Problem 25 - Thomas Calculus

Question

Motion along a parabola A particle moves along the top of the parabola \(y^{2}=2 x\) from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point \((2,2) .\)

Step-by-Step Solution

Answer

Velocity at (2,2) is \( (2\sqrt{5}, \sqrt{5}) \).

Step 1: Parameterize the Motion

The problem requires us to find the velocity of a particle moving along the parabola described by the equation \( y^2 = 2x \). First, let's express both \(x\) and \(y\) as functions of a parameter \(t\), which represents time. We calculate \(x\) and \(y\) using the given speed of 5 units.Use the relationship \( y = \sqrt{2x} \). By differentiating with respect to time \( t \), the velocity vector in parametric form is \( \frac{dy}{dt} \) along the y-direction. From the condition that speed is 5, i.e., \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = 5 \).

Step 2: Find Derivatives

To find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we use the derivative of the curve:1. Differentiate both sides of \( y^2 = 2x \): \( 2y \frac{dy}{dt} = 2 \frac{dx}{dt} \).2. Simplifying gives \( y \frac{dy}{dt} = \frac{dx}{dt} \).The velocity components can be expressed in terms of \( y \).

Step 3: Use the Speed Condition

Remember the speed equation, \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = 5 \). Substitute \( \frac{dx}{dt} = y \frac{dy}{dt} \) into the speed equation:\[ \sqrt{(y \frac{dy}{dt})^2 + (\frac{dy}{dt})^2} = 5 \]Simplifying:\[ |\frac{dy}{dt}| \sqrt{y^2 + 1} = 5 \]

Step 4: Solve for \( \frac{dy}{dt} \) at \( (2, 2) \)

At the point \( (2,2) \), substitute \( y=2 \) into the equation:\[ |\frac{dy}{dt}| \sqrt{2^2 + 1} = 5 \] \[ |\frac{dy}{dt}| \cdot \sqrt{5} = 5 \] Solving gives:\[ |\frac{dy}{dt}| = \frac{5}{\sqrt{5}} = \sqrt{5} \]

Step 5: Find \( \frac{dx}{dt} \) at \( (2, 2) \)

Substitute \( \frac{dy}{dt} = \sqrt{5} \) back into \( y \frac{dy}{dt} = \frac{dx}{dt} \): \[ \frac{dx}{dt} = 2 \cdot \sqrt{5} = 2\sqrt{5} \]

Step 6: Write the Velocity Vector

The velocity vector \( \mathbf{v} \) of the particle at the point \( (2,2) \) is \( \mathbf{v} = (\frac{dx}{dt}, \frac{dy}{dt}) = (2\sqrt{5}, \sqrt{5}) \). Both components have units of units per second, so the velocity vector of the particle at this point is: \[ \mathbf{v} = (2\sqrt{5}, \sqrt{5}) \]

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that make up a geometric object, such as a curve, using one or more parameters. In the context of our problem, the motion of a particle along a parabola is better expressed using parametric equations.
To do this, we introduce a parameter, often denoted as \( t \), which represents time. Each coordinate, \( x \) and \( y \), is then written as a function of \( t \).

• For the given parabola \( y^2 = 2x \), we can express \( y \) in terms of \( x \) as \( y = \sqrt{2x} \).
• This allows us to describe the coordinates of the particle as functions of time, i.e., \( x(t) \) and \( y(t) \).

Parametric equations are especially useful when we want to analyze motion or paths, as they allow us to track a point's position with respect to time.

Velocity

Velocity refers to the rate of change of an object's position with respect to time, not just in terms of how fast the particle is moving, but also its direction. In our problem, we find the velocity of a particle moving along a parabola.
To determine the velocity of the particle at a specific point, we need to calculate both components:

• \( \frac{dx}{dt} \): the change in \( x \) with respect to time.
• \( \frac{dy}{dt} \): the change in \( y \) with respect to time.

The velocity vector \( \mathbf{v} \) at any point \( (x,y) \) is given by \( \mathbf{v} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) \). It provides a comprehensive description of the state of motion of the particle.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a quantity changes. In the context of particle motion along a curve, differentiation allows us to derive expressions for velocity components.
To calculate the velocity of the particle moving along the parabola, we perform the following:

• Differentiating the equation of the parabola, \( y^2 = 2x \), with respect to time \( t \) yields \( 2y \frac{dy}{dt} = 2 \frac{dx}{dt} \).
• This simplifies to \( y \frac{dy}{dt} = \frac{dx}{dt} \), which relates the derivatives of \( x \) and \( y \) with respect to \( t \).

Finding these derivatives at a specific point provides the components needed to form the velocity vector.

Particle Motion

Particle motion studies describe the path that a particle takes and its various associated properties, such as speed and velocity, over time. In the given problem, we examine the motion of a particle along the specific path, which is the top of a parabola.
For this particle, the constant speed of 5 units per second is given, which implies that the magnitude of the vector defined by its velocity components must equal 5:
\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = 5 \]
The solution involves using this condition to compute the exact values of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) at the point \((2,2)\). Consequently, this enables us to describe how fast and in which direction the particle is moving at the moment it passes through this point on the parabola.