Rectangular - Polar Coordinate

Category: Analytic Geometry, Algebra, Trigonometry

"Published in Newark, California, USA"

Convert the equation of the hyperbola from Rectangular Coordinate to Polar Coordinate for

Solution:

There are two types of coordinate system in Analytic Geometry. The first type is the Rectangular Coordinate System where the coordinates are x and y. This is a very commonly used in plotting the points and sketching the graphs. The other type is the Polar Coordinate System where the coordinates are r and θ. The radius, r is always positive in the equation unlike the angle, θ can be a positive or a negative. If θ is positive, then the angle is measured in counterclockwise direction and if θ is negative, then the angle is measured in clockwise direction.

To convert the Rectangular Coordinate System to Polar Coordinate System, let's consider a point P(x, y) in a Rectangular Coordinate System. From point P, draw a vertical line perpendicular to x-axis. The resulting figure is a right triangle with r as the hypotenuse or a distance from a point to the origin and θ is the adjacent angle of the hypotenuse from the x-axis. Use Pythagorean Theorem and trigonometric functions to convert the values of x and y into their equivalent value of r and θ as follows

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Consider the given equation above

Substitute the values of x and y as follows

Graphical Sketch - Trigonometric Functions

Category: Trigonometry

"Published in Newark, California, USA"

Draw the graphs of y₁ = Sin x and y₂ = Cos x for ranging from 0 to 2π, and then by the addition of ordinates, obtain the graph of y = Sin x + Cos x. What is the period of Sin x + Cos x and also the amplitude?

Solution:

The first thing that we have to do is to create the table so that it is easier for us to compute their values of y. By the way, all the units are unitless and so their angles must be expressed in radians. You can use calculator so that you can get the accurate value of y.

Note: The conversion of degrees to radians is given by the formula

Next, use your scientific calculator to compute for their values of trigonometric functions as follows

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Since we have now the values of x and y, we can start now the graphing of the given trigonometric functions. First, let's start with the Sine function. As you notice that the trend of the graph is a wave because of the periodically repeating the values of y but the highest and the lowest values are always 1 and -1. 

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For the Cosine function, the trend of the graph is almost the same but it didn't pass through the origin and the highest and the lowest values are always 1 and -1 also.

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To draw the Sin x + Cos x function, you need a compass or divider for this. Using a divider, measure the height or the value of y of the cosine graph from the x-axis and then transfer it or plot the point to the sine function from the curve proper. Connect all the plotted points and the trend of the graph will be like this. 

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If you don't have a compass or a divider, you can use the values of y = Sin x + Cos x from the table to plot the points directly and then connect all the points. As you noticed that the resulted graph is also a repeated wave because of the periodically repeating the values of y also. 

From the trend of a graph and the table, the amplitude or the height of the curve is 1.4142 (+ and -) which is also equal to √2. 

To compute for the period, you need the graph proper to see the trend of a curve. The trend of a curve that we need is a one complete wave or one cycle which is started from y = 0 and then ends at y = 0 that passed the minimum and maximum amplitudes. By looking at the graph, the approximate value of a period is 2.3 - (- 3.9) = 2.3 + 3.9 = 6.2 radians. 

There's another way to compute for the period. You can also use the table and look for the value of y = Sin x + Cos x. As you can see that if y = 0 then x = 2.3562 radians and if y = 0 then x = 5.4978 radians. Therefore, the value of a period is 2(5.4978 - 2.3562) = 6.2852 radians which also equal to 360°.

Rate, Distance, Time - Problem

Category: Algebra

"Published in Newark, California, USA"

A bus bound for Baguio left Manila at exactly 5:00 am traveling at a uniform speed. Thirty minutes later, a car left Manila in pursuit of the bus. Traveling also at a uniform rate, the car overtook the bus 100 miles from the city. Had it increased its speed by 10 miles per hour, it would have overtaken the bus only 50 miles from Manila. Find the rate at which the bus was traveling.

Solution:

Well, this is a complicated problem solving about the rate, distance, and time problem. We have to analyze the given word problem as follows

Let  x = be the speed of a bus
      y = be the speed of a car
       t = time

Let's draw a simple figure to illustrate the problem

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As you noticed that we subtract 30 minutes or ½ hour for the travel time of a car because there's a 30 minute gap between a bus and a car. Since a bus left Manila first, then we have to subtract ½ hour for the travel time of a car. 

                                         Distance = speed x time

From the word statement "the car overtook the bus 100 miles from the city", the working equation for the said statement is

                              Distance of a Bus = Distance of a Car

The total time traveled by a car at uniform speed is 

If the speed of a car is increased by 10 miles per hour, then the total time will be

If you drive faster, you will arrive to your destination earlier. In this case, you save time for commuting or driving if you drive faster and you will have an excess time as well. Excess time is calculated as follows

Excess Time = Total Time at Uniform Speed - Total Time at Increased Speed

From the word statement "Had it increased its speed by 10 miles per hour, it would have overtaken the bus only 50 miles from Manila", then the working equation will be

                                           Distance = Rate x Time

If you choose a positive sign,

Total time traveled by a car

Therefore, the speed of a bus is

If you choose a negative sign,

Total time traveled by a car

Therefore, the speed of a bus is