Derivative - Chain Rule

Category: Differential Calculus, Algebra

"Published in Newark, California, USA"

Given the following functions:

                      

                    
Find dy/dx in terms of x and y. 

Solution:  

The first thing that we have to do is to get the derivative of each functions with respect to their independent variables. 

For the first function

For the second function

For the third function

Therefore,

The above equation is called the Derivative by Chain Rule. As you noticed that the differentials like du and dv will be cancelled. Multiply the derivative of three functions, we have 

The final answer must be free from other variables like u and v. We have to eliminate u and v by substituting their values with x. We know that 

Therefore,

Solving Equations - Homogeneous Functions

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for the equation:

Solution:

The equation above is Differential Equation because it has dx and dy. The type of equation is Homogeneous because the functions and variables cannot be separated by Separation of Variables. There's a method to solve the Homogeneous Functions. Consider the given equation

Let   y = vx
      dy = vdx + xdv

Substitute y and dy to the given equation, we have

The above equation can now be separated by Separation of Variables. Divide both sides of the equation by x². 

Integrate both sides of the equation

but y = vx and v = ^y/_x, therefore,

Multiply both sides of the equation by 4x,

Finding Equation, Circle - Given 3 Points

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the equation of a circle that passes through the points (2, 3), (6, 1), and (4, -3).

Solution:

To illustrate the problem, it is better to sketch the circle with three points as follows:

Photo by Math Principles in Everyday Life

We know that the equation of a circle in standard form is 

                        (x - h)² + (y - k)² = r² 

where (h, k) is the coordinate of the center of a circle and r is the radius of a circle. In this problem, the center and radius of a circle are not given and we cannot use the equation of a circle in standard form. 

The equation of a circle in general form is 

                      x² + y² + Dx + Ey + F = 0

where D, E, and F are the coefficients. Since the three points of a circle are given, we can use the equation of a circle in general form to solve for D, E, and F. 

For (2, 3), if x = 2 and y = 3, the equation becomes

                 (2)² + (3)² + D(2) + E(3) + F = 0

                  4 + 9 + 2D + 3E + F = 0

                  2D + 3E + F = -13    (equation 1)    

For (6, 1), if x = 6 and y = 1, the equation becomes

                 (6)² + (1)² + D(6) + E(1) + F = 0

                 36 + 1 + 6D + E + F = 0

                 6D + E + F = -37      (equation 2) 

For (4, -3), if x = 4 and y = -3, the equation becomes

                 (4)² + (-3)² + D(4) + E(-3) + F = 0

                 16 + 9 + 4D - 3E + F = 0

                  4D - 3E + F = -25    (equation 3)

Since the three points of a circle are given, there are three equations and three unknowns. We can now solve for D, E, and F by elimination method.

If you subtract equation 2 from equation 1, 

 2D + 3E + F = -13                 2D + 3E + F = -13
                              --------->
-(6D + E + F = -37)                   -6D - E - F = 37
                                            ---------------------------
                                                  -4D + 2E = 24

                                                    -2D + E = 12     (equation 4)

If you subtract equation 3 from equation 2, 

   6D + E + F = -37                  6D + E + F = -37
                               --------->
-(4D - 3E + F = -25)              -4D + 3E - F = 25
                                           ----------------------------
                                                  2D + 4E = - 12    (equation 5)

Add equation 4 and equation 5,

     -2D + E = 12
    2D + 4E = -12
  ----------------------
            5E = 0

              E = 0

Substitute the value of E to equation 4 or equation 5,

    2D + 4E = -12
  2D + 4(0) = -12
            2D = -12

              D = -6

Substitute the value of D and E to equation 1, equation 2, or equation 3,

         2D + 3E + F = -13
    2(-6) + 3(0) + F = -13
                -12 + F = -13  

                         F = -1 

Substitute the value of D, E, and F to the equation of a circle in general form,

                             x² + y² + Dx + Ey + F = 0

                     x² + y² + (-6)x + (0)y + (-1) = 0

Therefore, the equation of a circle is

                             x² + y² - 6x - 1 = 0