Differential Equations
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03-27-2017, 06:46 AM
Post: #7
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RE: Differential Equations
The order of a derivative refers to the order of rate of change. For example, if \( s(t) \) is a function that measures the distance \( s \) traveled after \( t \) hours, then the first (order) derivative of \( s(t) \) is the rate of change of \( s \) with respect to \( t \). Conceptually, this is the instantaneous speed, which (for convenience), we will call the velocity function \( v(t) \). Mathematically, it is written as \( v(t) = s'(t) \). We may then consider the rate of change of velocity over time, i.e. the first (order) derivative of \( v(t) \). Conceptually, this is the instantaneous acceleration, which may be written as \( a(t) = v'(t) \). Now \( a(t) \) is the first (order) derivative of \( v(t) \), but it is a second (order) derivative of \( s(t) \) -- it is the rate of change of the rate of change of \( s \) with respect to \( t \) -- and we write \( a(t) = s''(t) \).
Mathematically speaking, we can compute higher order derivatives by simply computing the derivative of previous (lower order) derivatives. These (higher order) derivatives may or may not have immediate interpretations in the real world beyond the first few orders, but mathematically they exist. The order of a differential equation, then, is simply the highest order of any derivative involved in the differential equation. So for example, the acceleration due to gravity is described by the second order differential equation \[ s''(t) = -9.8 \] since \( s''(t) \) is the second (order) derivative of \( s(t) \) -- which we mentioned earlier was the acceleration function -- and is the highest order derivative (in fact the only derivative) involved in the equation above. Graph 3D | QPI | SolveSys |
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