Calculus III 11 Exam 1

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Chapter 11 Exam

From Calculus 10e by Larson and Edwards, p. 811. Exercises 11, 20, 22, 30, 36, 39, 46, 50, 54, 60, 70.

Exercise 11 Find the Equation for a Sphere

Complete the square and write the equation for the sphere in standard form with the center and radius.

\(x^{2}+y^{2}+z^{2}-4x-6y+4=0\)

Solution Group like elements and complete the square

\( (x^{2}-4x+\: )+(y^{2}-6y+\: )+(z^{2}+\:) \) \(= -4\)
\( (x^{2}-4x+4 )+(y^{2}-6y+9 )+(z^{2}+0) \) \(= -4+4+9+0\)
\( (x-2)^{2}+(y-3)^{2}+z^{2} \) \(= 9\) Equation for the Sphere
$$ \frac{(x-2)^{2}}{9} +\frac{(y-3)^{2}}{9}+\frac{z^{2}}{9} $$ \(=1\) Standard Form

Centered at \((2,3,0)\).
The Radius is 3.

Exercise 20 Find the Dot Product

Let \(\textbf{u}=\vec{PQ}\) and \(\textbf{v}=\vec{PR}\) find
a. The component forms for \(\textbf{u}\) and \(\textbf{v}\)
b. \(\textbf{u} \cdot \textbf{v}\)
c. \(\textbf{v} \cdot \textbf{v}\)

\(P=(2,-1,3), \: Q=(0,5,1), \: R=(5,5,0)\)

Solution
a. The component form for \(\textbf{u}\) is

\(\textbf{u} = \langle 0-2, 5-(-1), 1-3 \rangle = \langle -2, 6, -2 \rangle \)

The component form for \(\textbf{v}\) is

\(\textbf{v} = \langle 5-2, 5-(-1), 0-3 \rangle = \langle 3, 6, -3 \rangle \)

b. \(\textbf{u} \cdot \textbf{v} = \langle -2+3, 6+ 6, -2-3 \rangle = \langle 1,12,-5 \rangle \)
c. \(\textbf{v} \cdot \textbf{v} = \langle 3+3, 6+6, -3-3 \rangle = \langle 6,12,-6 \rangle\)

Exercise 30 Find the projection for \(\textbf{u}\) onto \( \textbf{v} \)

\(\textbf{u} =5\textbf{i} +\textbf{j}+3\textbf{k}\), \( \textbf{v} = 2\textbf{i} +3\textbf{j}+\textbf{k}\)

Solution The projection for \( \textbf{u} \) onto \( \textbf{v} \) is

$$ \text{proj}_{v}\textbf{u} = \left ( \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \textbf{v}$$
\(\textbf{u} \cdot \textbf{v} = \langle 2-5, 3-1, 1-3 \rangle = \langle -3,2,-2 \rangle \)
\( \| \textbf{v} \|^{2} = (\sqrt{2^{2}+3^{2}+1^{1}})^{2} = (\sqrt{4+9+1})^{2} = (\sqrt{14})^{2} = 14 \)
$$=\langle \frac{-3}{7}, \frac{3}{7}, \frac{-1}{7} \rangle $$

Exercise 36 Cross Product

Find the cross product for \(\textbf{u} = \langle 0,2,1 \rangle\), \(\textbf{v} = \langle 1,-3,4 \rangle\)

Solution

$$ \textbf{u} \times \textbf{v} $$ $$= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 0 & 2 & 1 \\ 1 & -3 & 4 \end{vmatrix} $$
\(= \begin{vmatrix} 2 & 1 \\ -3 & 4 \end{vmatrix} \textbf{i} - \begin{vmatrix} 0 & 1 \\ 1 & 4 \end{vmatrix} \textbf{j} + \begin{vmatrix} 0 & 2 \\ 1 & -3 \end{vmatrix} \textbf{k} \)
\(= (8-(-3))\textbf{i} -(0-1)\textbf{j}+(0-2)\textbf{k} \)
\(= 11\textbf{i} + 1\textbf{j}-2\textbf{k} \)

Exercise 39 Torque

The specifications for a tractor state that the torque on a bolt with head size 7/8 inch cannot exceed 200 foot-pounds. Determine the maximum force \( \|\textbf{F}\| \) that can be applied to the wrench in Figure 1.

Figure 1

Solution
The \( \textbf{F} \) is where I got it wrong.

Figure 2

Exercise 46 Find the Parametric Equations

Find the parametric equations for the line that passes through the point (0,1,4) and is perpendicular to \( \textbf{u}=\langle 2,-5,1 \rangle \) and \( \textbf{v}=\langle -3,1,4 \rangle \).
Solution
A vector is needed that is perpendicular, orthogonal, to both \( \textbf{u} \) and \( \textbf{v} \).

$$ \textbf{u} \times \textbf{v} $$ $$= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 2 & -5 & 1 \\ -3 & 1 & 4 \end{vmatrix} $$
\(= \begin{vmatrix} -5 & 1 \\ 1 & 4 \end{vmatrix} \textbf{i} - \begin{vmatrix} 2 & 1 \\ -3 & 4 \end{vmatrix} \textbf{j} + \begin{vmatrix} 2 & -5 \\ -3 & 1 \end{vmatrix} \textbf{k} \)
\(= (-5 \cdot 4-1 \cdot 1)\textbf{i} -(2 \cdot 4 - (-3)(1))\textbf{j}+(2-(-3)(-5))\textbf{k} \)
\(= -21\textbf{i} -11\textbf{j}-13\textbf{k} \)

The vector \( \textbf{w} = -21\textbf{i} -11\textbf{j}-13\textbf{k} \) is perpendicular to both \( \textbf{u} \) and \( \textbf{v} \). Now for a line that is parallel to it and passes through the point (0,1,4).
The coordinates are \(x_{1}=0\), \(y_{1}=1\), and \(z_{1}=4\).
The direction numbers are \(a=-21\), \(b=-11\), and \(c=-13\).

\(x=0-21t\), \(y=1-11t\), \(z=4-13t\) Parametric equations
$$ \frac{x-0}{-21}=\frac{y-1}{-11}=\frac{z-4}{-13} \text{Symmetric equations}$$

Exercise 50 Find an Equation for a Plane

The plane passes through the points (5,1,3) and (2,-2,1) and is perpendicular to the plane \(2x+y-z=4\).
Solution
Let \( textbf{u} \) be the plain \(2x+y-z=4\). Let \( textbf{v} \) be the plain perpendicular to \( textbf{u} \).

\( textbf{u} \cdot textbf{v} \) \(= 0\)
\( \langle 2,1,-1 \rangle \cdot textbf{v} \) \(= 0 \)
\( \langle 2,1,-1 \rangle \cdot \angle 1,1,3 \rangle \) \(= 0 \)

We have a vector \( textbf{v} = \angle 1,1,3 \rangle \).
Now multiply the points together for the next vector.
\( (5,1,3) \times (2,-2,1) = \langle 2-5, -2-1, 1-3 \rangle = \langle -3,-3,-2 \rangle \)

$$= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 1 & 1 & 3 \\ -3 & -3 & -2 \end{vmatrix} $$
\(= \begin{vmatrix} 1 & 3 \\ -3 & -2 \end{vmatrix} \textbf{i} - \begin{vmatrix} 1 & 3 \\ -3 & -2 \end{vmatrix} \textbf{j} + \begin{vmatrix} 1 & 1 \\ -3 & -3 \end{vmatrix} \textbf{k} \)
\(= 7\textbf{i} +7\textbf{j}+0\textbf{k} \)

The plane is \( 7x+7y=0 \)

Exercise 54 Distance

Find the distance between the point (-5,1,3) and the line given by \(x=1+t\), \(y=3-2t\), and \(z=5-t\).
Solution
Let \(Q(-5,1,3)\). The direction numbers are 1, -2, and -1. The direction vector is \(\textbf{u}=\langle 1, -2,-1 \rangle\). Let \(t=0\) to produce \(P=(1,3,5)\).

\(\vec{PQ} = \langle -5 -1 , 1 -3 , 3 -5 \rangle = \langle -6, -2, -2 \rangle \)

$$\vec{PQ} \times \textbf{u} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ -6 & -2 & -2 \\ 1 & -2 & -1 \end{vmatrix} = \langle -2, -8,14 \rangle $$ \(Distance = \sqrt{ 4+64+196} / \sqrt{ 1+4+1} = \sqrt{264}/\sqrt{6} \)

Exercise 60 Sketch and Describe the Surface

\(16x^{2}=16y^{2}-9z^{2}=0\)
The surface is a double cone along the \(x\)-axis with the origin at (0,0,0).

Figure 3

Exercise 70 Cylindrical-to-Spherical Conversion

$$ \left ( 81, -\frac{5\pi}{6}, 27\sqrt{3} \right )$$ Solution
A cylinder is represented by the ordered triple \((r,\theta,z)\). This yields

\(r=81\)
\(\theta = -\frac{5\pi}{6} \)
\(z=27\sqrt{3}\)

A sphere is represented by the ordered triple \((\rho, \theta, \phi)\).
Now for the conversion.

\(\rho = \sqrt{ 81^{2} + ( 27\sqrt{3} )^{2} } = \sqrt{ 6561 + 2187} = \sqrt{ 8748} = 54\sqrt{3} \)

\( \theta = -\frac{5\pi}{6} \)

$$ \phi = \arccos \left ( \frac{27\sqrt{3}}{ \sqrt{ 81^{2} + ( 27\sqrt{3} )^{2} }} \right ) = \arccos \left ( \frac{27\sqrt{3}}{54\sqrt{3}} \right ) = \arccos \left ( \frac{1}{2} \right ) = \frac{\pi}{3} $$

The spherical point is:

$$ \left ( 54\sqrt{3}, -\frac{5\pi}{6}, \frac{\pi}{3} \right ) $$

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Parent Article: Calculus III Advanced (Course)