Problem 39 - 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

The velocity of the particle at point (2, 2) is \( \langle 2, 1 \rangle \).

Step 1: Understanding the Path Equation

We have the equation of the path that the particle is moving along: \( y^2 = 2x \). This is a parabolic curve. We need to determine the component form of velocity on this path.

Step 2: Implicit Differentiation

Differentiate both sides of the equation \( y^2 = 2x \) with respect to time \( t \). The differentiation yields \( 2y \frac{dy}{dt} = 2 \frac{dx}{dt} \). Simplifying, we have \( y \frac{dy}{dt} = \frac{dx}{dt} \) which relates the x and y components of velocity.

Step 3: Evaluating at the Given Point

At the point (2, 2), substitute \( y = 2 \) into the differentiated equation. Then \( 2 \frac{dy}{dt} = \frac{dx}{dt} \).

Step 4: Using Constant Speed

The magnitude of the velocity vector is given as 5 units per second. So \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = 5 \).

Step 5: Solving the System of Equations

Substitute \( \frac{dx}{dt} = 2 \frac{dy}{dt} \) into \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = 5 \). You get:\[\sqrt{(2 \frac{dy}{dt})^2 + (\frac{dy}{dt})^2} = 5\]This simplifies to \( \sqrt{5 (\frac{dy}{dt})^2} = 5 \), leading to \( \sqrt{5} |\frac{dy}{dt}| = 5 \). Thus, \( |\frac{dy}{dt}| = 1 \), resulting in \( \frac{dy}{dt} = 1 \) or \( \frac{dy}{dt} = -1 \).

Step 6: Determine Correct Sign for Dy/Dt

Since the particle moves from left to right, \( y \) increases as \( x \) increases. Thus \( \frac{dy}{dt} = 1 \).

Step 7: Calculate Velocity Components

Use \( \frac{dy}{dt} = 1 \) in \( \frac{dx}{dt} = 2 \frac{dy}{dt} \) to get \( \frac{dx}{dt} = 2 \). Hence, the velocity vector is \( (\frac{dx}{dt}, \frac{dy}{dt}) = (2, 1) \).

Step 8: Finalize the Particle's Velocity

Combine results for the velocity components. Therefore, the velocity at the point (2, 2) is \( \textbf{v} = \langle 2, 1 \rangle \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique used to differentiate equations where the variables are intertwined, rather than being isolated. In this exercise, the path of the particle is given by the equation \( y^2 = 2x \). This equation involves both \( y \) and \( x \) in a way that makes solving for one in terms of the other challenging. To find how each coordinate changes with time as the particle moves, we apply implicit differentiation with respect to \( t \).

• Differentiate both sides of the equation: \( 2y \frac{dy}{dt} = 2 \frac{dx}{dt} \).
• Solve for the relationship between the derivatives: \( y \frac{dy}{dt} = \frac{dx}{dt} \).

By understanding this relationship, we are able to establish a connection of the rate of change for both coordinates in terms of time.

Velocity Vector

The velocity vector is crucial because it tells us the direction and speed at which the particle is moving. For this parabola problem, once we have the derivatives, \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), we can form the velocity vector. The velocity vector \( \textbf{v} \) is represented as \( \langle \frac{dx}{dt}, \frac{dy}{dt} \rangle \). Here is how it works in this example:

• We already found that \( \frac{dy}{dt} = 1 \) and \( \frac{dx}{dt} = 2 \).
• The velocity vector becomes \( \textbf{v} = \langle 2, 1 \rangle \).

This vector makes it clear that for every second, \( x \) increases twice as fast as \( y \), illustrating the direction and speed of the particle's motion along the parabola.

Constant Speed

The concept of constant speed refers to the magnitude of the velocity vector remaining uniform over time. In this exercise, the particle moves at a constant speed of 5 units per second. Finding the velocity components that satisfy this condition is essential. Using the formula for speed, we have:

• \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = 5 \)
• We simplify using \( \frac{dx}{dt} = 2 \frac{dy}{dt} \) which leads to \( \sqrt{5 (\frac{dy}{dt})^2} = 5 \).

Solving this gives two possible solutions for \( \frac{dy}{dt} \), but we choose \( \frac{dy}{dt} = 1 \) based on the movement direction left to right. By ensuring this criteria is met, we confirm the motion is indeed at the constant speed stated.

Differential Equations

Differential equations are equations that involve the derivatives of a function. They are pivotal in modeling the parabolic motion of the particle in this problem. Here, we derive a differential equation when performing implicit differentiation:

• From differentiating \( y^2 = 2x \), we get \( 2y \frac{dy}{dt} = 2 \frac{dx}{dt} \).
• The differential equation \( \frac{dx}{dt} = y \frac{dy}{dt} \) then connects the rates of change of \( x \) and \( y \).

This equation helps establish how \( x \) and \( y \) both change as the particle moves. It shows a deep link between the parabolic path and the velocity of the particle, allowing us to solve for exact motion details given any point or speed.