/*! 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 51 Describe the level curves of the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 51

Describe the level curves of the function. Sketch the level curves for the given \(c\) -values. $$ z=\sqrt{25-x^{2}-y^{2}}, \quad c=0,1,2,3,4,5 $$

Short Answer

Expert verified
The level curves are six concentric circles with centres at the origin and decreasing radii from \(\sqrt{25}\) to \(\sqrt{0}\) as the \(c\) value increases.

Step by step solution

01

Solve for x and y

In this step, we want to solve the equation \(z = \sqrt{25 - x^2 - y^2} = c\) for \(x\) and \(y\). By squaring both sides of the equation we remove the square root from the right side: \(z^2 = 25 - x^2 - y^2\) or \(x^2 + y^2 = 25 - z^2\).
02

Use the constant values

Now we can substitute the given constants for \(c = 0, 1, 2, 3, 4, 5\) into the equation obtained from step 1. We get the following set of equations: \[\] 1) When \(c = 0\), \(x^2 + y^2 = 25\); \[\] 2) When \(c = 1\), \(x^2 + y^2 = 24\); \[\] 3) When \(c = 2\), \(x^2 + y^2 = 21\); \[\] 4) When \(c = 3\), \(x^2 + y^2 = 16\); \[\] 5) When \(c = 4\), \(x^2 + y^2 = 9\); \[\] 6) When \(c = 5\), \(x^2 + y^2 = 0\).
03

Sketch the level curves

Each of the equations found in step 2 correspond to circles centred on the origin with radii of \(\sqrt{25}, \sqrt{24}, \sqrt{21}, \sqrt{16}, \sqrt{9}\) and \(\sqrt{0}\) respectively. Now the task is to draw these circles in the x-y plane. The resulting graph would have 6 concentric circles with decreasing radii.

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 91影视!

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

The parametric equations for the paths of two projectiles are given. At what rate is the distance between the two objects changing at the given value of \(t ?\) \(x_{1}=48 \sqrt{2} t, y_{1}=48 \sqrt{2} t-16 t^{2}\) \(x_{2}=48 \sqrt{3} t, y_{2}=48 t-16 t^{2}\) \(t=1\)

Show that \(\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}=0\) for \(w=f(x, y), x=u-v,\) and \(y=v-u\)

Find \(\partial w / \partial r\) and \(\partial w / \partial \theta\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(r\) and \(\boldsymbol{\theta}\) before differentiating. \(w=\sqrt{25-5 x^{2}-5 y^{2}}, x=r \cos \theta, \quad y=r \sin \theta\)

Let \(w=f(x, y)\) be a function where \(x\) and \(y\) are functions of two variables \(s\) and \(t\). Give the Chain Rule for finding \(\partial w / \partial s\) and \(\partial w / \partial t\)

In Exercises 19 and \(20,\) use the gradient to find the directional derivative of the function at \(P\) in the direction of \(Q\). $$ g(x, y)=x^{2}+y^{2}+1, \quad P(1,2), Q(3,6) $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation
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.