91Ó°ÊÓ

Accept as true that if a particle moves along a graph of \(y=f(x)\) then the speed of the particle along the graph is \(\sqrt{\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}}\). (The notation means that \(x^{\prime}\) is the rate at which the \(x\) -coordinate is increasing, \(y^{\prime}\) denotes the rate at which the \(y\) -coordinate is increasing and the speed of the particle is the rate at which the particle is moving along the curve.) a. Suppose a particle moves along the graph of \(y=x^{2}\) so that \(y^{\prime}=2 .\) What is \(x^{\prime}\) when \(y=2 ?\) How fast is the particle moving? b. Suppose a particle moves along the circle \(x^{2}+y^{2}=1\) so that its speed is \(2 \pi\). Find \(x^{\prime}\) when \(x=1\) c. Suppose a particle moves along the circle \(x^{2}+y^{2}=1\) so that its speed is \(2 \pi\). Find \(x^{\prime}\) when \(x=0 .\) (Two answers.)

Step by step solution

01

Differentiate the curve equation (a)

The particle moves along the graph of \(y = x^2\). Differentiate the equation with respect to time \(t\) to find \(y'\): \[ y' = \frac{d}{dt}(x^2) = 2x \cdot x' \]Since \(y' = 2\), we have:\[ 2 = 2x \cdot x' \]

02

Solve for \(x'\) (a)

We rearrange the equation \(2 = 2x \cdot x'\) to solve for \(x'\):\[ x' = \frac{1}{x} \] When \(y = 2\), substitute in \(x\) from \(y = x^2\) which gives \(x^2 = 2\), so \(x = \sqrt{2}\).

03

Calculate speed of the particle (a)

Substitute \(x = \sqrt{2}\) into the equation for \(x'\):\[ x' = \frac{1}{\sqrt{2}} \]The speed is given by the magnitude of the velocity vector:\[ \text{speed} = \sqrt{(x')^2 + (y')^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + 2^2} = \sqrt{\frac{1}{2} + 4} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \]

04

Derive the expression (b)

For the circle \(x^2 + y^2 = 1\), differentiate both sides with respect to \(t\):\[ 2x x' + 2y y' = 0 \Rightarrow x x' + y y' = 0 \]Additionally, solve for speed \(2\pi = \sqrt{(x')^2 + (y')^2}\).

05

Solve for \(x'\) when \(x = 1\) (b)

With \(x = 1\), and on the circle, \(y = 0\). From the speed equation:\[ 2\pi = |x'| \Rightarrow x' = \pm 2\pi \]Using \(x=1\) and \(y=0\), \(y'\) does not affect the equation for speed, so no additional info needed.

06

Solve for \(x'\) when \(x = 0\) (c)

For \(x = 0\), from the circle equation, \(y = \pm 1\). Substitute each \(y\) into the derived speed equation:For \(y = 1\) or \(y=-1\):- Use \(2\pi = |y'|\), since \(x' = 0\), \(x' = 0\) solves both \(x'\) alternatives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Particle Motion

When we discuss particle motion in calculus, we are typically referring to the movement of a particle along a curve defined by a function, such as a graph. This movement can occur in two-dimensional space, where the particle has both an \(x\)-coordinate and a \(y\)-coordinate that change over time.

• Each coordinate can change at different rates, which are often represented by derivatives with respect to time \(t\), known as \(x'\) and \(y'\).
• The path of the particle can be along any curve described by an equation such as \(y = x^2\) or a circle like \(x^2 + y^2 = 1\).
• Understanding how these coordinates change allows us to find the speed and direction of the particle at any point on its path.

Being able to describe particle motion mathematically is essential in various fields, such as physics and life sciences, where understanding motion is key to analyzing systems, like how organisms move or how particles interact within a biological system.

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how functions change. It gives us the tools to calculate the rate at which one quantity changes with respect to another. When applied to particle motion:

• You find the derivatives of the position coordinates \(x\) and \(y\) with respect to time \(t\), typically denoted as \(x'\) and \(y'\).
• For example, if a particle moves along \(y = x^2\), differentiating this equation with respect to \(t\) yields \(y' = 2x \cdot x'\).
• This process helps determine how quickly the \(y\)-coordinate is increasing relative to time and how it depends on \(x\) and its rate of change \(x'\).

By applying differentiation, we can derive formulas that help us analyze velocities, accelerations, and other dynamic changes in particle motion, which are crucial for predicting future behavior of systems modeled by these equations.

Velocity Vector

The velocity vector is a pivotal component in analyzing particle motion. It combines the changes in position in both the \(x\) and \(y\) directions into a single entity, making it easier to calculate the overall speed and direction of the particle.

• The velocity of a particle in two dimensions is expressed as a vector \((x', y')\).
• The magnitude of this vector, or the speed of the particle, is given by \(\text{speed} = \sqrt{(x')^2 + (y')^2}\).
• For example, if \(x' = \frac{1}{\sqrt{2}}\) and \(y' = 2\) for a particle moving along \(y = x^2\), the speed is \(\sqrt{\frac{1}{2} + 4} = \frac{3}{\sqrt{2}}\).

Understanding the velocity vector provides insight into how quickly and in which direction the particle is moving. This is invaluable in fields like biology and medicine, where you might model the movement patterns of cells or organisms to study their behavior.