Question

Find parametric equations for the given rectangular equation using the parameter \(t=\frac{d y}{d x} .\) Verify that at \(t=1,\) the point on the graph has a tangent line with slope of 1. \(y=3 x^{2}-11 x+2\)

Short answer

Parametric equations are \( x = \frac{t+11}{6} \) and \( y = \frac{3(t + 11)^2}{36} - \frac{11(t + 11)}{6} + 2 \). At \( t=1 \), the tangent slope is indeed 1.

Step-by-step solution

Find dy/dx

First, we differentiate the rectangular equation with respect to x to find \( \frac{dy}{dx} \). Given the equation \( y = 3x^2 - 11x + 2 \), the derivative is \( \frac{dy}{dx} = 6x - 11 \).

Set Up Expression for x in Terms of t

Since \( t = \frac{dy}{dx} \), we have \( t = 6x - 11 \). Solving for x gives \( x = \frac{t + 11}{6} \).

Express y in Terms of t

Use the expression for x derived in Step 2 to express y in terms of t. Substitute \( x = \frac{t + 11}{6} \) into the original equation \( y = 3x^2 - 11x + 2 \). This leads to:\[y = 3\left(\frac{t + 11}{6}\right)^2 - 11\left(\frac{t + 11}{6}\right) + 2\]

Simplify y(t)

Expand and simplify the expression for y:\[y = \frac{3(t + 11)^2}{36} - \frac{11(t + 11)}{6} + 2\]Simplify further to get:\[ y = \frac{(t + 11)^2}{12} - \frac{11(t + 11)}{6} + 2 \].

Verify Tangent Slope at t=1

Check that at \( t=1 \), the slope of the tangent line is 1. From \( x = \frac{t+11}{6} \), when \( t=1 \), we have \( x = 2 \). Evaluate \( y \) using the simplified expression from Step 4 to find the corresponding y-coordinate. Substitute x back into \( \frac{dy}{dx} = 6x - 11 \) to confirm the slope of the tangent line at this point is \( t = \frac{dy}{dx} = 1 \).

Key concepts

Rectangular Equation

A rectangular equation, also known as a Cartesian equation, is a common way of representing a function or relation between two variables in the XY-plane using algebraic expressions. The equation is referred to as "rectangular" because it is described by standard rectangular coordinates (x, y) in the Cartesian plane.
A typical example, like the equation given in the exercise, is of the form:

• \(y = 3x^2 - 11x + 2\)

Here, \(y\) is expressed solely in terms of \(x\), forming a traditional function format. Rectangular equations are useful for plotting graphs and solving super straightforward algebraic problems.
They are the starting point from which other forms of equations, like parametric equations or polar equations, can be derived to explore the properties and behaviors of curves more fully.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. Specifically, it provides the slope of the tangent line to the curve of the function at any given point. In other words, the derivative tells us how steep the graph is at any point.
In the exercise, the derivative of our rectangular equation \(y = 3x^2 - 11x + 2\) with respect to \(x\) is calculated as:

• \(\frac{dy}{dx} = 6x - 11\)

This means, for any specific value of \(x\), the slope of the function at that point can be calculated by plugging \(x\) into this expression. The derivative is essential for understanding the instantaneous rate of change and is crucial in finding the tangent lines to curves.

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point and has the same slope as the curve at that point. The tangent line is vital in calculus, as it approximates the curve locally, providing insight into the curve's behavior around that particular point.
For the parametric equations derived from the original rectangular form, we needed to confirm the slope of the tangent line at the parameter\( t = 1 \). We knew that this line's slope should equal 1, making it crucial to determine if our parametric derivation aligns with this.

• At \(t=1\), we found \(x = 2\).
• Verifying using the derivative, \(\frac{dy}{dx} = 6 \times 2 - 11 = 1\).

This equality confirms the slope of the tangent line at \(t=1\) indeed equals 1, indicating a successful calculation of the tangent line for the derived parametric equations.

Parametrization

Parametrization involves representing a curve by a set of equations for each coordinate in terms of a third variable, usually \(t\), known as the parameter. This technique is particularly useful in capturing the direction and motion along curves.
In our exercise, we started with the rectangular equation and converted it into parametric form using:

• \(t = \frac{dy}{dx}\)
• Solving gave us \(x = \frac{t + 11}{6}\)
• We then expressed \(y\) using \(x\) in terms of \(t\):
• \(y = \frac{(t + 11)^2}{12} - \frac{11(t + 11)}{6} + 2\)

Parametrization is beneficial because it simplifies analyzing complex curves, allows for easier graph representation, and helps understand the behavior of a curve over an interval. By converting a function into parametric form, you gain concise insight into how the x and y coordinates change with respect to an independent third variable, \(t\).