-1)(t+1)}{t-2} \). Critical points are at \( t = -1, 1, 2 \), corresponding to points \((3, 4), (-3, -4), (0, 0)\) on the graph.

Step-by-step solution

Parametric Derivatives Setup

Identify the parametric equations for the curve: \( x(t) = t^2 - 4t \) and \( y(t) = 2t^3 - 6t \). We need to find \( \frac{dy}{dx} \) by first finding the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Calculate dx/dt

Differentiate \( x(t) = t^2 - 4t \) with respect to \( t \). The derivative is \( \frac{dx}{dt} = 2t - 4 \).

Calculate dy/dt

Differentiate \( y(t) = 2t^3 - 6t \) with respect to \( t \). The derivative is \( \frac{dy}{dt} = 6t^2 - 6 \).

Find dy/dx using Chain Rule

The derivative \( \frac{dy}{dx} \) is found using \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substituting the derivatives:\( \frac{dy}{dx} = \frac{6t^2 - 6}{2t - 4} \).

Simplify dy/dx

Factor out common terms from \( \frac{dy}{dt} \) to simplify: \( 6(t^2 - 1) = 6(t - 1)(t + 1) \).Substitute back into the expression, \( \frac{dy}{dx} = \frac{6(t-1)(t+1)}{2(t-2)} \). This simplifies to \( 3 \cdot \frac{(t-1)(t+1)}{t-2} \).

Determine Critical Points

Critical points occur when \( \frac{dy}{dx} = 0 \) or is undefined.\( \frac{dy}{dx} = 0 \) when \( (t-1)(t+1) = 0 \), giving \( t = 1 \) and \( t = -1 \).\( \frac{dy}{dx} \) is undefined when \( t - 2 = 0 \), giving \( t = 2 \).The critical points are at \( t = -1, 1, 2 \).

Locate Critical Points on Graph

Substitute \( t = -1, 1, 2 \) back into the original parametric equations to get the points on the plane curve:- For \( t = -1 \): \((x, y) = (3, 4) \)- For \( t = 1 \): \((x, y) = (-3, -4) \)- For \( t = 2 \): \((x, y) = (0, 0) \)

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. When dealing with parametric equations, it is an essential tool. Instead of expressing a curve in the traditional \(x\) and \(y\) form, we define each variable as a function of another variable, usually denoted as \(t\).
The chain rule helps us find the derivative of one variable with respect to another by linking the derivatives with respect to \(t\). Specifically, if we have parametric equations \(x(t)\) and \(y(t)\), the derivative of \(y\) with respect to \(x\), \(\frac{dy}{dx}\), can be computed using:

• Find \(\frac{dx}{dt}\) by differentiating \(x(t)\) with respect to \(t\).
• Find \(\frac{dy}{dt}\) by differentiating \(y(t)\) with respect to \(t\).
• Use the chain rule formula \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

This formula aligns the rates of change of the parametric functions, allowing us to study how \(y\) changes with respect to \(x\).

Critical Points

Critical points on a curve are where the derivative \(\frac{dy}{dx}\) is either undefined or equal to zero. These points are crucial because they often indicate where the curve has a horizontal tangent or a vertical cusp.
To find these critical points:

• Set \(\frac{dy}{dx}\) equal to zero and solve for \(t\). This gives us the points where the tangent is horizontal.
• Find when \(\frac{dy}{dx}\) is undefined, typically when the denominator \(\frac{dx}{dt}\) is zero. This gives us possible vertical tangents or cusps.

After determining the \(t\) values for critical points, we substitute them back into the original parametric equations to find their corresponding \(x, y\) coordinates on the plane, giving us a complete picture of the behavior of the curve at these points.

Parametric Equations

Parametric equations are a powerful way to represent plane curves, especially when the curve is complicated or does not define \(y\) as a single function of \(x\).
Unlike the standard format, where \(y\) is typically a function of \(x\), in parametric equations both \(x\) and \(y\) are expressed in terms of a parameter \(t\).

• The equation \(x(t) = t^2 - 4t\) provides the \(x\)-coordinates for any given \(t\).
• Similarly, \(y(t) = 2t^3 - 6t\) determines the \(y\)-coordinates.

This representation allows more flexibility, such as representing curves that loop back on themselves or have cusps, which would be difficult to express in a standard function form. Parametric equations are particularly useful in animations or simulations where the curve's traversal is time-dependent.

Derivatives

In calculus, derivatives represent the rate of change of a function. When working with parametric equations, we need to determine how one variable changes with another, often as \(\frac{dy}{dx}\).
The derivatives for parametric curves involve two main steps:

• Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to get how \(x\) and \(y\) change with \(t\).
• Combine these using the chain rule: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

This gives us the slope of the tangent to the curve at any point \(t\). Understanding derivatives in this context helps us analyze motion along a curve, such as speed and direction. Calculating derivatives accurately is key to understanding the underlying dynamics of any parametrically defined curve.

Plane Curves

Plane curves are curves that lie on a single plane, described using mathematical equations. These curves can be represented in different forms such as implicit, explicit, or parametric.
Using parametric equations, plane curves can capture complex behaviors that are hard to express otherwise. They can show more intricate paths, like loops and intersections, within a clear, mathematical framework. By choosing a range for the parameter \(t\), we can define what portion of the curve is visible.
The study of plane curves involves:

• Finding derivatives to understand tangent lines and slopes.
• Identifying critical points to learn about the curve's features.
• Graphing the curve using calculated \(x, y\) points.

Understanding plane curves is essential in fields ranging from physics to computer graphics, providing critical insights into both natural phenomena and engineered systems.