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Chapter 13: Problem 2

(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 \(t\) \(w=x^{2}+y^{2}, \quad x=\cos t+\sin t, \quad y=\cos t-\sin t ; \quad t=0\)

Short Answer

Expert verified
\(\frac{dw}{dt} = 0\) at \(t = 0\).

Step by step solution

01

Express w in terms of t

To express \( w \) in terms of \( t \), substitute the given expressions for \( x \) and \( y \) into \( w = x^2 + y^2 \). Since \( x = \cos t + \sin t \) and \( y = \cos t - \sin t \), we have: \[ w = (\cos t + \sin t)^2 + (\cos t - \sin t)^2 \] Expanding both squares gives: \[ w = (\cos^2 t + 2 \cos t \sin t + \sin^2 t) + (\cos^2 t - 2 \cos t \sin t + \sin^2 t) \] Combine like terms: \[ w = 2 \cos^2 t + 2 \sin^2 t \] Using the identity \( \cos^2 t + \sin^2 t = 1 \), it simplifies to \[ w = 2 \].
02

Differentiate w directly with respect to t

Taking the derivative of \( w \) with respect to \( t \), since \( w = 2 \), the derivative is simply: \[ \frac{dw}{dt} = 0 \] because \( w \) is a constant.
03

Use the Chain Rule to express dw/dt

Use the Chain Rule formula: \[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} \] First find the partial derivatives: \[ \frac{\partial w}{\partial x} = 2x, \quad \frac{\partial w}{\partial y} = 2y \] Then, find the derivatives: \[ \frac{dx}{dt} = -\sin t + \cos t \] and \[ \frac{dy}{dt} = -\sin t - \cos t \]. Substitute all into the Chain Rule: \[ \frac{dw}{dt} = 2x(-\sin t + \cos t) + 2y(-\sin t - \cos t) \].
04

Substitute values for x and y in the Chain Rule

From previous substitutions: \( x = \cos t + \sin t \) and \( y = \cos t - \sin t \). Substitute \( t = 0 \) into \( x \), \( y \), and their derivatives: \( x(0) = \cos(0) + \sin(0) = 1 \), \( y(0) = \cos(0) - \sin(0) = 1 \), \( \frac{dx}{dt}(0) = -\sin(0) + \cos(0) = 1 \), \( \frac{dy}{dt}(0) = -\sin(0) - \cos(0) = -1 \). Substitute these into the chain rule expression: \( \frac{dw}{dt} = 2(1)(1) + 2(1)(-1) = 2 - 2 = 0 \).
05

Evaluate dw/dt at t = 0

Both methods - differentiating directly and using the Chain Rule - give the same result. Substitute \( t = 0 \) into either expression already evaluated. Since we have previously found \( \frac{dw}{dt} = 0 \) for any \( t \), the value at \( t = 0 \) is simply: \[ \frac{dw}{dt} \Bigg|_{t=0} = 0 \].

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Key Concepts

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

Differentiation
Differentiation is one of the fundamental concepts in calculus. It refers to the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. In the context of the exercise, differentiation is used to find \( \frac{dw}{dt} \), which shows how the function \( w \) changes with respect to the change in variable \( t \).

Direct Differentiation
- When \( w = 2 \) as a constant, its derivative with respect to \( t \) is 0 because the function does not change.
- This is an example of a simple differentiation, which highlights a key rule: the derivative of a constant is always zero.

Differentiation can also involve more complex functions where after expressing a variable in terms of another, we apply rules such as the product rule, quotient rule, or the power rule. These help us in evaluating functions that are composed of multiple parts or terms. In this exercise, while \( w \) becomes a constant, the crucial understanding lies in simplifying expressions until reaching that determination.
Partial Derivatives
Partial derivatives are used when dealing with functions of multiple variables. They allow us to see how a function changes with respect to one variable while keeping others constant. In this exercise, partial derivatives are used to handle the function \( w = x^2 + y^2 \) where \( x \) and \( y \) are both dependent on \( t \).

To compute partial derivatives:
  • For \( \frac{\partial w}{\partial x} = 2x \), we differentiate \( w \) concerning \( x \) while treating \( y \) as a constant.
  • Similarly, for \( \frac{\partial w}{\partial y} = 2y \), we differentiate \( w \) concerning \( y \) while treating \( x \) as a constant.
Partial derivatives are combined using the chain rule to determine the full derivative \( \frac{dw}{dt} \). This involves computing separate rates of change with respect to each variable and then summing them accordingly.

This principle is critical in fields like multivariable calculus and vector calculus, where analyzing multidimensional surfaces or fields is common.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are always true for any value of the variables involved. They are essential tools in simplifying the expressions before differentiating.

One of the most commonly used identities is:
  • \( \cos^2 t + \sin^2 t = 1 \)
This identity was crucial in simplifying the expression \( w = 2 \) in the problem. By recognizing this identity when expanding and combining terms, we not only simplified \( w \) but also made differentiation much easier.

Trigonometric identities come in handy in many areas of math and physics:
  • They assist in transforming complex trigonometric expressions into simpler ones.
  • Help solve equations and evaluate integrals that would otherwise be difficult to manage.
Understanding how these identities function and how they can be applied is invaluable in calculus and beyond, providing shortcuts and insights into deeper mathematical interpretations.

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Most popular questions from this chapter

Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6\)

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right)\) then$$\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L$$ Use this result to support your answers to the questions. Does knowing that \(|\cos (1 / y)| \leq 1\) tell you anything about $$\lim _{(x, y) \rightarrow(0,0)} x \cos \frac{1}{y} ?$$ Give reasons for your answer.

Gives a function \(f(x, y, z)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y, z)\) $$\sqrt{x^{2}+y^{2}+z^{2}}<\delta \Rightarrow|f(x, y, z)-f(0,0,0)|<\epsilon$$ Show that \(f(x, y, z)=x+y-z\) is continuous at every point \(\left(x_{0}, y_{0}, z_{0}\right)\)

If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r \cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting expression as \(r \rightarrow 0 .\) In other words, try to decide whether there exists a number \(L\) satisfying the following criterion: $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is, we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2} /\left(x^{2}+y^{2}\right)\) does not exist. In each of these instances, the existence or nonexistence of the limit as \(r \rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) \(\theta=\) constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r \neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is \(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2} \theta\) and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$f(x, y)=\tan ^{-1}\left(\frac{|x|+|y|}{x^{2}+y^{2}}\right)$$

If \(f\left(x_{0}, y_{0}\right)=3,\) what can you say about \(\lim _{(x, y) \rightarrow\left(x_{0}, y_{\mathrm{a}}\right)} f(x, y)\) if \(f\) is continuous at \(\left(x_{0}, y_{0}\right) ?\) If \(f\) is not continuous at \(\left(x_{0}, y_{0}\right) ?\) Give reasons for your answers.
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