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Chapter 13: Problem 49

Sketch the \(x y\) -trace of the sphere. $$ x^{2}+y^{2}+z^{2}-6 x-10 y+6 z+30=0 $$

Short Answer

Expert verified
The xy-trace of the sphere is a circle with center at (3, 5) and radius 4.

Step by step solution

01

Set z equal to 0

First, we need to set \(z = 0\) in the given equation. This produces the following equation: \(x^{2}+y^{2}-6 x-10 y+30=0\)
02

Reformulate the equation

Try to rewrite our equation and group the x terms and y terms together: \((x^{2}-6x)+(y^{2}-10y)=-30\)
03

Completing the square

For both x and y terms we complete the square by adding \((\frac{-6}{2})^{2}=9\) for the x-terms and \((\frac{-10}{2})^{2}=25\) for the y-terms, both to the left side and the right side of the equation. We get then \((x-3)^{2}+(y-5)^{2}=16\)
04

Recognize the form

We can see now that this equation is of a circle in standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\), where h = 3 and k = 5 are the center of the circle and r = 4 is the radius

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Most popular questions from this chapter

A firm's weekly profit in marketing two products is given by \(P=192 x_{1}+576 x_{2}-x_{1}^{2}-5 x_{2}^{2}-2 x_{1} x_{2}-5000\) where \(x_{1}\) and \(x_{2}\) represent the numbers of units of each product sold weekly. Estimate the average weekly profit if \(x_{1}\) varies between 40 and 50 units and \(x_{2}\) varies between 45 and 50 units.

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Plot the points and determine whether the data have positive, negative, or no linear correlation (see figures below). Then use a graphing utility to find the value of \(r\) and confirm your result. The number \(r\) is called the correlation coefficient. It is a measure of how well the model fits the data. Correlation coefficients vary between \(-1\) and 1, and the closer \(|r|\) is to 1, the better the model. $$ (1,3),(2,6),(3,2),(4,3),(5,9),(6,1) $$

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