Calculus III 11 Exam 1
Contents
- 1 Chapter 11 Exam
- 1.1 Exercise 11 Find the Equation for a Sphere
- 1.2 Exercise 20 Find the Dot Product
- 1.3 Exercise 30 Find the projection for \(\textbf{u}\) onto \( \textbf{v} \)
- 1.4 Exercise 36 Cross Product
- 1.5 Exercise 39 Torque
- 1.6 Exercise 46 Find the Parametric Equations
- 1.7 Exercise 50 Find an Equation for a Plane
- 1.8 Exercise 54 Distance
- 1.9 Exercise 60 Sketch and Describe the Surface
- 1.10 Exercise 70 Cylindrical-to-Spherical Conversion
- 2 Internal Links
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.
Solution
The \( \textbf{F} \) is where I got it wrong.
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).
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:
$$ \left ( 54\sqrt{3}, -\frac{5\pi}{6}, \frac{\pi}{3} \right ) $$
Internal Links
Parent Article: Calculus III Advanced (Course)