Derivative - Algebraic Functions, Radicals

Category: Differential Calculus, Algebra

"Published in Newark, California, USA"

Find the derivative for

Solution:

Consider the given equation

Express the above equation in terms of an exponential power as follows

First, get the derivative of a function in terms of a power

Next, get the derivative of a function inside the exponential function in terms of a rational function

Simplify the above equation

We can make a negative exponent into a positive exponent by getting the reciprocal of a rational function inside the exponential function as follows

As a rule in Mathematics, we have to rationalize the denominator in order to eliminate the cube root sign at the denominator as follows

Therefore,

Two Coincident Lines

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the points of intersection of the following lines:

                                2x - y = 8

                              4x - 2y = 16

Solution:

Since the given equations are all first degree, then they are linear equations. They are straight lines. We can graph the two lines by getting their slope and y-intercept. 

For 2x - y = 8

                                  2x - y = 8

                                        -y = -2x + 8

                                         y = 2x - 8

                                  slope (Δy/Δx), m = 2

                                  y-intercept, b = -8

To trace the graph, plot -8 at the y-axis. This is your first point of the line (0, -8). Next, use the slope to get the second point. From the first point, count 1 unit to the right and then 2 units upward. 

For 4x - 2y = 16

                                  4x - 2y = 16

                                       -2y = -4x + 16

                                          y = 2x - 8

                                 slope (Δy/Δx), m = 2

                                  y-intercept, b = -8


To trace the graph, plot -8 at the y-axis. This is your first point of the line (0, -8). Next, use the slope to get the second point. From the first point, count 1 unit to the right and then 2 units upward.

Photo by Math Principles in Everyday Life

From the graph, the two lines are coincide to each other because their slopes and y-intercepts are the same which are 2 and -8. The two lines will have an infinite number of points of intersection. When you solve for x and y from the two given equations, their x, y, and constant will be equal to zero. From the two given equations,

                                  2x - y = 8

                                4x - 2y = 16

Multiply the first equation by 2 and -1 at the second equation. Add the two equations and let's see what will happen to x, y, and constant.

        2 (2x - y = 8)                             4x - 2y = 16
                                         →
     -1 (4x - 2y = 16)                         -4x + 2y = -16
                                                   _______________

                                                                  0 = 0

Since everything in the equation are all equal to zero, then there's no way that we can solve for x and y. Therefore, the two lines are coincide to each other.

Graphical Sketch - Parabola

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the vertex, focus, and end points of the latus rectum for parabola with an equation 

and sketch the graph.

Solution:

The first thing that we have to do is to write the equation of a parabola in standard form as follows

Transpose -3y and 10 to the right side of the equation

Complete the square at the left side of the equation by adding 1 on both sides of the equation

Therefore, the vertex of a parabola is V(1, 3). 

Since the second degree in the equation is x, the parabola is either concave upward or downward. 

Next, let's solve for the coordinates of the focus and the ends points of the latus rectum. Since 4a = 3 is positive, then the parabola is concave upward. a is the distance between the focus and the vertex of a parabola. Solve for the value of a as follows

                                       4a = 3
                                    
                                         a = ¾

To get the coordinates of the focus, count ¾ units upward from the vertex. Therefore, the coordinates of the focus is F(1, 3¾). 

To get the coordinates of the first end point of the latus rectum, count 2a = 2(¾) = 1½ units to the right from focus. The first end point of the latus rectum is B(2½, 3¾). 

To get the coordinates of the other end point of the latus rectum, count 2a = 2(¾) = 1½ units to the left from focus. The other end point of the latus rectum is A(-½, 3¾).

Finally, we have the coordinates of the vertex, focus, and the end points of the latus rectum. We can sketch the graph of the parabola as follows

Photo by Math Principles in Everyday Life