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

Graphical Sketch - Plane

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Sketch the graph for 

Also, find the intercepts and the distance of the plane from the origin.

Solution:

The given equation above represents a plane in the x, y, and z axes. To sketch the graph of a plane, let's consider the following procedure as follows

For 

  
Set x = 0 and the equation becomes 

or

If y = 0, then z = 2 and if z = 0, then y = 4.

Set y = 0 and the equation becomes

or

If x = 0, then z = 2 and if z = 0, then x = 6.

Set z = 0 and the equation becomes

If x = 0, then y = 4 and if y = 0, then x = 6. 

Therefore the intercepts are:

                       x-intercept = 6
                       y-intercept = 4
                       z-intercept = 2

From the values of intercepts, we can plot the points in the x, y, and z axes and sketch the graph by connecting all the points as follows

Photo by Math Principles in Everyday Life

The distance from a point to a plane is given by the formula

where the sign of the radical is the opposite sign of D.

In this case if the equation of a plane is 

and a point is the origin, therefore the distance of a plane to the origin is 

The negative sign means that the plane is above the given point or origin.