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!