Indeterminate Form - Zero Raised Zero

Category: Differential Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate 

Solution:

Consider the given equation

Substitute the value of x to the above equation

Since the answer is 0⁰, then it is also another type of Indeterminate Form and it is not accepted as final answer in Mathematics. We know that any number raised to zero power is always equal to one except for zero that why it is also Indeterminate Form. In this type of Indeterminate Form, we cannot use the L'Hopital's Rule because the L'Hopital's Rule is applicable for the Indeterminate Forms like 0/0 and ∞/∞. Since the given equation is exponential equation, let's consider the following procedure

let

Take natural logarithm on both sides of the equation, we have

Substitute the value of x to the above equation

Since the Indeterminate Form is 0∙∞, then we have to rewrite the above equation as follows

Substitute the value of x to the above equation

Since the Indeterminate Form is ∞/∞, then we can apply the L'Hopital's Rule as follows

Substitute the value of x to the above equation

Take the inverse natural logarithm on both sides of the equation, we have

Therefore,

Solving 2 x 2 Determinants

Category: Algebra

"Published in Newark, California"

Solve the following systems of equations by determinants:

                                        x - 2y = 7

                                        3x - y = 11

Solution:

The first thing that we have to do is to write the determinants for D_x, D_y, and D from the given two linear equations. Determinant is a value associated with square matrix. Matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. Consider the given equations


                                        x - 2y = 7

                                        3x - y = 11

To write the determinant of D, consider the coefficients of x and y as follows

To write the determinant of D_x, replace the coefficients of x with the coefficients of the right side of the equation as follows

To write the determinant of D_y, replace the coefficients of y with the coefficients of the right side of the equation as follows

Next, solve for the value of x as follows

Finally, solve for the value of y as follows

Note: To get the value of a 2 x 2 Matrix, principal diagonal (top left term times bottom right term) minus secondary diagonal (bottom left term times top right term).

Check: To see if you got the correct answers, substitute the values of x and y to either of the two given equations as follows

                                        x - 2y = 7

                                     3 - 2(-2) = 7

                                         3 + 4 = 7

                                               7 = 7

Graphical Sketch - Ellipse

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the center, foci, and semi-axes for the ellipse and sketch the graph

Solution:

From the given equation above, the curve is ellipse because both x² and y² are positive but their coefficients are different.  The general equation of an ellipse must be simplified in standard form in order to get its center, foci, and semi-axes as follows

Group the equation according to their variables and transpose the coefficient to the right side of the equation

Complete the square of each group

Divide both sides of the equation by 6

The above equation is now simplified in standard form. Since the denominator at x group is greater than the denominator at y group, then the major axis is parallel to x-axis. 

To solve for the coordinates of the center:

Equate x + 2 = 0                                     Equate y + 1 = 0
            x = -2                                                      y = -1

Therefore, the coordinates of the center is C(-2, -1).

Next, solve for the value of semi-major axis

Therefore, the coordinates of the ends of major axis are V₁(-4.4495, -1) and V₂(0.4495, -1).

Next, solve for the value of semi-minor axis

Therefore, the coordinates of the ends of minor axis are V₃(-2, 0.4142) and V₄(-2, -2.4142). 

Next, solve for the distance of a focus to the center

Therefore, the coordinates of the foci are F₁(-4, -1) and F₂(0, -1). 

Next, solve for the length of a semi-latus rectum 

Add and subtract this value to the y values of foci in order to get the coordinates of the ends of latera recta. Therefore, the coordinates of the ends of latera recta are L₁(-4, 0.1835), L₂(-4, -1.8165), L₃(0, 0.1835), and L₄(0, -1.8165). 

Since we know the coordinates of the center, ends of major axis, ends of minor axis, foci, and ends of latera recta, then we can sketch the graph of an ellipse as follows

Photo by Math Principles in Everyday Life