Quadratic growth

Unlimited growth: Linear and quadratic growth

Quadratic growth

For quadratic growth of $y(t)$ at time $t$ :

the derivative of quantity $y(t)$ changes with the same value per unit of time;
$y(t)$ can be described by a quadratic function $y(t)=a\cdot t^2+b\cdot t + c$ , where $2a$ is the increase or decrease of the derivative $y'(t)$ per unit of time, $b$ the value of the derivative at time $t=0$ and $c$ the initial value at time $t=0$ ;
the corresponding $(y,t)$ -graph is a parabola;
the second derivative of the quantity $y(t)$ is a constant.

The first two properties mean that for quadratic growth, the derivative $y'(t)$ is a linear function and equal to $2a\cdot t + b$ . This expression is the same as the formula of the straight line that represents the graph of $y'$ .

Quadratic growth of a quantity $y(t)$ is defined as: $y''(t)=2a$ , for some constant $a$ . This is a differential equation because there is a derivative in it. In this case, the second derivative plays a role; in mathematical jargon, this is a second order differential equation.

The general solution of the differential equation

$y''(t)=2a$ is equal to

$y(t)=a\cdot t^2 + b\cdot t + c$ with parameters $a$ , $b$ and $c$ .

For math enthusiasts we give the proof of the assertion that $y(t)=a\cdot t^2 + b\cdot t + c$ is the general solution of the differential equation $y''(t)=2a$ .
In this proof we use the following lemma: if $f'(t)=b$ is a certain function $f(t)$ , then $f(t)=b\cdot t +c$ , in which $b$ and $c$ are constants. In particular, it holds that $f(t)$ is a constant function when $f'(t)=0$ .
Suppose $y(t)$ is a solution of the differential equation $y''(t)=2a$ is. We can now define a new feature $f(t)$ by $f(t)=y(t)-a\cdot t^2$ . The difference rule for differentiating yields: $f'(t) = y'(t)-(a\cdot t^2)'= y'(t) - 2a\cdot t$ . Differentiating again gives the comparison $f''(t) = y''(t)-(2a\cdot t)'= a - a = 0$ . The derivative of $f'(t)$ therefore equals 0, and it follows from the lemma that function $f'(t)$ is constant, say $f'(t)=b$ . But apply the lemma again and we find: $f(t)=y(t)-a\cdot t^2 = b\cdot t+ c$ , or $y(t)=a\cdot t^2 + b\cdot t +c$ .
This is what had to be proved!