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PAGE 1

Vector Perpendicularity Problem

Find \( a \) such that \( \mathbf{u} = 2\mathbf{i} - \mathbf{j} + a\mathbf{k} \) and \( \mathbf{v} = \mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \) are perpendicular.

A. 3
B. 2
C. 1
D. -1
E. -2

Solution

If \( \vec{u} \perp \vec{w} \), then \( \vec{u} \cdot \vec{w} = 0 \)

\[ \vec{u} \cdot \vec{w} = \langle 2, -1, a \rangle \cdot \langle 1, 4, 2 \rangle = 0 \]

\[ (2)(1) + (-1)(4) + (a)(2) = 0 \]

\[ 2 - 4 + 2a = 0 \]

\[ 2a = 2 \]

\[ a = 1 \]

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Vector Projection Magnitude

If \( \mathbf{v} = \mathbf{i} + \mathbf{j} + \mathbf{k} \) and \( \mathbf{w} = 2\mathbf{i} - \mathbf{k} \), find \( |\text{proj}_{\mathbf{v}}(\mathbf{w})| \).

A. \( 1/\sqrt{3} \)
B. \( \sqrt{3} \)
C. \( \sqrt{3}/5 \)
D. \( 2\sqrt{3} \)
E. \( \sqrt{3}/2 \)

Geometric Interpretation

The magnitude of the projection of \( \vec{w} \) onto \( \vec{v} \) is equal to the scalar projection of \( \vec{w} \) onto \( \vec{v} \).

\[ |\text{proj}_{\vec{v}} \vec{w}| = \text{scal}_{\vec{v}} \vec{w} \]
Vector diagram showing vector w at an angle to vector v, with proj_v(w) along v.
Right triangle with hypotenuse magnitude of w, angle theta, and base labeled scal_v(w).

Derivation

\[ \cos \theta = \frac{\text{scal}_{\vec{v}} \vec{w}}{|\vec{w}|} \]

\[ \text{scal}_{\vec{v}} \vec{w} = |\vec{w}| \cos \theta \]

\[ \vec{v} \cdot \vec{w} = |\vec{v}| |\vec{w}| \cos \theta \]

\[ \cos \theta = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| |\vec{w}|} \]

Calculation

\[ \text{scal}_{\vec{v}} \vec{w} = |\vec{w}| \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| |\vec{w}|} = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}|} \]

\[ = \frac{\langle 1, 1, 1 \rangle \cdot \langle 2, 0, -1 \rangle}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}} \]

PAGE 3

Area of a Triangle in 3D Space

Find the area of the triangle with vertices \( P = (0, 0, 0) \), \( Q = (1, 2, 1) \), and \( R = (2, 1, -1) \).

A. \( \sqrt{27} \)
B. \( \frac{\sqrt{27}}{2} \) (Correct)
C. \( \frac{\sqrt{11}}{2} \)
D. \( \sqrt{19} \)
E. \( \frac{\sqrt{3}}{2} \)

Vector Representation

We define vectors \( \vec{u} \) and \( \vec{v} \) from the vertex \( P \):

\[ \vec{u} = \vec{PR} = \langle 2-0, 1-0, -1-0 \rangle = \langle 2, 1, -1 \rangle \]\[ \vec{v} = \vec{PQ} = \langle 1, 2, 1 \rangle \]
Sketch of vectors u and v forming a triangle with area 1/2 |u x v| and a parallelogram with area |v x u|.

Cross Product Calculation

The area of the parallelogram formed by \( \vec{u} \) and \( \vec{v} \) is given by the magnitude of their cross product \( |\vec{u} \times \vec{v}| \).

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = \vec{i} \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} - \vec{j} \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} + \vec{k} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} \]\[ = \vec{i}(1 - (-2)) - \vec{j}(2 - (-1)) + \vec{k}(4 - 1) \]\[ = \langle 3, -3, 3 \rangle \]

Final Area Calculation

\( |\vec{u} \times \vec{v}| = \sqrt{9 + 9 + 9} = \sqrt{27} \) = area of parallelogram

Triangle Area: \( \frac{\sqrt{27}}{2} \)

PAGE 4

Area Between Curves

The area of the region enclosed by the curves \( y = x^2 + 1 \) and \( y = 2x + 9 \) is given by:

A. \( \int_{-2}^{4} (x^2 + 1 - 2x - 9) \, dx \)
B. \( \int_{-2}^{4} (2x + 9 - x^2 - 1) \, dx \) (Correct)
C. \( \int_{-2}^{2} (2x + 9 - x^2 - 1) \, dx \)
D. \( \int_{-4}^{2} (2x + 9 - x^2 - 1) \, dx \)
E. \( \int_{-4}^{2} (x^2 + 1 - 2x - 9) \, dx \)

Finding Intersection Points

Set the equations equal to find the limits of integration:

\[ x^2 + 1 = 2x + 9 \]\[ x^2 - 2x - 8 = 0 \]\[ (x - 4)(x + 2) = 0 \]\[ x = -2, \, x = 4 \]
Coordinate graph showing the region between line y=2x+9 and parabola y=x^2+1 from x=-2 to x=4.

Integral Setup

The area is calculated by integrating the top function minus the bottom function over the interval:

\[ \text{area} = \int_{\text{left}}^{\text{right}} (\text{top} - \text{bottom}) \, dx \]\[ = \int_{-2}^{4} [(2x + 9) - (x^2 + 1)] \, dx \]\[ = \int_{-2}^{4} (2x + 9 - x^2 - 1) \, dx \]
PAGE 5

Volume of Solids of Revolution

Let \( R \) be the region between the graphs of \( y = x^2 \) and \( y = x \). Find the volume of the solid generated by revolving \( R \) about the \( x \)-axis.

A. \( \frac{\pi}{6} \)
B. \( \frac{\pi}{12} \)
C. \( \frac{\pi}{4} \)
D. \( \frac{\pi}{15} \)
E. \( \frac{2\pi}{15} \)
Coordinate graph showing the region R between y=x and y=x^2 from x=0 to x=1.
3D sketch of the solid generated by revolving the region R around the x-axis.

Two Methods:

  • Disk / Washer
  • Shell

Disk Method: Rectangle Perpendicular to Axis of Revolution

\( \text{disk volume} = \pi r^2 dx \)

\( [\pi(r_{\text{out}})^2 - \pi(r_{\text{in}})^2] dx \)

\( = [\pi(x)^2 - \pi(x^2)^2] dx \)

\( = \pi(x^2 - x^4) dx \)

Diagram of a washer element with thickness dx, outer radius r_o and inner radius r_i.

\( \text{Total Volume} = \int_{0}^{1} \pi(x^2 - x^4) dx \)

\( = \pi \left( \frac{x^3}{3} - \frac{x^5}{5} \right) \Big|_0^1 \)

\( = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2}{15}\pi \)

PAGE 6

If \( R \) is the region bounded by the curves \( x = 0 \) and \( x = y - y^2 \), and if \( R \) is revolved around the \( y \)-axis, then the volume of the solid is

A. \( \frac{\pi}{15} \)
B. \( \frac{\pi}{30} \)
C. \( \frac{\pi}{12} \)
D. \( \frac{\pi}{3} \)
E. \( \frac{\pi}{6} \)
Graph of x = y - y^2 from y=0 to y=1 with a vertical representative rectangle.
Sketch of the solid formed by revolving the parabolic region around the y-axis.

\( x = y - y^2 \) parabola opening left/right

intercepts: \( 0 = y - y^2 \)

\( = y(1 - y) \)

\( y = 0, y = 1 \)

at \( y = \frac{1}{2}, x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \)

Method Selection

This time, let's use shell: rectangle \( \parallel \) axis

But the same curve is at the ends of the rectangle, not convenient. Difficult to express height.

So, back to disk then:

A horizontal disk element with radius r and thickness dy.
Graph of the region with a horizontal representative rectangle of thickness dy.

\( \text{disk volume} = \pi (r)^2 dy \)

\( = \pi (y - y^2)^2 dy \)

PAGE 7

Integration Calculation

Integrate bottom (␀y = 0␁) to top (␀y = 1␁)

\[ \int_{0}^{1} \pi (y - y^2)^2 dy = \pi \int_{0}^{1} y^2 - 2y^3 + y^4 dy \]\[ = \pi \left( \frac{y^3}{3} - \frac{y^4}{2} + \frac{y^5}{5} \right) \bigg|_0^1 \]\[ = \pi \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = \pi \left( \frac{10 - 15 + 6}{30} \right) = \pi \left( \frac{1}{30} \right) \]
PAGE 8

Volume by Shell Method

\( R \) bounded by \( y = x^2 \), \( y = x \), around x-axis but shell this time.

rectangle \( \parallel \) axis of rev

Graph of y=x and y=x^2 on the xy-plane, showing a horizontal rectangle at height y with radius y.

\( \text{volume} = 2\pi (\text{radius}) (\text{height}) dy \)

\( = 2\pi (y) (\sqrt{y} - y) dy = 2\pi (y^{3/2} - y^2) dy \)

Diagram of a cylindrical shell with radius y, height sqrt(y)-y, and thickness dy.

\( \text{volume of whole thing} \)

\[ \int_{\text{bottom } 0}^{\text{top } 1} 2\pi (y^{3/2} - y^2) dy = 2\pi \left( \frac{2}{5} y^{5/2} - \frac{1}{3} y^3 \right) \bigg|_0^1 \]\[ = 2\pi \left( \frac{2}{5} - \frac{1}{3} \right) = 2\pi \left( \frac{1}{15} \right) \]\[ = \frac{2\pi}{15} \]
A rectangle representing the unrolled shell with length 2 pi times radius and height.
PAGE 9

A force of 4 lb. is required to stretch a spring 1/2 ft. beyond its natural length. How much work is required to stretch the spring from its natural length to 2 ft.

A. 8 ft-lbs.
B. 12 ft-lbs.
C. 16 ft-lbs.
D. 24 ft-lbs.
E. 32 ft-lbs.
Note: 2 ft. beyond natural

Spring Force (Hooke's Law)

\[ F = kx \]
  • F: force
  • k: spring constant
  • x: change from natural length

Finding the Spring Constant

Given the initial conditions:

\[ 4 = k \cdot \frac{1}{2} \]

Stretch \( \frac{1}{2} \) ft beyond natural (\( -\frac{1}{2} \) if compressed)

So, \( k = 8 \)

Calculating Work

Work is the integral of force over the distance stretched:

\[ \text{work} = \int \text{force} = \int_{\text{start-natural}}^{\text{end-natural}} kx \, dx = \int_{0}^{2} 8x \, dx = 4x^2 \Big|_0^2 = 16 \]

The total work required is 16 ft-lbs.