# Combustion Analysis

Definition: The analysis which is made to determine the amounts of different elements present in a compound by burning a known amount of organic compound containing C, H and O in the presence of oxygen is called” combustion analysis.” Procedure: A weighed sample of the organic compound is placed in the combustion tube. This combustion […]

About Soli: Project Soli is Google ATAP’s attempt to get rid of external input devices and make use of the human body as the only input device. As devices get smaller, interacting with them via a display become difficult. Project Soli makes use of your hand and finger movments to interact with wearables or devices […]

# Newton’s Third Law of Motion

Definition: “To every action there is an equal and opposite reaction.” OR “When two bodies exert mutual forces on one another, the two forces are always equal in magnitude and opposite in direction.” Mathematically we can write; FAB = – FBA it is important to remember that the action and reaction forces always act on […]

# Newton’s Second Law of Motion

Definition: “If a net force acts on a body the body accelerates. The direction of acceleration is the same as the direction of net force. The magnitude of acceleration is in inversely proportional to the mass of the body.” Mathematical: [latex s=2]a\propto F[/latex] [latex s=2]a\propto \frac{1}{m}[/latex] Combining [latex s=1]F = K ma[/latex] where k=1 [latex s=1]F […]

# Newton’s First Law of Motion

DEFINITION: “In the absence of unbalanced force (net forces), a body at rest will remain at rest and a body in uniform motion will remain in uniform motion in a straight line.” EXPLANATION: Experiments shows that when no not force acts on a body the body either remain at rest or moves with uniform velocity […]

# Case 4. Repeated quadratic term in denominator

[latex s=2]F(x) = \frac{2x^3+5x}{(x^2+3)^2}[/latex] [latex s=3]F(x) = \frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+3)^2}[/latex] Multiply an divide $(x^2+3)^2$ with $Ax+B$ [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)(x^2+3)}+ \frac{Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)^2}+ \frac{Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)+Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3Ax+3B+Cx+D}{(x^2+3)^2}[/latex] [latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3(A+C)x+(3B+D)}{(x^2+3)^2}[/latex] Comparing the coefficient we get $\framebox[1.1\width]{A=2}$ $\framebox[1.1\width]{B=0}$ $3A+C=5 \longmapsto \framebox[1.1\width]{C=-1}$ $3B+D=0 \longmapsto \framebox[1.1\width]{D=0}$ Put the value […]

# Case 3. Quadratic term in denominator.

[latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)}[/latex] [latex s=2]F(x) = \frac{8x^2-12}{x(x^2+2x-6)} = \frac{A}{(x)}+\frac{Bx+C}{(x^2+2x-6)}[/latex] Multiply by  [latex s=1]x(x^2+2x-6) [/latex] with above equation on both sides we get [latex s=1]\Big(x(x^2+2x-6)\Big) \Big(\frac{8x^2-12}{x(x^2+2x-6)}\Big) = \Big(x(x^2+2x-6)\Big) \Big(\frac{A}{(x)}\Big)+ \Big(x(x^2+2x-6)\Big) \Big(\frac{B}{x(x^2+2x-6)}\Big)[/latex] $8x^2-12 = A(x^2+2x-6)+x(Bx+C)$ We could choose $x=0$ to get the value of A, but that’s the only constant that we could get using this method […]

# Case 2. Repeated linear factor in denominator.

[latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2}[/latex] [latex s=2]F(x) = \frac{4x^2}{(x-1)(x-2)^2} = \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}[/latex] Multiply by $(x-1)(x-2)^2$ with above equation on both sides we get [latex s=1]\Big((x-1)(x-2)^2\Big) \Big(\frac{4x^2}{(x-1)(x-2)^2}\Big) = \Big((x-1)(x-2)^2\Big) \Big(\frac{A}{(x-1)}\Big)+ \Big((x-1)(x-2)^2\Big) \Big(\frac{B}{(x-2)}\Big)+ \Big((x-1)(x-2)^2\Big) \Big(\frac{C}{(x-2)^2}\Big)[/latex] [latex s=1]4x^2 = A(x-2)^2+B(x-1)(x-2)+C(x-1)[/latex]  …………(1) Now we have $(x-1)$ and $(x-2)^2$ $x-1=0 \longmapsto x=1$ $(x-2)^2=0 \longmapsto x=2$ Put $\textbf{x=1}$ and $\textbf{x=2}$ in equation (1) […]

# Case 1. Distinct linear factor in denominator

Let us see the how to slove when distinct linear factor is in denominator. Distinct Linear Factor: [latex s= 2]F(x) = \frac{8x-42}{x^2+3x-18}[/latex] [latex s=2]F(x) = \frac{8x-42}{x^2+3x-18} = \frac{8x-42}{(x+6)(x-3)} = \frac{A}{x+6}+\frac{B}{x-3}[/latex] Multiply by $\textbf(x+6)(x-3)$ on both sides with above equation we get [latex s=-1]\Big((x+6)(x-3)\Big)\Big(\frac{8x-42}{x^2+3x-18}\Big) = \Big(\Big((x+6)(x-3)\Big)\Big(\frac{A}{x+6}\Big)\Big) + \Big(\Big((x+6)(x-3)\Big)\Big(\frac{B}{x-3}\Big)\Big)[/latex]   $8x-42=A(x-3)+B(x+6)$  …………(1) Now we have $\textbf{(x+6)}$ and […]

# Partial Fraction Method

First Method The method for computing partial fraction decomposition applies to all rational functions with one qualification: ‘The degree of the numerator must be less than the degree of the denominator’ Works for Proper Rational Expressions, where the degree of the numerator is less than the bottom. Note: When order/degree of numerator is not less […]