Graphical Sketch - Ellipse, 2

Category: Analytic Geometry, Algebra

"Published in Suisun City, California, USA"

Sketch the graph for the conic section

Solution:

The given equation above represents a conic section. The general equation of any conic sections is given by the equation

To identify the type of a conic section, let's consider the following conditions, if

A = C and B = 0, then it is a Circle
B² - 4AC = 0, then it is a Parabola
B² - 4AC < 0, then it is an Ellipse
B² - 4AC > 0, then it is a Hyperbola

Now, let's identify the type of conic section for the given equation above as follows

B² - 4AC = (192)² - 4(153)(97)
               = 36864 - 59364
               = - 22500

Since B² - 4AC < 0, then the given equation is an Ellipse.

Since the equation of an Ellipse has an xy term, then the major and minor axes of an Ellipse are not parallel to x-axis and y-axis. First, let's get the angle of inclination of an Ellipse as follows

Next, get the value of Cos 2θ as follows

Photo by Math Principles in Everyday Life

Use the Half-Angle Formula to get the values of Sin θ and Cos θ as follows

Next, we need to simplify the given equation in order to eliminate the xy term. The equations that we will use are the following

and

Substitute the values of Sin θ and Cos θ to the two equations above

and

Substitute the two equations above to the given equation, we have

Divide both sides of the equation by 225, we have

Since the value of a² is at y'², then the major axis is parallel to y' axis.

To draw the x' axis, we will use the values of Sin θ and Cos θ. Count 4 units to the right from the origin since the center of an Ellipse is C(0, 0) and then 3 units upward. Connect that point with the origin and we have now x' axis.

To draw the y' axis, we will use the values of Sin θ and Cos θ. Count 4 units upward from the origin since the center of an Ellipse is C(0, 0) and then 3 units to the left. Connect that point with the origin and we have now y' axis.

In the next procedure, we will use the x' axis and y' axis to sketch the graph of an Ellipse.

If

then

If

then

The distance of the two foci from the center of an ellipse, c is calculated as follows

The distance of the four end points of latera recta from the two foci of an ellipse, p is calculated as follows

Label the center, four vertices, two foci, and the four ends of the latera recta using the x' axis and y' axis. The graph of an Ellipse will be like this

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Newtons Law - Cooling

Category: Chemical Engineering Math, Algebra

"Published in Suisun City, California, USA"

A cup of coffee has a temperature of 200 ºF and is placed in a room that has a temperature of 70 ºF. After 10 minutes, the temperature of the coffee is 150 ºF. Find a formula for the temperature of the coffee at time t. Find the temperature of the coffee after 15 minutes. When will the coffee have cooled to 100 ºF?

Solution:

Since the above word problem is about the cooling of a coffee at a certain period of time at a constant temperature of a surrounding, then we will use the Newton's Law of Cooling. The Newton's Law of Cooling is given by the formula as follows

where

          T(t) = final temperature at time t
           T_s = temperature of a surrounding
           D_o = initial temperature difference of the object and the surrounding
             k = constant of cooling
              t = time of cooling

In this case, the temperature of a room is T_s = 70 ºF. The initial temperature difference of a coffee in a room will be

The Newton's Law of Cooling of a coffee at time t is

To solve for the value of k, we need to substitute T = 150 ºF and t = 10 mins to the above equation as follows

Take Natural Logarithm on both sides of the equation

Therefore, the working equation is

After 15 minutes, the temperature of a coffee will be

The temperature of a coffee will be cooled down to 100 ºF at

Take Natural Logarithm on both sides on the equation

Finding Equation - Hyperbola

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the equation of hyperbola if the asymptotes are x + y = 1 and x - y = 1, and passing through through the point (3, 1).

Solution:

The first thing that we have to do is to get the point of intersection of the asymptotes as follows

       x + y = 1                            x + y = 1
                        --------------->      
        x - y = 1                             x - y = 1
                                              -----------------
                                                    2x = 2
                      
                                                      x = 1

Substitute the value of x to either one of the asymptotes to get the value of y, we have

                                x + y = 1
                                1 + y = 1
                                      y = 0

The center of the hyperbola is C(1, 0).

To illustrate the problem, it is better if you sketch the graph of the two asymptotes and a point as follows

Photo by Math Principles in Everyday Life

Next, we need to get the values of a and b which are the values of semi-transverse axis and semi-conjugate axis of hyperbola. Since the slopes of two asymptotes of the hyperbola are +1 and -1, we can solve for the values of a and b as follows

Consider the positive sign in getting the values of a and b and the above equation becomes

The equation of the hyperbola in standard form if the transverse axis is parallel to x-axis is

But C(1, 0) and a = b, the above equation becomes

To solve for the value of a, substitute the values of x and y from the given point P(3, 1) as follows

Therefore, the equation of hyperbola in standard form is

We can also express the equation of hyperbola in general form as follows

Multiply both sides of the equation by 3, we have