Problem 20 - Thomas Calculus

Question

An object of mass \(m\) travels along the parabola \(y=x^{2}\) with a constant speed of 10 units/ sec. What is the force on the object due to its acceleration at \((0,0) ?\) at \(\left(2^{1 / 2}, 2\right) ?\) Write your answers in terms of \(\mathbf{i}\) and \(\mathbf{j}\) . (Remember Newton's law, \(\mathbf{F}=m \mathbf{a} . )\)

Step-by-Step Solution

Answer

The force is \(2m\mathbf{j}\) at both \((0,0)\) and \((\sqrt{2}, 2)\).

Step 1: Understand the Motion

The object moves along the parabola described by the equation \(y = x^2\). We know the speed of the object is constant at 10 units/sec.

Step 2: Find Derivative of the Path

To determine the direction of motion, derive the expression for the tangent to the curve. The derivative \(\frac{dy}{dx} = 2x\) gives us the slope of the tangent to the parabola at any point \((x, y)\).

Step 3: Parametrize the Motion

Consider the parametric equations \(x = t\) and \(y = x^2 = t^2\). So, the position vector as a function of time \(t\) becomes \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}\).

Step 4: Calculate Velocity

Find velocity as the derivative of position: \(\mathbf{v}(t) = \frac{d}{dt}(t\mathbf{i} + t^2\mathbf{j}) = \mathbf{i} + 2t\mathbf{j}\). The magnitude of this velocity is \(\sqrt{1 + (2t)^2} = 10\). Solve for \(t\).

Step 5: Calculate Acceleration

Acceleration is the derivative of velocity: \(\mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t\mathbf{j}) = 2\mathbf{j}\).

Step 6: Evaluate Acceleration at (0,0)

At \((0,0)\), when \(t = 0\), the acceleration is \(\mathbf{a}(0) = 0\mathbf{i} + 2\mathbf{j}\). Thus, \(\mathbf{F}(0) = m(0\mathbf{i} + 2\mathbf{j}) = 2m\mathbf{j}\).

Step 7: Evaluate Acceleration at \((\sqrt{2}, 2)\)

At \((\sqrt{2}, 2)\), when \(t = \sqrt{2}\), the acceleration remains \(2\mathbf{j}\) because it does not depend on \(t\). So, \(\mathbf{F}(\sqrt{2}) = m(0\mathbf{i} + 2\mathbf{j}) = 2m\mathbf{j}\).

Key Concepts

Parametric Equations

Parametric equations provide a powerful way to represent paths or curves using a separate parameter, often denoted as "t". This allows us to describe both x and y coordinates as functions of time or another independent variable.

• For this problem, the movement along the parabola is represented by the equations: \( x = t \) and \( y = t^2 \).
• These equations define a clear pathway the object follows as time progresses.

Tracking pathways with parameters can simplify the calculation of velocity and acceleration.
When talking about curves like a parabola, using a parameter helps break down complex motion into more manageable steps. Instead of directly dealing with the curve's relationship \( y = x^2 \), you solve it by focusing on how "t" influences both x and y over time.

Derivatives

In calculus, derivatives are fundamental to understanding changes and rates. They help us express how a quantity varies based on another variable.

• The derivative \( \frac{dy}{dx} = 2x \) tells us the slope or direction of the parabola at any specific point \( (x, y) \).
• When you derive parametric equations by time: \( \mathbf{v}(t) = \frac{d}{dt}(t\mathbf{i} + t^2\mathbf{j}) = \mathbf{i} + 2t\mathbf{j} \), it reveals the velocity vector.

This interpretation provides insight into how quickly and in which direction the object moves along the path.
Knowing the derivative offers a peek into dynamic behavior—a necessary step for calculating further properties like acceleration.

Newton's Laws

Newton's laws describe the relationship between the motion of an object and the forces acting upon it. Specifically, Newton's second law states that \( \mathbf{F} = m \mathbf{a} \), where \( \mathbf{F} \) is force, \( m \) is mass, and \( \mathbf{a} \) is acceleration.

• In this exercise, you determine the force on the object by evaluating its acceleration at given points.
• The force at points \( (0,0) \) and \( (\sqrt{2}, 2) \) is computed as \( 2m\mathbf{j} \), indicating that it's directed vertically upwards.

Newton's second law provides the essential link needed to apply physical principles in mathematical form. This connection allows us to translate mathematical results into real-world insights about force.

Velocity and Acceleration Analysis

Velocity and acceleration are key concepts in describing motion.

• Velocity informs us of the speed and direction of movement. In this example, the constant speed is 10 units/sec.
• The velocity vector is \( \mathbf{v}(t) = \mathbf{i} + 2t\mathbf{j} \), capturing changes due to the parabola's slope and shape.
• Acceleration, derived as \( \mathbf{a}(t) = 2\mathbf{j} \), points out the variation in velocity over time.

In contexts like this, analyzing both velocity and acceleration is crucial.
It helps us understand not just where the object is heading, but also how it's speeding up or slowing down as it moves along its path.