Problem 58 - Advanced Engineering Mathematics

Question

A particle moves in 3-space so that its coordinates at any time are \(x=4 \cos t, y=4 \sin t, z=5 t, t \geq 0\). Use the Chain Rule to find the rate at which its distance $$ w=\sqrt{x^{2}+y^{2}+z^{2}} $$ from the origin is changing at \(t=5 \pi / 2\) seconds.

Step-by-Step Solution

Answer

At \(t=5\pi/2\), the rate of distance change is approximately 6.9.

Step 1: Parameterize the Distance Function

First, express the distance function in terms of the parameter \(t\). Since the position of the particle is given by \(x=4 \cos t\), \(y=4 \sin t\), and \(z=5t\), we have:\[w = \sqrt{(4 \cos t)^2 + (4 \sin t)^2 + (5t)^2}\]Simplifying the expression, we get:\[w = \sqrt{16(\cos^2 t + \sin^2 t) + 25t^2}\]Using the Pythagorean identity, \(\cos^2 t + \sin^2 t = 1\), it reduces to:\[w = \sqrt{16 + 25t^2}\].

Step 2: Differentiate the Distance Function with Respect to Time

To find the rate at which the distance \(w\) is changing with respect to time, differentiate the expression for \(w\) with respect to \(t\). Starting with:\[w = \sqrt{16 + 25t^2}\]Using the chain rule, the derivative \( \frac{dw}{dt} \) can be found as follows:\[\frac{dw}{dt} = \frac{1}{2\sqrt{16 + 25t^2}} \cdot \frac{d}{dt}(16 + 25t^2)\].Since \( \frac{d}{dt}(16 + 25t^2) = 50t\), it follows that:\[\frac{dw}{dt} = \frac{50t}{2\sqrt{16 + 25t^2}} = \frac{25t}{\sqrt{16 + 25t^2}}\].

Step 3: Evaluate the Derivative at t = 5π/2

Substitute \(t = \frac{5\pi}{2}\) into the derivative expression:\[\frac{dw}{dt}\Big|_{t=\frac{5\pi}{2}} = \frac{25 \times \frac{5\pi}{2}}{\sqrt{16 + 25\left(\frac{5\pi}{2}\right)^2}}\].Calculate \(16 + 25\left(\frac{5\pi}{2}\right)^2\):\[16 + 25 \times \frac{25\pi^2}{4} = 16 + \frac{625\pi^2}{4}\].So,\[\frac{dw}{dt}\Big|_{t=\frac{5\pi}{2}} = \frac{125\pi}{2} \times \frac{4}{\sqrt{64 + 625\pi^2}}\].Simplifying:\[\frac{dw}{dt}\Big|_{t=\frac{5\pi}{2}} = \frac{250\pi}{\sqrt{64 + 625\pi^2}}\].

Step 4: Compute the Numerical Value

Calculate the numerical value:\(\frac{dw}{dt}\Big|_{t=\frac{5\pi}{2}} \approx \frac{250\pi}{\sqrt{64 + 625\times9.8696}}\).This simplifies to:\(\frac{250\pi}{\sqrt{64 + 6160}}\).Approximating the denominator and simplifying gives:\(\frac{250\pi}{\sqrt{6224}}\).Thus, the rate of change of distance at \(t = \frac{5\pi}{2}\) is approximately 6.9.

Key Concepts

Parametric Equations

When dealing with motion in 3-space, parametric equations are incredibly helpful. They allow us to express the coordinates of a particle using a parameter, typically time. This makes it easier to analyze the particle's movement over time.

In our exercise, the particle’s path is given in terms of the parameter \(t\):

• \(x = 4 \cos t\)
• \(y = 4 \sin t\)
• \(z = 5t\)

These equations describe a helical or spiral path around the z-axis as time progresses.

Using parametric equations simplifies the process of finding other related quantities, like the particle's velocity, acceleration, and distance from a point (e.g., the origin). Since each component \(x\), \(y\), and \(z\) depends on time, you can conveniently compute how these components change as time changes.

Rate of Change

The concept of rate of change is central to understanding motion and dynamics in calculus.

To find how quickly something is changing, we calculate its derivative. In this exercise, we want to know how the distance of the particle from the origin is changing over time. The distance is represented by the variable \(w\) and given by:

• \(w = \sqrt{x^2 + y^2 + z^2}\)

To find the rate at which \(w\) changes with time, \(\frac{dw}{dt}\), we need to differentiate \(w\) with respect to \(t\).

This involves applying the Chain Rule, a powerful tool in calculus.The Chain Rule helps us find the derivative of variables that depend indirectly on another variable, in this case, \(t\). The rate of change tells us the speed of the particle's movement away from the origin at a specific time.

Derivative Calculations

Calculating derivatives accurately is crucial when working with rates of change.

• First, express the function in a way that's easy to differentiate.
• In this task, we have \(w = \sqrt{16 + 25t^2}\).

Now, using the Chain Rule, differentiating this expression with respect to \(t\) involves these steps:

• Identify the outer function \(w = \sqrt{u}\), where \(u = 16 + 25t^2\).
• Derive the inner function with respect to \(t\), which is \(\frac{d}{dt}(16 + 25t^2) = 50t\).

Combining these, the derivative becomes:

• \(\frac{dw}{dt} = \frac{1}{2\sqrt{16 + 25t^2}} \times 50t\).

Putting \(t = \frac{5\pi}{2}\) in the derivative formula gives us the rate of distance change at a particular moment.

This step-by-step calculation ensures we correctly capture how quickly the distance variable \(w\) is changing at any given time.