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. \(y=\sqrt{x}\) on \([0, \infty)\)

Short answer

The parametric equations are \( x(t) = \frac{1}{4t^2} \) and \( y(t) = \frac{1}{2t} \). At \( t=1 \), the slope is 1.

Step-by-step solution

Identify Derivative

First, we find the derivative of the rectangular equation to express the parameter \( t \). Given \( y = \sqrt{x} \), we differentiate both sides with respect to \( x \). The derivative \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).

Express x in terms of t

Since \( t = \frac{dy}{dx} \), set \( t = \frac{1}{2\sqrt{x}} \). Solve for \( x \) in terms of \( t \): \( 2\sqrt{x} = \frac{1}{t} \) implies \( \sqrt{x} = \frac{1}{2t} \), and squaring both sides gives \( x = \frac{1}{4t^2} \).

Express y in terms of t

From the relationship \( y = \sqrt{x} \), substitute the expression for \( x \) from the previous step: \( y = \sqrt{\frac{1}{4t^2}} = \frac{1}{2t} \).

Parametric Equations

The parametric equations are now fully described as \( x(t) = \frac{1}{4t^2} \) and \( y(t) = \frac{1}{2t} \).

Verify Slope at t=1

Substitute \( t = 1 \) into the parametric equations: \( x(1) = \frac{1}{4(1)^2} = \frac{1}{4} \), \( y(1) = \frac{1}{2(1)} = \frac{1}{2} \). The point is \( (\frac{1}{4}, \frac{1}{2}) \). Calculate the slope by differentiating the parametric equations and evaluating at \( t=1 \). \( \frac{dx}{dt} = -\frac{1}{2t^3} \) and \( \frac{dy}{dt} = -\frac{1}{2t^2} \). The slope of the tangent line is \( \frac{dy/dt}{dx/dt} = t \). At \( t=1 \), the slope is \( 1 \), confirming the solution.

Key concepts

Rectangular Equation

A rectangular equation is a standard way of representing a relation between variables, typically in the form of an equation like \( y = f(x) \). This standard form allows us to easily visualize the relationship on a Cartesian plane.
For instance, with the equation \( y=\sqrt{x} \) on \([0, \infty)\), this describes the curve of one half of a sideways parabola that opens to the right.
Such equations can be limiting because they express a dependent variable \( y \) in terms of an independent variable \( x \). To allow for more dynamic expressions, we use parametric forms in mathematics.

Derivative

The derivative of a function measures how a function's value changes as the input changes. It gives the rate of change or the slope of the function at any point.
For the rectangular equation \( y = \sqrt{x} \), the derivative \( \frac{dy}{dx} \) is \( \frac{1}{2\sqrt{x}} \).
This derivative tells us how steep the curve is at each point. Differentiation is a critical step before transitioning from a rectangular equation to its parametric form since it assists in defining the parameter \( t \).

Tangent Line

A tangent line is a straight line that just "touches" a curve at a particular point, matching the curve's direction (or slope) exactly at that spot.

• When we say a tangent line to the curve \( y = \sqrt{x} \) has a slope of 1 at a specific point, it means that the incline of the line matches the steepness of the curve exactly at that point.

Establishing the tangent’s slope involves using derivatives computed from the parametric equations. In this exercise, when \( t = 1 \), the tangent has a slope of 1, reflecting that it perfectly aligns with the curve at that point.

Slope

The slope of a line, or curve at a point, considers how much the value of \( y \) (output) changes when \( x \) (input) changes. It is fundamental for describing the steepness and direction.
For our parametric scenario, \( t \) is related to the slope, with \( t = \frac{dy}{dx} \). This implies the rate at which \( y \) changes compared to \( x \) can be expressed with parameter \( t \).

• The specific calculation of slope in this instance utilizing parametric equations involves the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).

Therefore, the slope \( \frac{dy/dt}{dx/dt} = t \) reveals the relationship between the rate changes of \( x \) and \( y \).

Parametrization

Parametrization involves the expression of equations in terms of a third variable, usually \( t \), known as the parameter. It provides a more flexible way to describe curves and shapes that may not be possible with simple rectangular equations.
By parametrizing, you can represent both \( x \) and \( y \) as independent functions of \( t \). For example, in this exercise, \( x(t) = \frac{1}{4t^2} \) and \( y(t) = \frac{1}{2t} \) express the same relationship originally indicated by \( y = \sqrt{x} \).
With parametrization, it's easier to describe complex shapes such as circles or ellipses, and it also simplifies computations in calculus, such as integration and finding tangents.