Introduction to linear growth equation:
A linear growth is defined as the growth of the dependent variable with direct proportion to independent variable. Example let y=mx+b where m>0. Then whenever say x increases by 1, x becomes x+1. y = f(x) = mx+b. f(x+1) = m(x+1)+b.
So increase in y = f(x+1)-f(x) = m(x+1)+b -(mx+b) = mx+m+b -mx-b = m. So increase of y/increase of x = m/1 =m. Since m>0, y increases by m for each increase of x. Hence a linear growth equation is defined as y=mx+b for all positive values of m. The slope of the line = positive as m is positive.
Examples are: y=x, y=2x+1, y =x/2+1, y =3x-2.
In these examples of linear relations slope is positive, When here we get `dy/dx` >0. In a functional language, we know that whenever first derivative is positive, the function is an increasing function. This is an additional proof for the increase in value of y for increase in value of x.
Since the growth is linear y is related to x by a linear relation or in other words, the graph of the function is always a straight line with positive slope,
The example below guides us to find out linear growth equation.
Find whether the following is a linear growth"
1. y =x-3 - This is the equation of straight line with slope 1>0, So linear growth equation.
2. y = -2x+5, This is the equation of straight line with slope -2<0, So though linear, not a growth.
3. y =x²+2x+3. This is not a linear equation hence not a linear growth equation.
Other forms of linear relation equation: The linear growth equations can also be of the other form as given below:
Standard form: ax+by+c =0.
Now we find the slope for this by differentiating. We get a + b `dy/dx` =0. So `dy/dx` = -`(a)/(b)` . Now we have this value positive if a and b have different signs. Or if a>0, b<0 then a/b is negative thus -a/b positive. So the equation will be linear growth.
So we come to conclusion that ax+by+c=0 where a and b have different signs, is a linear growth equation.
Intercept form: In intercept form we have equation `(x)/(a)` + `(y)/(b)` =1. Here coefficient of x =1/a and coefficient of y is 1/b.
So we get by the previous deduction that if a and b have different signs, this equation is a linear growth equation.
Linear Growth Equation with Diagram
linear growth equation
Equations and Graphs of positive slope Straight Lines
Problems on Linear Growth Equations
We solve some problems here on linear growth equation.
Prob 1: Find the equation of line passing through (2,3) and (5,9) is a linear growth equation or not,
Sol: Now we have the slope formula for finding the slope of line passing through two points.
Slope of the line = `(y2-y1)/(x2-x1)` = `(9-3)/(5-2)` = 2 >0.
Since slope >0 this is a linear growth equation.
Prob 2: Find a if the equation represented by ax+2y+3 =0 is a linear growth equation.
Sol: For finding whether the equation is a linear growth one or not, we find the slope.
Differentiating this equation wrt x, a + 2 `dy/dx` =0
`dy/dx` = -`(a)/(2)` .>0 if a<0,
So the solution for a is (0,`oo` ).
Prob 3; Find the linear growth equation between x and y if linear growth is 2 and initial dependent variable is 2.
Sol: Here we have m=2 and b =2 = y intercept (initial dependent variable)
so y =2x+2 is the equation.
Prob 4: In a business, variable cost = 5 per unit, sales =10 per unit, Fixed cost = 3000 then find the linear growth equation for costs and sales.
Sol: Let c be the total cost, and s total sales.
Then we have S = C + P So if x is production in number of units, sales =10x, variable cost =5x, fixed cost =3000.
So we have S=10x =5x+3000+P
Or P = 5x -3000 where P is the profit, x the no of units produced.
Prob 4: Find the velocity of a vehicle at t seconds if initial velocity is u and acceleration is a.
Sol: We have velocity at time t = v = initial velocity + acceleration * time
or v=u+at. where u is a constant.
Thus v and t are linearly related. So v=u+at is a linear growth equation for all a >0.
Conclusion: In this article, we found out linear growth equation. This has many applications in Physics, as to accleration-velocity. This principle is also used to find Break even point in businesses, business plans in times of competitions etc.