PAGE 1

Orthogonal Vectors Problem

For what value of \( a \) is the vector \( \vec{v} = \langle 5, 8, 3 \rangle \) orthogonal to the vector \( \vec{w} = \langle a, -3, 7 \rangle \)?

A. 9
B. \( \frac{5}{3} \)
C. \( \frac{9}{5} \)
D. \( \frac{1}{9} \)
E. \( \frac{3}{5} \) (Selected)

Figure: Sketch of two vectors v and w with an angle theta between them.

Solution Steps

Angle between vectors:

Dot product: \( \vec{v} \cdot \vec{w} = |\vec{v}| |\vec{w}| \cos \theta \)

Orthogonal: \( \vec{v} \cdot \vec{w} = 0 \)

\[ \vec{v} \cdot \vec{w} = 5a - 24 + 21 = 0 \]\[ 5a = 3 \]\[ a = 3/5 \]

PAGE 2

Vector Projection Problem

Let \( \vec{u} = \langle 10, 5 \rangle \) and \( \vec{v} = \langle 2, 6 \rangle \). Find \( \text{proj}_{\vec{v}}(\vec{u}) \), the orthogonal projection of \( \vec{u} \) onto \( \vec{v} \).

A. \( \langle 4, 2 \rangle \)
B. \( \langle \frac{1}{2}, \frac{3}{2} \rangle \)
C. \( \langle 5\sqrt{10}, 15\sqrt{10} \rangle \)
D. \( \langle \frac{5}{2}, \frac{15}{2} \rangle \) (Selected)
E. \( \langle 20\sqrt{5}, 10\sqrt{5} \rangle \)

Figure: Geometric diagram of vector u being projected onto vector v, showing the resulting projection vector.

Solution Steps

Vector projection:

\[ \text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} \frac{\vec{v}}{|\vec{v}|} \]

Where \( \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} \) is the scalar projection \( \text{scal}_{\vec{v}} \vec{u} \).

\( \vec{u} \cdot \vec{v} = 20 + 30 = 50 \)

\( |\vec{v}| = \sqrt{2^2 + 6^2} = \sqrt{40} \)

\[ \frac{50}{\sqrt{40}} \frac{\langle 2, 6 \rangle}{\sqrt{40}} = \frac{50}{40} \langle 2, 6 \rangle = \langle \frac{5}{2}, \frac{15}{2} \rangle \]

PAGE 3

Area Between Curves

Find the area enclosed by the curves \( x = 2y - y^2 \) and \( x = y^2 \).

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

Figure: Coordinate graph showing parabolas x=y^2 and x=2y-y^2 intersecting at (0,0) and (1,1) with shaded area.

parabola opening left or right

y-int: \( x = 0 \)

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

\( 0 = y(2 - y) \)

\( y = 0, y = 2 \)

Solution Steps

integrate in \( y \) since equations are \( x \) as functions of \( y \)

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

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

\( 2y(y - 1) = 0 \) → \( y = 0, y = 1 \)

\( \int_{0}^{1} [ (2y - y^2) - (y^2) ] dy = \int_{0}^{1} 2y - 2y^2 dy \)

\( = y^2 - \frac{2}{3}y^3 \Big|_0^1 = 1 - \frac{2}{3} = \frac{1}{3} \)

PAGE 4

Volume of a Solid of Revolution

Find a formula for the volume of the following solid: The region bounded by \( y = \sin(x) \), the \( x \)-axis, \( x = 0 \) and \( x = \frac{\pi}{2} \), is revolved around the line \( x = -1 \).

A. \( \int_{0}^{\pi/2} 2\pi(x - 1) \sin(x) dx \)
B. \( \int_{0}^{1} \pi \{[\sin^{-1}(y)]^2 - 1\} dy \)
C. \( \int_{0}^{1} \pi [\sin^{-1}(y) - 1]^2 dy \)
D \( \int_{0}^{\pi/2} 2\pi(x + 1) \sin(x) dx \)
E. \( \int_{0}^{\pi/2} 2\pi x \sin(x + 1) dx \)

Figure: Diagram showing the region under y=sin(x) and a representative shell at distance x+1 from the axis x=-1.

Method: Cylindrical Shells

Shell: rectangle parallel to axis of revolution

\( \int_{0}^{\pi/2} 2\pi (\text{radius}) (\text{height}) (\text{thickness}) \)

Radius: \( 1 + x \)
Height: \( \sin(x) \)
Thickness: \( dx \)

\( = \int_{0}^{\pi/2} 2\pi (1 + x) \sin(x) dx \)

PAGE 5

Disk/Washer Method

When using the disk or washer method, we consider a rectangle perpendicular to the axis of revolution. The volume is calculated by integrating the difference between the outer and inner circular areas over the thickness of the slice.

\[ \int [\pi (\text{outer radius})^2 - \pi (\text{inner radius})^2] (\text{thickness}) \]

Figure: Coordinate graph showing a region in the first quadrant bounded by y=sin(x) and x=pi/2, with a horizontal slice.

Based on the geometry shown in the figure:

Outer radius = \( \frac{\pi}{2} + 1 \)
Inner radius = \( 1 + \sin^{-1}(y) \)
Thickness = \( dy \)

Substituting these values into the integral from \( y = 0 \) to \( y = 1 \):

\[ = \int_{0}^{1} [\pi (\frac{\pi}{2} + 1)^2 - \pi (1 + \sin^{-1}(y))^2] dy \]

PAGE 6

If the method of washers is used, the volume of a solid obtained by revolving a certain region about the \( x \)-axis is given by:

\[ \pi \int_{0}^{1} (x - x^4) dx \]

What integral below represents the same volume if the method of cylindrical shells is used?

A. \( 2\pi \int_{0}^{1} (y)(\sqrt{y} - y^2) dy \)
B. \( 2\pi \int_{0}^{1} (y)(y^2 - \sqrt{y}) dy \)
C. \( 2\pi \int_{0}^{1} (x)(x - x^4) dx \)
D. \( 2\pi \int_{0}^{1} (x)(x^4 - x) dx \)
E. \( 2\pi \int_{0}^{1} (y)(\sqrt{y} - y) dy \)

Analysis

From the given washer integral, we identify the components for vertical rectangles:

\( \pi \int_{0}^{1} ((\text{outer})^2 - (\text{inner})^2) dx \)
\( x = (\text{outer})^2 \implies \text{outer} = \sqrt{x} \implies y = \sqrt{x} \)
\( x^4 = (\text{inner})^2 \implies \text{inner} = x^2 \implies y = x^2 \)

Figure: Graph showing the region between y=sqrt(x) and y=x^2 with a horizontal shell radius labeled as y.

Converting to horizontal shells for integration with respect to \( y \):

Right curve: \( y = x^2 \implies x = \sqrt{y} \)
Left curve: \( y = \sqrt{x} \implies x = y^2 \)
Radius = \( y \)

The shell method integral is:

\[ 2\pi \int_{0}^{1} (y)(\sqrt{y} - y^2) dy \]

Correct Answer: A

PAGE 7

A swimming pool has a rectangular base that is 5 m long and 6 m wide. The sides are 2 m high and the pool is half full of water. How much work will it take to empty the pool by pumping the water out over the top of the pool? Write your answer in terms of the gravitational acceleration constant \(g\) and the density of water \(\rho\).

A. \(75\rho g\) N·m
B. \(90\rho g\) N·m
C. \(60\rho g\) N·m
D. \(30\rho g\) N·m
E \(45\rho g\) N·m

Figure: 3D sketch of a rectangular pool with base 5 by 6 and height 2. A horizontal slice dy is shown at height y.

Calculations

\(\text{mass} = \text{density} \cdot \text{volume}\)

\(= \rho \cdot 5 \cdot 6 \cdot dy = 30\rho \, dy\)

\(\text{weight} = \text{mass} \cdot \text{gravity}\)

\(= 30\rho g \, dy\)

\(\text{work} = \text{weight} \cdot \text{distance}\)

\(= 30\rho g (2-y) \, dy\)

Note: The distance to the top is \(2-y\) because the pool is 2m high.

Total Work Calculation:

\[ \text{total work} = \int_{0}^{1} 30\rho g (2-y) \, dy \]\[ = 30\rho g \int_{0}^{1} (2-y) \, dy \]\[ = 30\rho g \left[ 2y - \frac{1}{2}y^2 \right]_{0}^{1} \]\[ = 30\rho g \cdot \frac{3}{2} = 45\rho g \]

Integration limits are from 0 (bottom of pool) to 1 (water surface) since the pool is half full.