Difference between revisions of "Calculus III Advanced (Course) (11.2) (Homework)"
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Latest revision as of 14:38, 7 February 2017
Contents
Section 11.2 Homework
From Calculus 10e by Larson and Edwards, p. 763. Exercises 6, 26, 44, 58, 84.
Exercise 11.2.6 Finding Coordinates for a Point
The point is located seven units in front of the \(yz\)-plane, two units to the left of the \(xz\)-plane, and one unit below the \(xy\)-plane.
Solution
- x = 7
- y = 4
- z = -1
- x = 7
Exercise 11.2.26 Finding the Distance Between Two Points in Three Dimensions
\((2,2,3),(4,−5,6)\)
Solution
Distance | \( =\sqrt{(2−4)^{2}+(2−(−5))^{2}+(3−6)^{2}}\) |
\(=\sqrt{4+49+9}\) | |
\(=\sqrt{62}\) |
Exercise 11.2.44 Finding the Equation for a Sphere
Complete the square to write the equation for the sphere in standard form. Given \(4x^{2}+4y^{2}+4z^{2}−24x−4y+8z−23=0\), find the center and radius.
Solution
\(0\) | \(=4x^{2}+4y^{2}+4z^{2}−24x−4y+8z−23\) | |
\(=(4x^{2}−24x) +(4y^{2}−4y)+(4z^{2}+8z)−23\) | Group Like Items | |
\(=4(x^{2}−6x) +4(y^{2}−y)+4(z^{2}+2z)−23\) | Extract the common multiplier, here 4. | |
\(=4(x^{2}-3x-3x+9−9) +4(y^{2}-1/2y-1/2y+1/4−1/4)+4(z^{2}+z+z +1-1)−23\) | Complete the Square per variable using the \([b*1/2]^{2}\) method. | |
\(=4[(x^{2}-3x-3x+9)−9] +4[(y^{2}-1/2y-1/2y+1/4)−1/4]+4[(z^{2}+z+z+1)-1]−23\) | Group by Squares. | |
\(=4[(x-3)^{2}−9] +4[(y-1/2)^{2}−1/4]+4[(z+1)^{2}-1]−23\) | Reduce | |
\(=4(x-3)^{2}−36+4(y-1/2)^{2}−1+4(z+1)^{2}-4−23\) | Distribute 4. | |
\(=4(x-3)^{2}+4(y-1/2)^{2}+4(z+1)^{2}−64\) | Sum the stray integers. | |
\(=(x-3)^{2}+(y-1/2)^{2}+(z+1)^{2}−16\) | Divide by 4. | |
\(16\) | \(=(x-3)^{2}+(y-1/2)^{2}+(z+1)^{2}\) | Set equal to the stray integer. |
\(4\) | \(=(x-3)+(y-1/2)+(z+1)\) | Take the root. |
The center is \((-3,-1/2,1)\) and the radius is 4.
Exercise 11.2.58 Finding a Vector
Find the vector \(\textbf{z} = 5\textbf{u}-3\textbf{v}-1/2\textbf{w}\), given that \(\textbf{u}= \langle 1,2,3 \rangle\), \(\textbf{v}= \langle 2,2,-1 \rangle\), and \(\textbf{w}= \langle 4,0,-4 \rangle\).
Solution
\(\textbf{z}\) | \( = 5\langle 1,2,3 \rangle-3\langle 2,2,-1 \rangle-1/2\langle 4,0,-4 \rangle\) |
\(= \langle 5,10,15 \rangle + \langle -6,-6,3 \rangle + \langle -2,0, 2 \rangle \) | |
\(= \langle -3, 4, 20 \rangle \) |
Exercise 11.2.84 Finding a Vector
Find the vector \(\textbf{v}\) with the given magnitude and the same direction as \(\textbf{u}\).
Magnitude | Direction |
---|---|
\(\left \| \textbf{v} \right \| = 3\) | \(\textbf{u}= \langle 1,1,1 \rangle\) |
Solution
Magnitude for \(\textbf{u}\) is
\( \left \| \textbf{u} \right \| = \sqrt{ 1^{2} + 1^{2} + 1^{2} } = \sqrt{3} \)
The answer is \( \langle 7/\sqrt{3},7/\sqrt{3},7/\sqrt{3} \rangle\).
Internal Links
Parent Article: Calculus III Advanced (Course)