Problem 41 - 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 \(\langle 2\sqrt{5}, \sqrt{5} \rangle\).

Step 1: Express y in terms of x

We have the equation of the parabola \(y^{2} = 2x\). We'll need \(y\) in terms of \(x\) to find derivatives later. Take the square root of both sides to get \(y = \sqrt{2x}\). Since the particle is moving along the top of the parabola and upwards, we take the positive root.

Step 2: Find derivative of y with respect to x

To find the velocity components, we'll need \(\frac{dy}{dx}\). Using implicit differentiation on \(y^2 = 2x\), differentiate both sides with respect to \(x\) to get \(2y \frac{dy}{dx} = 2\). Solving for \(\frac{dy}{dx}\), we have \(\frac{dy}{dx} = \frac{1}{y}\).

Step 3: Evaluate dy/dx at the given point

Now substitute the point \((2,2)\) into \(\frac{dy}{dx} = \frac{1}{y}\) to find the slope at that point. Thus \(\frac{dy}{dx} = \frac{1}{2}\). This indicates the direction of the slope of the tangent line to the parabola at the point \((2,2)\).

Step 4: Find the unit tangent vector

Given \(\frac{dy}{dx} = \frac{1}{2}\), the direction vector for the tangent can be expressed as \(\langle 1, \frac{1}{2} \rangle\). To convert this to a unit vector, calculate its magnitude \(\sqrt{1^2 + (\frac{1}{2})^2} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\). The unit tangent vector is \(\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \rangle\).

Step 5: Find the velocity vector

The velocity vector \(\mathbf{v}\) is in the direction of the unit tangent vector with magnitude 5 units/sec. Thus, \(\mathbf{v} = 5 \cdot \langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \rangle = \langle 2\sqrt{5}, \sqrt{5} \rangle\).

Step 6: Simplify the velocity vector

Since \(\mathbf{v} = \langle 2\sqrt{5}, \sqrt{5} \rangle\), we'll express each component. The \(x\)-component is \(2\sqrt{5}\) and the \(y\)-component is \(\sqrt{5}\). So, the velocity vector is \(\langle 2\sqrt{5}, \sqrt{5} \rangle\).

Key Concepts

Velocity Analysis

Velocity analysis is crucial when studying the movement of objects. It helps us understand the speed and direction of a moving particle. In this scenario, we're examining a particle traveling along a parabola with a constant speed of 5 units per second.

To determine the velocity, we use a vector that indicates both magnitude (speed) and direction. Here, the velocity vector is calculated by finding the unit tangent vector at a given point and then scaling it by the particle's speed.

Steps for Velocity Calculation:

• Find the derivative of the curve's equation to determine the slope of the tangent.
• Convert the direction vector into a unit vector by dividing by its magnitude.
• Multiply by the speed of the particle to obtain the velocity vector.

The result is a velocity vector that describes how quickly and in what direction the particle moves at a specific point on the parabola.

Parabola Motion

Parabola motion refers to the path traced by an object traveling along a parabolic trajectory. A parabola, in mathematics, is a curve represented by a quadratic equation, such as the one in this exercise: \(y^2=2x\).

Key Features of Parabolic Motion:

• It can be either symmetrical or asymmetrical concerning its axis.
• The focus and directrix influence the shape and position of the parabola.
• When an object moves along a parabola, its path is curved, unlike linear motion.

In our exercise, the particle is constrained to the upper half of the parabola. This means its \(y\)-value increases as it "moves to the right." Understanding the parabola's attributes helps when analyzing the particle's trajectory and velocity at specific points.

This exercise illustrates how velocity components can vary along different sections of a parabola.

Implicit Differentiation

Implicit differentiation is a powerful calculus technique used when it is not feasible or easy to solve biological equations for one variable in terms of the other. In this exercise, we have \(y^2 = 2x\), a situation where both variables are entangled.

Steps in Implicit Differentiation:

• Differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\).
• Apply the chain rule where necessary, as seen with \(\frac{dy}{dx}\).
• Solve the resulting equation for \(\frac{dy}{dx}\) to find the slope at any point (e.g., \(\frac{dy}{dx} = \frac{1}{y}\) in this case).

By implicitly differentiating, we derive relationships that we can't see immediately. This gives us the rate at which \(y\) changes with \(x\), which is vital when analyzing the dynamics of curves in more complex scenarios.