Solving Vector Problems

Category: Trigonometry

"Published in Newark, California, USA"

A pilot flew in the direction of 140° from A to B and then in the direction of 240° from B to C. A is 500 miles from B and 800 miles from C. Find the distance of the last flight and the direction from A to C. 

Solution:

The first thing that we have to do is to analyze the problem very well. The given angles are expressed as course. Typically course is measured in degrees from 0° clockwise to 360° in compass convention (0° being north, 90° being east).  
Let's draw the figure to analyze the given word problem as follows

Photo by Math Principles in Everyday Life

Complete further the labeling of the above figure as follows

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Next, use Cosine Law to solve for the distance of the last flight from A to C

Therefore, the distance of the last flight from A to C is 1,040.0154 miles.

Finally, use Sine Law to solve for the direction of the last flight from A to C

or

Therefore, the direction of the last flight from A to C is 140° + 49°14'50" = 189°14'50".

Right Circular Cylinder - Right Circular Cone, 2

Category: Solid Geometry

"Published in Newark, California, USA"

An ink bottle is in the form of a right circular cylinder with a large conical opening as shown in the figure. When it is filled with the bottom of the opening, it can just be turned upside down without any ink spilling. Prove that the depth of the cone is ³/₅ the depth of the bottle.

Photo by Math Principles in Everyday Life

Solution:

When you flipped the ink bottle, there should be no spills of ink as shown below

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The vertical section of the above figure will be like this

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As you can see in the figure that the level of the ink is exactly at the tip of the right circular cone. If it is above the tip of the right circular cone, then there will be the spillage of the ink. Let's label further the given figure as shown below

Photo by Math Principles in Everyday Life

                  Volume before Inversion = Volume after Inversion

but

The above equation becomes

Therefore

Graphical Sketch - Hyperbola

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Sketch the graph for

Solution:

Consider the given equation above

The above equation represents a hyperbola because one of the 2nd degree variable is negative. If y² is negative, then the curve of the hyperbola opens to the left and to the right. If x² is negative, then the curve of the hyperbola opens upward and downward. 

The first step is to reduce the given equation of the hyperbola to its standard form as follows

Group the above terms according to its variables

Complete the square of the grouped terms

Divide both sides of the equation by 36

Since y is negative, then the curve of the hyperbola opens to the left and to the right. The transverse axis is parallel to the x-axis where the center, two foci, and two vertices are located. 

The center of the hyperbola is C(-1, 2).

The semi-major axis of the hyperbola is 

Therefore, the main vertices of the hyperbola are 

                      V₁(-1 + 2, 2) = V₁(1, 2)
                      V₂(-1 - 2, 2) = V₂(-3, 2)

The semi-minor axis of the hyperbola is

Therefore, the other vertices of the hyperbola are

                    V₃(-1, 2 + 1½) = V₃(-1, 3½)
                    V₄(-1, 2 - 1½) = V₄(-1, ½)

These two vertices of the hyperbola will not be used to draw the curve proper but it is helpful to draw the imaginary rectangle. To draw the imaginary rectangle, draw a light vertical line that passes through V₁, another light vertical line that passes through V₂, a light horizontal line that passes through V₃, and another light horizontal line that passes through V₄. The resulting imaginary rectangle will be a guide to draw the two asymptotes of the hyperbola in which you will connect and extend the two opposite vertices of a rectangle. The two asymptotes of the hyperbola should meet or intersect at the center of the hyperbola. The two equations of the asymptotes of the hyperbola are

Change the right side of the equation to zero, we have

Factor the above equation as the sum and the difference of two squares

Equate each factor to zero to get the equations of the two asymptotes.

For the first asymptote of the hyperbola

Multiply both sides of the equation by 6, we have

For the second asymptote of the hyperbola

Multiply both sides of the equation by 6, we have

Next, we need to get the coordinates of the foci and the ends of the latera recta of the hyperbola. First, we need to solve for the value of c which is the distance of the focus to the center of the hyperbola as follows

Therefore, the two foci of the hyperbola are

                    F₁(-1 + 2½, 2) = F₁(1½, 2)
                    F₂(-1 - 2½, 2) = F₂(-3½, 2)

Next, solve for the length of the latus rectum as follows

Therefore, the end points of the latera recta are

                    L₁(1½, 2 + 1⅛) = L₁(1½, 3⅛)
                    L₂(1½, 2 - 1⅛) = L₂(1½, ⅞)
                    L₃(-3½, 2 + 1⅛) = L₃(-3½, 3⅛)
                    L₄(-3½, 2 - 1⅛) = L₄(-3½, ⅞)

Finally, connect the points L₃, V₂, and L₄ to draw a curve that opens to the left and connect the points L₁, V₁, and L₂ to draw a curve that opens to the right as follows

Photo by Math Principles in Everyday Life