/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 (a) express \(d w / d t\) as a f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(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
\(\frac{dw}{dt}\) is 0 regardless of the approach used.

Step by step solution

01

Express w in terms of t

We know that \(w = x^2 + y^2\). Substituting the expressions for \(x\) and \(y\) in terms of \(t\), we have:\[ w = (\cos t + \sin t)^2 + (\cos t - \sin t)^2 \]Expanding both squares, we get:\[ w = (\cos^2 t + 2\cos t\sin t + \sin^2 t) + (\cos^2 t - 2\cos t\sin t + \sin^2 t) \]By simplifying, we add up the terms to get:\[ w = 2\cos^2 t + 2\sin^2 t \]Using the Pythagorean identity \(\cos^2 t + \sin^2 t = 1\), we can simplify this further to:\[ w = 2 \]
02

Differentiate w with respect to t

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

Verify differentiation using Chain Rule

Using the chain rule, we first need \(\frac{dw}{dx}\) and \(\frac{dw}{dy}\):\[ \frac{dw}{dx} = 2x \quad \text{and} \quad \frac{dw}{dy} = 2y \]Next, compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):\[ \frac{dx}{dt} = -\sin t + \cos t \quad \text{and} \quad \frac{dy}{dt} = -\sin t - \cos t \]According to the chain rule:\[ \frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} \]Substitute for \(x\) and \(y\): \[ \frac{dw}{dt} = 2(x)(-\sin t + \cos t) + 2(y)(-\sin t - \cos t) \] Calculating at \(x = \cos t + \sin t\) and \(y = \cos t - \sin t\), the expression simplifies to 0 (as shown in earlier calculations of \(w\) being constant).
04

Evaluate at t=0

At \(t=0\), since \(\frac{dw}{dt} = 0\) both when differentiating directly or applying the chain rule as calculated. Thus the value at \(t = 0\) is:\[ \frac{dw}{dt} = 0 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differentiation
Differentiation is the mathematical process of finding the rate at which a function changes at any given point. It's a core concept in calculus and essential when dealing with functions that describe physical phenomena. By understanding differentiation, you can determine how variables relate to each other and change over time.

In our exercise, we have the function for a circle: \(w = x^2 + y^2\). This expression represents the equation of a circle centered at the origin in the coordinate plane. Differentiating \(w\) with respect to \(t\), either directly or using the Chain Rule, helps us find the rate at which \(w\) changes when \(x\) and \(y\) themselves are functions of \(t\). For this specific case, since \(w\) simplifies to a constant \(2\), the derivative \( \frac{dw}{dt} \) straightforwardly evaluates to zero. This means that \(w\) does not change as \(t\) changes, consistent with the nature of constants.

Differentiation provides a powerful tool for analyzing dynamic systems and relationships in various fields of study, including physics, engineering, and economics.
Pythagorean identity
The Pythagorean identity is an important trigonometric identity used often in calculus and trigonometry. It states that for any angle \(t\), the identity \(\cos^2 t + \sin^2 t = 1\) holds. This identity is derived from the Pythagorean theorem and is fundamental in simplifying trigonometric expressions.

In our exercise, we used this identity to simplify the expression for \(w\). Initially, \(w = (\cos t + \sin t)^2 + (\cos t - \sin t)^2\), and after expanding the squares, we obtained \(\cos^2 t + 2\cos t \sin t + \sin^2 t\) and \(\cos^2 t - 2\cos t \sin t + \sin^2 t\).

When combined,
  • the \(\cos^2 t\) and \(\sin^2 t\) terms add up to 1 based on the identity, leading to \(w = 2\).
  • This simplification shows the power of trigonometric identities in reducing complex expressions to more recognizable forms.
Pythagorean identities are particularly useful when dealing with trigonometric equations and can simplify calculations significantly.
Functional expressions
Functional expressions are equations that define functions in terms of variables, often linking dependencies between different variables. They play an essential role in calculus and mathematical modeling of real-world phenomenons.

In the problem, we began with \(w = x^2 + y^2\), where \(x\) and \(y\) are both expressed in terms of \(t\): \(x = \cos t + \sin t\) and \(y = \cos t - \sin t\). These are functional expressions because they clearly define how \(x\) and \(y\) depend on \(t\).
  • Substituting \(x\) and \(y\), and subsequently simplifying \(w\) using trigonometric identities, we expressed \(w\) solely in terms of \(t\).
  • This transformation allowed for direct differentiation with respect to \(t\), as seen through the constant nature of \(w\).
Understanding how to work with functional expressions enables us to see the intricate relationships between variables and to calculate derivatives representing rates of change.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}, \quad \quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right), \quad \quad (3)$$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically. Figure cannot copy

Minimum distance to the origin \(\quad\) Find the points on the surface \(z^{2}=x y+4\) closest to the origin.

Find and sketch the domain for each function. $$f(x, y)=\frac{\sin (x y)}{x^{2}+y^{2}-25}$$

Rectangular box of largest volume in a sphere \(\quad\) Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere.

Constrained minimum Find the points on the curve \(x y^{2}=54\) nearest the origin.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.