Integration - Hyperbolic Functions

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

There are two functions in the given equation which are trigonometric and hyperbolic functions. Since there are two functions in the integration, then we have to integrate the given equation by using Integration by Parts. 

If

then 

If

then

Using Integration by Parts,

Since the second term of the above equation have two functions, then we have to use the Integration by Parts again. 

Consider

If

then

If 

then

Again, using by Integration by Parts,

From the first integration by Integration by Parts,

but

Therefore, the final answer is

Two Intersecting Lines, 2

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the equation of the line passing through the point of intersection of the given lines 

and containing the point (-1, 3).

Solution:

To illustrate the problem, it is better to draw the figure as follows:

For x - 2y = 3,

                             x - 2y = 3
                             2y = x - 3
                             y = ½ x - ³/₂

                             slope (∆y/∆x), m = ½
                             y-intercept = - ³/₂

To trace the graph, plot - ³/₂ at the y-axis. This is your first point of the line (0, - ³/₂). Next, use the slope to get the second point. From the first point, count 2 units to the right and then 1 unit upward. 

For 3x + y = 5,

                             3x + y = 5
                             y = - 3x + 5 

                             slope (∆y/∆x), m = - 3
                             y-intercept = 5

To trace the graph, plot 5 at the y-axis. This is your first point of the line (0, 5). Next, use the slope to get the second point. From the first point, count 1 unit to the left and then 3 units upward. 

Photo by Math Principles in Everyday Life

To get their point of intersection, we have to use the two given equations and solve for x and y, we have

Multiply the second equation by 2 and then add in order to eliminate y and solve for the value of x.

                                          ────────────────

Substitute x to either of the two equations,

Therefore, their point of intersection is P(¹³/₇, - ⁴/₇).

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Finally, we can get the equation of a line using the Two Point Form, we have

Volume - Cube, Cutting Plane

Category: Solid Geometry

"Published in Newark, California, USA"

The plane section ABCD shown in the figure is cut from a cube of edge a. Find the area of this section if D and C are each at the midpoint of an edge.

Photo by Math Principles in Everyday Life

Solution:

The given problem above is asking for the area of a cutting plane ABCD. If C and D are the midpoints of the parallel edges of a cube, then the parallel edges of a cube will bisect into equal parts as shown in the figure below

Photo by Math Principles in Everyday Life

As you notice that there are two right triangles in a cube that are parallel to each other. We need to solve for their hypotenuse using Pythagorean Theorem, as follows

Since AD ║ BC, AB ║ CD, AD ≅ BC, and AB ≅ CD, then quadrilateral ABCD is a rectangle which is a cutting plane. Therefore, the area of a rectangle is