Question

For the following exercises, find points on the curve at which tangent line is horizontal or vertical.\(x=t\left(t^{2}-3\right), \quad y=3\left(t^{2}-3\right)\)

Short answer

Horizontal at (0, -9); Vertical at (-2, -6) and (2, -12).

Step-by-step solution

Understand Tangent Line Conditions

The tangent line to a curve is horizontal when its slope is zero. For a parametric curve, this happens when the derivative of the y-component with respect to the parameter is zero while the x-derivative is non-zero. It is vertical when the derivative of the x-component is zero and the y-derivative is non-zero.

Calculate Derivatives

Calculate the derivatives of the given parametric equations with respect to parameter \(t\):\[\frac{dx}{dt} = \frac{d}{dt}[t(t^2 - 3)] = 3t^2 - 3\]\[\frac{dy}{dt} = \frac{d}{dt}[3(t^2 - 3)] = 6t\]

Find Horizontal Tangents

Set \(\frac{dy}{dt}\) to zero and solve for \(t\):\[6t = 0 \Rightarrow t = 0\]To ensure it's horizontal, check that \(\frac{dx}{dt} eq 0\) for \(t = 0\):\[\frac{dx}{dt} = 3(t^2 - 1) = 3(0^2 - 1) = -3 eq 0\]

Find Vertical Tangents

Set \(\frac{dx}{dt}\) to zero and solve for \(t\):\[3(t^2 - 1) = 0 \Rightarrow t^2 = 1 \Rightarrow t = \pm 1\]To ensure it's vertical, check that \(\frac{dy}{dt}\) is non-zero:For \(t = 1\), \[\frac{dy}{dt} = 6(1) = 6 eq 0\]For \(t = -1\), \[\frac{dy}{dt} = 6(-1) = -6 eq 0\]

Calculate Points on the Curve

Find the coordinates for each \(t\):For \(t = 0\), \[x = 0(0^2 - 3) = 0, \quad y = 3(0^2 - 3) = -9\]For \(t = 1\), \[x = 1(1^2 - 3) = -2, \quad y = 3(1^2 - 3) = -6\]For \(t = -1\), \[x = -1(-1^2 - 3) = 2, \quad y = 3(-1^2 - 3) = -12\]

Key concepts

Parametric Equations

Parametric equations express a set of related quantities, where each quantity is defined as a function of an independent parameter, usually denoted as \( t \). In the context of the problem given, the equations describe a curve in the plane through two expressions: \( x = t(t^2 - 3) \) and \( y = 3(t^2 - 3) \). Here, \( t \) acts as the parameter that varies the position on the curve.

• Understanding the Structure: Unlike Cartesian equations which directly relate \( x \) and \( y \), parametric equations describe each coordinate separately as functions of \( t \).
• Benefits: Parametric forms are especially useful in modeling scenarios where the path of motion is complex, making it easy to describe curves and dynamics.
• Applications: These equations often appear in physics and engineering to explain trajectories, where time \( t \) acts as a natural parameter.

Tangent Lines

A tangent line to a curve is a straight line that just "touches" the curve at a specific point. It matches the slope or direction of the curve at that point and does not cross the curve about that point.

• Horizontal and Vertical Tangents: By examining the slope of the tangent line, we can identify points where the tangent transitions from sloped to horizontal or from sloped to vertical.
• Slope in Parametric Curves: The slope of the tangent line at any point is given by the division of the change in \( y \) by the change in \( x \), expressed as \( \frac{dy/dt}{dx/dt} \). This ratio finds the instantaneous rate of change in \( y \) relative to \( x \) at the parameter \( t \).
• Interpreting Conditions: For a horizontal tangent, \( \frac{dy}{dt} = 0 \) and \( \frac{dx}{dt} eq 0 \) must hold. For a vertical tangent, \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} eq 0 \).

Derivatives

Derivatives represent the rate at which a function is changing at any given point and are fundamental in calculus, especially when dealing with curves.

• Calculating Derivatives: In our exercise, to determine the derivatives for \( x \) and \( y \) with respect to \( t \), we differentiate the expressions for \( x(t) \) and \( y(t) \). This gives us \( \frac{dx}{dt} = 3t^2 - 3 \) and \( \frac{dy}{dt} = 6t \).
• Purpose: These derivatives help identify points where the curve changes direction or shape, such as where horizontal or vertical tangents occur.
• Solving for Critical Points: By setting \( \frac{dy}{dt} \) or \( \frac{dx}{dt} \) to zero, we determine \( t \) values where these transitions occur, thus finding horizontal and vertical tangent points on the curve.