Question

Parametric equations for a curve are given. (a) Find \(\frac{d y}{d x}\). (b) Find the equations of the tangent and normal line(s) at the point(s) given. (c) Sketch the graph of the parametric functions along with the found tangent and normal lines. \(x=t^{2}-1, y=t^{3}-t ; \quad t=0\) and \(t=1\)

Short answer

\(\frac{d y}{d x}\) at \(t=0\) is undefined; at \(t=1\), it is 1. Tangents: \(x=-1\) (\(t=0\)), \(y=x\) (\(t=1\)); Normals: \(y=0\) (\(t=0\)), \(y=-x\) (\(t=1\)).

Step-by-step solution

Differentiate x and y with respect to t

We start by differentiating the parametric equations with respect to the parameter \( t \). For \( x = t^2 - 1 \), the derivative is \( \frac{dx}{dt} = 2t \).For \( y = t^3 - t \), the derivative is \( \frac{dy}{dt} = 3t^2 - 1 \).

Use the chain rule to find \( \frac{dy}{dx} \)

To find \( \frac{dy}{dx} \), we use the chain rule, which gives the formula: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 1}{2t}. \]

Evaluate \( \frac{dy}{dx} \) at the given t-values

Substitute \( t = 0 \) into \( \frac{dy}{dx} \):\[ \frac{dy}{dx} \bigg|_{t=0} = \frac{3(0)^2 - 1}{2(0)} = \text{undefined (since division by zero)}. \]Substitute \( t = 1 \) into \( \frac{dy}{dx} \):\[ \frac{dy}{dx} \bigg|_{t=1} = \frac{3(1)^2 - 1}{2(1)} = \frac{2}{2} = 1. \]

Find the equations of the lines at t = 0 and t = 1

First, find the point on the curve at \( t = 0 \):\( x = 0^2 - 1 = -1 \) and \( y = 0^3 - 0 = 0 \)Since \( \frac{dy}{dx} \) is undefined, the tangent line is vertical, so the equation is \( x = -1 \).At \( t = 1 \):\( x = 1^2 - 1 = 0 \) and \( y = 1^3 - 1 = 0 \)The slope \( \frac{dy}{dx} = 1 \), thus the equation of the tangent line is:\[ y - 0 = 1(x - 0), \text{ which reduces to } y = x. \]

Find the equations of the normal lines at t = 0 and t = 1

The normal line at any point is perpendicular to the tangent line, with slope \( -\frac{1}{m} \) where \( m \) is the slope of the tangent line.At \( t = 0 \), the normal line has an undefined slope (horizontal line), so the equation is \( y = 0 \).At \( t = 1 \):Since the slope of the tangent line is \( 1 \), the slope of the normal line is \( -1 \).Thus, the equation of the normal line is:\[ y - 0 = -1(x - 0), \text{ which reduces to } y = -x. \]

Sketch the graph

Sketch the parametric curve defined by \( x = t^2 - 1 \) and \( y = t^3 - t \).Locate and draw the point at \( t = 0 \) at \((-1, 0)\) with a vertical tangent line \( x = -1 \) and horizontal normal line \( y = 0 \).Draw the point at \( t = 1 \) at \((0, 0)\) with the tangent line \( y = x \) and normal line \( y = -x \).

Key concepts

Tangent Line

The tangent line to a curve at a given point is a straight line that "just touches" the curve at that point. In mathematical terms, it represents the limit of the slope of the curve as it approaches the point of tangency.

To find the equation of a tangent line when dealing with parametric equations, first find the derivative \( \frac{dy}{dx} \). For a given parameter \( t \), compute the x and y coordinates, and then use \( \frac{dy}{dx} \) to find the slope of the tangent line. At \( t = 0 \), if the derivative \( \frac{dy}{dx} \) is undefined, the tangent line is vertical, such as \( x = -1 \). At \( t = 1 \), the slope is defined, leading to the equation \( y = x \) for the tangent line, since \( \frac{dy}{dx} = 1 \).

Checking the tangent line's behavior is crucial when you’re sketching graphs or understanding the distinct properties of curves at specific points.

Normal Line

The normal line to a curve at a particular point is perpendicular to the tangent line at that same point. If you know the slope of the tangent line, the slope of the normal line is the negative reciprocal, unless the tangent line is vertical or horizontal.

For parametric equations, once you find the slope of the tangent line, you can easily derive the equation for the normal line. At \( t = 0 \), with a vertical tangent, the normal line is horizontal, meaning the slope is zero, giving \( y = 0 \). At \( t = 1 \), with a tangent line slope of \( 1 \), the normal line has a slope of \( -1 \), leading to the equation \( y = -x \).

This perpendicular relationship ensures that tangent and normal lines provide a comprehensive view of the curve's behavior at crucial points.

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating compositions of functions. It is particularly useful in working with parametric equations where one differentiates both x and y with respect to a different, common parameter, often denoted by \( t \).

For example, if x is defined in terms of \( t \) as \( x(t) \) and y as \( y(t) \), the derivative \( \frac{dy}{dx} \) can be obtained by dividing the derivative \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \). This uses the formula: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] In the given exercise, applying the chain rule allows us to find \( \frac{dy}{dx} = \frac{3t^2 - 1}{2t} \).

By mastering the chain rule, you gain the ability to handle more complex rate-of-change problems across different fields of mathematics and physics.

Derivative

A derivative represents the rate at which one quantity changes with respect to another. In parametric equations, it reflects how the curve described by the parameter \( t \) behaves as \( t \) changes.

In the context of the exercise, finding the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) is the first step. For the functions \( x = t^2 - 1 \) and \( y = t^3 - t \), these derivatives help in evaluating \( \frac{dy}{dx} \) which is \( \frac{3t^2 - 1}{2t} \). At particular points, like \( t = 0 \) and \( t = 1 \), the derivative helps determine specific behavior, such as vertical tangency at \( t = 0 \).

Understanding derivatives and their calculation is crucial for interpreting and predicting the behavior of curves, which forms the base for advanced calculus problems.