Problem 13 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14 , find the total derivative \(d u / d t\) by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) or for \(x, y\), and \(z\) before differentiating. $$ u=\sqrt{x^{2}+y^{2}+z^{2}} ; x=\tan t ; y=\cos t ; z=\sin t ; 0

Step-by-Step Solution

Answer

\( \frac{du}{dt} = \sec t \, \tan t \)

Step 1: Define the problem

We need to find the total derivative \( \frac{du}{dt} \) of \( u = \sqrt{x^2 + y^2 + z^2} \) where \( x = \tan t \), \( y = \cos t \), and \( z = \sin t \) using two methods.

Step 2: Method (a) - Use the chain rule

Apply the chain rule for the function \( u = \sqrt{x^2 + y^2 + z^2} \):\[ \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt} \]

Step 3: Compute partial derivatives

First, compute the partial derivatives of \( u \) with respect to \( x, y, \) and \( z \):\[ \frac{\partial u}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \]\[ \frac{\partial u}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \]\[ \frac{\partial u}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \]

Step 4: Compute the derivatives of x, y, z with respect to t

Next, compute the derivatives of the parametric equations with respect to \( t \):\[ \frac{dx}{dt} = \sec^2 t \]\[ \frac{dy}{dt} = -\sin t \]\[ \frac{dz}{dt} = \cos t \]

Step 5: Combine the derivatives

Combine the derivatives using the chain rule formula:\[ \frac{du}{dt} = \frac{x \, \sec^2 t}{\sqrt{x^2 + y^2 + z^2}} + \frac{y \, (-\sin t)}{\sqrt{x^2 + y^2 + z^2}} + \frac{z \, \cos t}{\sqrt{x^2 + y^2 + z^2}} \]

Step 6: Method (b) - Substitute before differentiating

Substitute the parametric equations into \( u \) before differentiating:\[ u = \sqrt{\tan^2 t + \cos^2 t + \sin^2 t} = \sqrt{\tan^2 t + 1} = \sqrt{\sec^2 t} = |\sec t| \]

Step 7: Differentiate with respect to t

Differentiate \( u = \sec t \) with respect to \( t \):\[ \frac{du}{dt} = \frac{d}{dt}(\sec t) = \sec t \, \tan t \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to compute the derivative of a composite function. It connects the rate of change of the outer function to the rate of change of its inner functions. This rule states that if you have a function that depends on several variables, and each of those variables is itself a function of another variable, you can find the total derivative by multiplying the derivatives of each inner function with the derivative of the outer function with respect to each inner variable. For example, if we have a function u = f(x, y, z) and each of x, y, z depends on \texttt{t}, the chain rule can be written as:

\[\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt}\]

• First, compute the partial derivatives \( \frac{\partial u}{\partial x} \), \( \frac{\partial u}{\partial y} \), and \( \frac{\partial u}{\partial z}\).
• Then, find the derivatives of x, y, z with respect to t, such as \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt}\).
• Lastly, multiply and sum these derivatives accordingly.

Partial Derivatives

Partial derivatives are used when dealing with functions of several variables. They show how the function changes as one of the variables changes while keeping the others constant. When analyzing a function u = f(x, y, z), it is essential to determine how u changes with respect to each of its variables x, y, and z independently. Here’s how to compute partial derivatives:

• To find \( \frac{\partial u}{\partial x} \), differentiate u with respect to x, treating y and z as constants.
• To find \( \frac{\partial u}{\partial y} \), differentiate u with respect to y, treating x and z as constants.
• To find \( \frac{\partial u}{\partial z} \), differentiate u with respect to z, treating x and y as constants.

For the given function u = \sqrt{x^2 + y^2 + z^2}, the partial derivatives are:

• \( \frac{\partial u}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \)
• \( \frac{\partial u}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \)
• \( \frac{\partial u}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \)

These derivatives are essential for applying the chain rule effectively.

Parametric Equations

Parametric equations represent a set of equations where the system’s state variables are expressed as functions of one or more independent parameters. In our exercise, we have:

• x = \tan t
• y = \cos t
• z = \sin t

These equations express x, y, and z in terms of the parameter t. Using parametric equations, you can easily describe more complex relationships between variables that would be tough to define using standard equations alone. Parametric equations are beneficial in physics and engineering to model motions and forces. For instance, the trajectory of a particle can be described using parametric equations that incorporate time t.

To differentiate a parametric equation with respect to time t, you apply differentiation rules to each equation separately:

• \( \frac{dx}{dt} = \sec^2 t \)
• \( \frac{dy}{dt} = -\sin t \)
• \( \frac{dz}{dt} = \cos t \)

Substitution Method

The substitution method involves replacing the variables in the function with their parametric equations before differentiating. This method simplifies the differentiation process as it reduces the function to one variable, typically time t. Here’s how we can use it in our problem:

u = \sqrt{x^2 + y^2 + z^2}

First, substitute x, y, and z with their parametric forms:

• x = \tan t
• y = \cos t
• z = \sin t

Then, the function u becomes:

u = \sqrt{\tan^2 t + \cos^2 t + \sin^2 t}

Use trigonometric identities to simplify this:

• \( \cos^2 t + \sin^2 t = 1\)
• \(\tan^2 t + 1 = \sec^2 t\)

So, u becomes:

u = \sqrt{\sec^2 t} = |\sec t|

Finally, differentiate with respect to t:

• \( \frac{du}{dt} = \frac{d(\sec t)}{dt} = \sec t \cdot \tan t \)

This method is often simpler and more intuitive when dealing with composite functions and parameterized variables.