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Chapter 11: Problem 44

Show that the ellipsoid \(3 x^{2}+2 y^{2}+z^{2}=9\) and the sphere \(x^{2}+y^{2}+z^{2}-8 x-6 y-8 z+24=0\) are tangent to each other at the point \((1,1,2) .\) (This means that they have a common tangent plane at the point.)

Short Answer

Expert verified
The ellipsoid and sphere are tangent at point (1,1,2).

Step by step solution

01

Write the equation of the planes tangent to the sphere

The equation for a sphere is in the form \[ x^{2} + y^{2} + z^{2} - 8x - 6y - 8z + 24 = 0. \]First, complete the square for each variable to find the equation of the sphere in its standard form:- For the x-term: \[ x^{2} - 8x = (x - 4)^{2} - 16 \]- For the y-term: \[ y^{2} - 6y = (y - 3)^{2} - 9 \]- For the z-term: \[ z^{2} - 8z = (z - 4)^{2} - 16 \]Thus, the equation becomes:\[ (x - 4)^{2} + (y - 3)^{2} + (z - 4)^{2} = 1. \]This represents a sphere with center \((4, 3, 4)\) and radius \(1\). The tangent plane at any point \((x_{0}, y_{0}, z_{0})\) on the sphere is given by:\[ (x_0 - 4)(x - x_0) + (y_0 - 3)(y - y_0) + (z_0 - 4)(z - z_0) = 0. \]In our case, at the point \((1, 1, 2)\), the equation of the tangent plane simplifies to:\[ (1 - 4)(x - 1) + (1 - 3)(y - 1) + (2 - 4)(z - 2) = 0 \].

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Key Concepts

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

Ellipsoid
An ellipsoid is a three-dimensional geometric figure that resembles an elongated or flattened sphere. It can be described by the general equation:
  • \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \)
This is similar to an ellipse but extended into three dimensions. In the given exercise, the ellipsoid is defined by the equation:
  • \( 3x^2 + 2y^2 + z^2 = 9 \)
This equation can be rearranged to standard form by dividing each term by 9, yielding:
  • \( \frac{x^2}{3} + \frac{y^2}{4.5} + \frac{z^2}{9} = 1 \)
An ellipsoid, depending on the values of \(a\), \(b\), and \(c\), can take the form of a sphere, a prolate spheroid, or an oblate spheroid. Here, it indicates that the object is stretched along different axes. The ellipsoid's axial lengths tell you how it scales in the \(x\), \(y\), and \(z\) directions.
Sphere
A sphere is a perfectly symmetrical 3D shape, all points on its surface being at the same distance from its center. The standard equation for a sphere is:
  • \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \)
where \((h, k, l)\) is the center and \(r\) is the radius. For the given sphere, the initial equation:
  • \( x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 0 \)
needs to be rewritten into its standard form. Completing the square for each variable gives us:
  • \( (x-4)^2 + (y-3)^2 + (z-4)^2 = 1 \)
This reveals the sphere has a center at \((4, 3, 4)\) and a radius of 1. Notice how completing the square helps unfold the sphere鈥檚 true geometry easily.
Complete the Square
Completing the square is a technique used for transforming a quadratic equation into a form that makes it easy to identify the geometric properties of curves such as circles and ellipses. The process involves:
  • Isolating the quadratic and linear terms together from a general quadratic expression,
  • Adding and subtracting the square of half the coefficient of the linear term within the parenthesis.
For example, given:
  • \( x^2 - 8x \), it becomes \( (x - 4)^2 - 16 \)
This process was applied for all variables in our sphere equation, resulting in its standard form. This step was crucial in allowing us to easily identify the sphere's center and radius, and set up the tangent plane equations.
Tangent Point
A tangent point is where a tangent plane just touches the surface of a 3D shape without cutting through it. In the context of the exercise, the ellipsoid and the sphere share this tangent point. At the point
  • \((1, 1, 2)\)
Both shapes share a common tangent plane indicating they are touching at precisely one point. To find this plane, we use the tangent plane equation for spheres:
  • \((x_0 - h)(x - x_0) + (y_0 - k)(y - y_0) + (z_0 - l)(z - z_0) = 0\)
For our shapes, substituting the coordinates of \((1,1,2)\) and the sphere鈥檚 center \((4,3,4)\), we arrive at:
  • \(-(x - 1) - 2(y - 1) - 2(z - 2) = 0\)
Thus, they not only touch at the defined point but also share the same tangent behavior there, confirming the common tangent plane.

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Most popular questions from this chapter

Find the absolute maximum and minimum values of \(f\) on the set \(D\). $$\begin{array}{l}{f(x, y)=x^{4}+y^{4}-4 x y+2} \\ {D=\\{(x, y) | 0 \leq x \leqslant 3,0 \leq y \leqslant 2\\}}\end{array}$$

Find the point on the plane \(x-2 y+3 z=6\) that is closest to the point \((0,1,1) .\)

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y(1-x-y)$$

Suppose that a scientist has reason to believe that two quantities \(x\) and \(y\) are related linearly, that is, \(y=m x+b,\) at least approximately, for some values of \(m\) and \(b\) . The scientist performs an experiment and collects data in the form of points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) and then plots these points. The points don't lie exactly on a straight line, so the scientist wants to find constants \(m\) and \(b\) so that the line \(y=m x+b\) "fits" the points as well as possible (see the figure). Let \(d_{i}=y_{i}-\left(m x_{i}+b\right)\) be the vertical deviation of the point \(\left(x_{i}, y_{i}\right)\) from the line. The method of least squares determines \(m\) and \(b\) so as to minimize \(\Sigma_{i-1}^{n} d_{i}^{2},\) the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when $$\begin{array}{c}{m \sum_{i=1}^{n} x_{i}+b n=\sum_{i=1}^{n} y_{i}} \\ {m \sum_{i=1}^{n} x_{i}^{2}+b \sum_{l=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i} y_{i}}\end{array}$$ Thus the line is found by solving these two equations in the two unknowns \(m\) and \(b .\)

(a) Maximize \(\sum_{i=1}^{n} x_{i} y_{i}\) subject to the constraints \(\sum_{i=1}^{n} x_{i}^{2}=1\) and \(\sum_{i=1}^{n} y_{i}^{2}=1\) (b) Put $$ x_{i}=\frac{a_{i}}{\sqrt{\sum a_{j}^{2}}} \quad \text { and } \quad y_{i}=\frac{b_{i}}{\sqrt{\sum b_{j}^{2}}} $$ to show that $$ \sum a_{i} b_{i} \leqslant \sqrt{\sum a_{j}^{2}} \sqrt{\Sigma b_{j}^{2}} $$ for any numbers \(a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n} .\) This inequality is known as the Cauchy-Schwarz Inequality.
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