Problem 16 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 18 , find the total derivative \(d u / d t\) by using the chain rule; do not express \(u\) as a function of \(t\) before differentiating. $$ u=x y+x z+y z ; x=t \cos t ; y=t \sin t ; z=t $$

Step-by-Step Solution

Answer

The total derivative \(du/dt\) is \((t \sin t + t)(\cos t - t \sin t) + (t \cos t + t)(\sin t + t \cos t) + (t \cos t + t \sin t)\).

Step 1 - Identify given functions

Given functions are: \[u = xy + xz + yz\]with \[x = t \, \cos t, \quad y = t \, \sin t, \quad z = t\]

Step 2 - Compute partial derivatives of u

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

Step 3 - Compute derivatives of x, y, and z with respect to t

Find the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\): \[\frac{d x}{d t} = \frac{d}{dt}(t \cos t) = \cos t - t \sin t\] \[\frac{d y}{d t} = \frac{d}{dt}(t \sin t) = \sin t + t \cos t\] \[\frac{d z}{d t} = \frac{d}{dt}(t) = 1\]

Step 4 - Apply the chain rule to find du/dt

The total derivative is given by the chain rule: \[\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt}\] Substitute the partial derivatives and the derivatives with respect to \(t\) calculated: \[\frac{du}{dt} = (y + z)(\cos t - t \sin t) + (x + z)(\sin t + t \cos t) + (x + y)\times 1\]

Step 5 - Substitute x, y, z back into the equation

Replace \(x\), \(y\), and \(z\) in the equation using the given relationships: \[\frac{du}{dt} = (t \sin t + t)(\cos t - t \sin t) + (t \cos t + t)(\sin t + t \cos t) + (t \cos t + t \sin t)\]Simplify the expression to get the final form.

Key Concepts

chain rule

The chain rule is an essential tool in calculus for finding the derivative of a function based on its composition with other functions. It helps to differentiate complex functions that are expressed as compositions of simpler functions. For instance, if you have a function that depends on other variables, and those variables depend on yet another variable, the chain rule allows you to find the derivative of the original function concerning the final variable. To apply the chain rule, you need to determine the partial derivatives of the function with respect to each independent variable and then multiply each partial derivative by the derivative of the respective variable. The total derivative is the sum of these products.

partial derivatives

Partial derivatives measure how a function changes as one of its variables changes while keeping the others constant. They are a core concept in multivariable calculus. For example, if we have a function \(u = xy + xz + yz\), its partial derivative with respect to \(x\) assumes \(y\) and \(z\) are constants:

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

Similarly, for \(y\) and \(z\):

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

These calculations let you understand how the function changes as each individual variable changes while the others are held constant.

differentiation with respect to t

When dealing with a function where the variables themselves are functions of another variable, such as time \(t\), you need to differentiate each of these inner functions concerning \(t\). In our example, the functions are \(x = t \cos t, y = t \sin t, z = t\). Differentiating them with respect to \(t\) gives:

• \(\frac{dx}{dt} = \cos t - t \sin t\)
• \(\frac{dy}{dt} = \sin t + t \cos t\)
• \(\frac{dz}{dt} = 1\)

This step is crucial before applying the chain rule, as it provides the derivatives needed to compute the total derivative effectively.

calculus

Calculus is the branch of mathematics that studies continuous change. Divided into two main parts, differential calculus focuses on finding the rates at which quantities change, while integral calculus deals with the accumulation of quantities. In the context of this exercise, differential calculus comes into play. The goal is to find how the function \(u\) changes with respect to time \(t\), using various tools such as the chain rule and partial derivatives. Understanding these concepts allows you to analyze and predict the behavior of complex systems, making calculus a foundational subject in mathematics and crucial for fields ranging from physics to engineering.