Algebraic Operations - Exponents

Category: Algebra

"Published in Suisun City, California, USA"

Simplify the expression and eliminate any negative exponents for

Solution:

Consider the given equation above

The first grouped term has an exponent which is 3. If you will cube the first grouped term, then the exponents of variables inside the parenthesis will be multiplied by 3 as follows

Since the exponents of both grouped terms are now equal to 1, then we can multiply the two fractions directly. Numerator times numerator and denominator times denominator. If you  multiply the two terms with the same variable, then their exponents will be added as follows

If you divide the two terms with the same variable, then their exponents will be subtracted as follows

Any number or any variable (except zero) raised to zero power is always equal to one. Since there's no y at the numerator, then the exponent of y at the numerator is equal to zero.  

In order to eliminate the negative exponents of b and y, we have to transfer the two terms in the denominator as follows

Therefore,

Some Substitution Methods

Category: Differential Equations, Algebra, Trigonometry

"Published in Newark, California, USA"

Find the general solution for

Solution:

Consider the given equation above

Since the above equation contains trigonometric functions, then it is considered as a complicated equation with differentials. To avoid the confusion in solving the equation, it is better to rewrite the equation by using the substitution method. We will substitute all trigonometric functions with another variables.

Let

so that 

Let

so that

Substitute all the above values to the given equation, as follows

Rewrite the above equation as a first order, first degree linear equation, we have

where

and

Since the above equation is already a first order, first degree linear equation in terms of v, then the integrating factor will be equal to

Therefore, the general solution for the above equation is

but

and

and the final answer for the above equation is

Volume - Solid Revolution, 3

Category: Integral Calculus, Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the volume of the solid generated by revolving about y = -1 for the area bounded by the given curves.

Solution:

To illustrate the problem, it is better to sketch the graph of the two equations above using the principles of Analytic Geometry as follows

Photo by Math Principles in Everyday Life

From the figure above, there are two points of intersection between the two curves. Solve the systems of two equations two unknowns in order to get the coordinates of the points of intersection as follows

Equate

and

we have,

Therefore, their points of intersection are (-2, 5) and (2, 5).

Next, from the given two equations above, it is better to use a vertical strip at the area bounded by two curves and label further the figure as follows

Photo by Math Principles in Everyday Life

If you rotate the shaded area about y = -1, the vertical strip becomes a ring as follows

Photo by Math Principles in Everyday Life

The volume of a ring formed by the rotation of a vertical strip about the x-axis is  

Integrate on both sides of the equation to get the volume of a solid formed by the rotation of the area about the x-axis as follows

Since the axis of rotation is NOT the x-axis as stated in the given problem, then we have to get the outer radius and inner radius first. The axis of rotation is one unit below the x-axis which is y = -1. The outer radius and inner radius of a disk will be increased by one unit as follows

Substitute the values of the variables to the above equation, we have

Simplify the fractions by getting their Least Common Denominator (LCD) and the volume of the solid of revolution along y = -1 is