Friday, February 10, 2017
Unizor - Indefinite Integrals - Speed
Notes to a video lecture on http://www.unizor.com
Indefinite Integral - Speed and Distance
Consider a function s(t) that describes the distance covered by a moving object as a function of time.
Recall the procedure of differentiation as it is applied to this function.
Since a ratio of increment of distance Δs covered during an increment of time from momentt to moment t+Δt is average speed during this interval of time, the limit of this ratio, as time interval Δt converges to zero, is the instantaneous speed of an object at moment t:
v(t) = limΔt→0[Δs/Δt] = ds/dt = sI(t)
Consider a situation when we monitor only the speed as a function of time v(t). Can we find the distance we have covered?
The operation of integration answers this.
Since indefinite integral of v(t)describes a function, whose derivative is v(t), it fits exactly what distance, as a function of time s(t), is. Additional constant, participating in the result of integration, can be interpreted as distance covered before we started our observation of speed and does not change the character of dependency between speed as a function of time v(t) and distance as a function of times(t).
This can be expressed as
∫ v(t) dt = s(t)+C
Let's analyze the above expression of an integral.
Since v(t) is an instantaneous speed at moment t, the result of its multiplication by an infinitesimal increment of time dt represents a distance covered during this infinitesimal interval from t to t+dt.
To get the overall distance, we have summarize infinite number of infinitesimal distances, so the sign of integration can be interpreted as such sum. Incidentally, a sign ∫reminds a letter S (from the word "sum") stretched vertically.
The point of this analysis is that the whole expression ∫ v(t) dt can be interpreted as "sum of infinite number of infinitesimal products of v(t) by dt".