Problem 3 (a) express $d w / d t$ as a f... [FREE SOLUTION]

Chapter 14: Problem 3

(a) express $d w / d t$ as a function of $t,$ both by using the Chain Rule and by expressing $w$ in terms of $t$ and differentiating directly with respect to $t .$ Then (b) evaluate $d w / d t$ at the given value of $t$. $$w=\frac{x}{z}+\frac{y}{z}, \quad x=\cos ^{2} t, \quad y=\sin ^{2} t, \quad z=1 / t ; \quad t=3$$

Short Answer

Expert verified

$ \frac{dw}{dt} = 1 $ at $ t = 3 $.

Step by step solution

- Express w as a function of t

The given expression for $ w $ is: \[ w = \frac{x}{z} + \frac{y}{z} \]Substitute the expressions for $ x $, $ y $, and $ z $ into this equation:\[ x = \cos^2 t, \qquad y = \sin^2 t, \qquad z = \frac{1}{t} \]\[ w = \frac{\cos^2 t}{1/t} + \frac{\sin^2 t}{1/t} = t(\cos^2 t + \sin^2 t) \]Since $ \cos^2 t + \sin^2 t = 1 $, then $ w = t $.

- Differentiate directly with respect to t

As derived, $ w = t $.Differentiate $ w $ with respect to $ t $:\[ \frac{dw}{dt} = \frac{d}{dt}(t) = 1 \].

- Use the Chain Rule

The original expression for $ w $ is:\[ w = \frac{x}{z} + \frac{y}{z} \]Using the Chain Rule,\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \].Calculate each partial derivative:\[ \frac{\partial w}{\partial x} = \frac{1}{z} = t \]\[ \frac{\partial w}{\partial y} = \frac{1}{z} = t \]\[ \frac{\partial w}{\partial z} = -\frac{x+y}{z^2} = -\frac{t}{t^{-2}} = -t^3 \]Calculate each derivative with respect to $ t $:\[ \frac{dx}{dt} = \frac{d}{dt}(\cos^2 t) = -2 \cos t \sin t \]\[ \frac{dy}{dt} = \frac{d}{dt}(\sin^2 t) = 2 \cos t \sin t \]\[ \frac{dz}{dt} = \frac{d}{dt}(t^{-1}) = -t^{-2} \]Substitute back into the Chain Rule:\[ \frac{dw}{dt} = t(-2 \cos t \sin t) + t(2 \cos t \sin t) - t^3(-t^{-2}) \]Simplifying:\[ \frac{dw}{dt} = 0 + t = 1 \].

- Evaluate at t = 3

From either method, differentiating directly or using the Chain Rule, we have found that:\[ \frac{dw}{dt} = 1 \]At $ t = 3 $, $ \frac{dw}{dt} $ is still:\[ \frac{dw}{dt} = 1 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation

Differentiation is a key concept in calculus used to find the rate at which a function is changing at any given point. It involves the process of finding the derivative, which represents the function's slope or rate of change.

To differentiate a function, you evaluate how small changes in the input (or independent variable) affect changes in the output (or dependent variable).

The derivative of a function is commonly denoted as $\frac{d}{dx}f(x)$, highlighting the change in $f$ with respect to $x$.
For simple functions like $f(x) = ax + b$, the derivative simplifies to $a$, the coefficient of $x$.

In this exercise, we differentiated by expressing $w$ in terms of $t$, where $w = t$. Deriving with respect to $t$, we found $\frac{dw}{dt} = 1$, indicating a constant rate of change for $w$ as $t$ increases.

Partial Derivatives

The concept of partial derivatives is essential when dealing with functions of multiple variables. A partial derivative represents the rate of change of a multivariable function with respect to one variable, keeping the others constant.

In the context of our exercise, $w$ is expressed as a function of $x$, $y$, and $z$. Each variable itself is a function of $t$. Let's delve into why partial derivatives are useful here:

A partial derivative is denoted by $\frac{\partial f}{\partial x}$, indicating the change in $f$ with only $x$ changing.
They are critical in applying the Chain Rule, which connects how $x$, $y$, and $z$ influence $w$ through $t$.

By calculating the partial derivatives $\frac{\partial w}{\partial x}$, $\frac{\partial w}{\partial y}$, and $\frac{\partial w}{\partial z}$, we capture the distinct individual impacts on $w$. This meticulous breakdown leads to understanding how changes in $t$ translate through each variable to ultimately affect $w$.

Trigonometric Functions

Trigonometric functions are fundamental in connecting geometry with algebra through angles and ratios of right-angled triangles. In calculus, they appear frequently in differentiation, especially with problems involving periodic behavior.

In this exercise, $x$ and $y$ utilized the trigonometric functions $\cos(t)$ and $\sin(t)$ respectively. These exhibit:

The Pythagorean identity $\cos^2 t + \sin^2 t = 1$, which simplified $w = t(\cos^2 t + \sin^2 t)$ to $w = t$.
The derivatives such as $\frac{d}{dt}(\cos^2 t) = -2 \cos t \sin t$, derived using the chain rule and standard derivatives of trigonometric functions $\frac{d}{dt}(\cos t) = -\sin t$ and $\frac{d}{dt}(\sin t) = \cos t$.

Understanding these concepts is crucial since trigonometric functions are omnipresent in practical applications, like electromagnetic waves or sound waves analysis. A strong grasp allows one to navigate complex differential equations involving periodic phenomena effectively.

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Short Answer

Step by step solution

- Express w as a function of t

- Differentiate directly with respect to t

- Use the Chain Rule

- Evaluate at t = 3

Key Concepts

Differentiation

Partial Derivatives

Trigonometric Functions

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