Problem 29 - Thomas Calculus

Question

Motion in the plane The coordinates of a particle in the metric \(x y\) -plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{sec}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{sec} .\) How fast is the particle's distance from the origin changing as it passes through the point \((5,12) ?\)

Step-by-Step Solution

Answer

The particle's distance from the origin is decreasing at a rate of 5 m/s.

Step 1: Find the Particle's Distance from the Origin

The particle's distance from the origin at any point \((x, y)\) is given by the distance formula: \( r = \sqrt{x^2 + y^2} \).

Step 2: Differentiate the Distance with Respect to Time

To find how fast the particle's distance from the origin is changing, we need to differentiate the distance \( r \) with respect to time \( t \). Use the chain rule: \( \frac{dr}{dt} = \frac{d}{dt}\sqrt{x^2 + y^2} = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}}(2x \frac{dx}{dt} + 2y \frac{dy}{dt}) \).

Step 3: Substitute Known Values

We know the particle passes through the point \((5, 12)\), so we substitute \(x = 5\), \(y = 12\), \(\frac{dx}{dt} = -1\), and \(\frac{dy}{dt} = -5\) into the differentiated distance formula: \( \frac{dr}{dt} = \frac{1}{\sqrt{5^2 + 12^2}}(5(-1) + 12(-5)) \).

Step 4: Simplify the Equation

Calculate the distance from the origin: \( r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)Substitute into the equation:\( \frac{dr}{dt} = \frac{1}{13}(5(-1) + 12(-5)) \).Simplify the expression:\( \frac{dr}{dt} = \frac{1}{13}(-5 - 60) = \frac{1}{13}(-65) = -5 \).

Step 5: Interpret the Result

A negative rate of change, \( \frac{dr}{dt} = -5 \), indicates that the distance from the particle to the origin is decreasing at a rate of 5 m/s as it passes through the point \((5,12)\).

Key Concepts

Motion in the Plane

Understanding motion in the plane is crucial for analyzing the path of an object, such as a particle moving on a coordinate plane. When a particle moves through the plane, its position can be described using coordinates

• The position is usually given as \((x, y)\), representing where the particle is along the x-axis and y-axis at any time \(t\).
• Each coordinate can be a function of time, meaning \(x\) and \(y\) change as time passes.

In the context of the exercise, the particle’s path and speed along each axis are determined by these coordinate functions. Understanding how these functions work is key to predicting how the particle moves.
To determine the velocity components \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we need to understand how quickly the x and y positions change over time.

• In our exercise, these are specified as \(-1 \, \mathrm{m/s}\) and\(-5 \, \mathrm{m/s}\), respectively.
• These values not only reflect the speed but also the direction of motion along the axes.

Differentiation

Differentiation is a powerful tool in calculus used to compute the rate at which one quantity changes with respect to another. Here, we are interested in how the particle's distance from the origin changes as time progresses.
To find this rate, we rely on differentiation to process the distance function:

• The given distance from the origin is described by \(r = \sqrt{x^2 + y^2}\), where \((x, y)\) are the current coordinates of the particle.
• With differentiation, we aim to find \(\frac{dr}{dt}\), representing the rate of change of \(r\) over time.

Using differentiation involves the application of certain rules, which, in our case, leads us to the Chain Rule. Differentiation simplifies complex relationships by breaking them down into manageable rates of change, crucial for understanding motion dynamics.

Chain Rule

The chain rule is an essential concept in calculus, particularly when dealing with functions that depend on other functions, like our distance function \(r = \sqrt{x^2 + y^2}\) with respect to time.
This rule helps us to differentiate composite functions, i.e., functions within functions. Here's how we apply it in this context:

• We want to differentiate the function \(\sqrt{x^2 + y^2}\), which means applying the chain rule to account for how both \(x\) and \(y\) change with time.
• Using the chain rule, we find:\(\frac{dr}{dt} = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}}(2x \frac{dx}{dt} + 2y \frac{dy}{dt})\).

This applies the derivatives of inside functions, showing how changes in each function contribute to the overall rate of change, an invaluable technique in analyzing complex motion scenarios.

Distance Formula

The distance formula is a fundamental concept that calculates the distance between two points in a plane. For a particle's motion on the \(xy\)-plane, it provides the necessary basis to evaluate its overall distance from a reference point, typically the origin.
In calculus, deriving this distance with respect to time enables us to understand the dynamic changes occurring as the particle travels.

• The formula \(r = \sqrt{x^2 + y^2}\) computes the hypotenuse in a right-angled triangle where \(x\) and \(y\) represent the opposite and adjacent sides.
• Using this formula, the exercise computes the particle’s distance as 13 m while passing through \((5, 12)\).

This value contributes to further calculations to find \(\frac{dr}{dt}\), enabling us to deduce how the distance is changing, essential for predicting motion behavior.