Instantaneous Velocity
Key Questions
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Provided that the graph is of distance as a function of time, the slope of the line tangent to the function at a given point represents the instantaneous velocity at that point.
In order to get an idea of this slope, one must use limits. For an example, suppose one is given a distance function
#x = f(t)# , and one wishes to find the instantaneous velocity, or rate of change of distance, at the point#p_0 = (t_0, f(t_0))# , it helps to first examine another nearby point,#p_1 = (t_0+a, f(t_0+a))# , where#a# is some arbitrarily small constant. The slope of the secant line passing through the graph at these points is:#[f(t_0+a)-f(t_0)]/a# As
#p_1# approaches#p_0# (which will occur as our#a# decreases), our above#difference quotient# will approach a limit, here designated#L# , which is the slope of the tangent line at the given point. At that point, a point-slope equation using our above points can provide a more exact equation.If instead one is familiar with differentiation, and the function is both continuous and differentiable at the given value of
#t# , then we can simply differentiate the function. Given that most distance functions are polynomial functions, of the form#x = f(t) = at^n + bt^(n-1) + ct^(n-2) + ... + yt + z,# these can be differentiated using the power rule which states that for a function#f(t) = at^n, (df)/dt# (or#f'(t)# ) =#(n)at^(n-1)# .Thus for our general polynomial function above,
#x' = f'(t) = (n)at^(n-1) + (n-1)bt^(n-2) + (n-2)ct^(n-3) + ... + y# (Note that since#t = t^1# (as any number raised to the first power equals itself), reducing the power by 1 leaves us with#t^0 = 1# , hence why the final term is simply#y# . Note also that our#z# term, being a constant, did not change with respect to#t# and thus was discarded in differentiation).This
#f'(t)# is the derivative of the distance function with respect to time; thus, it measures the rate of change of distance with respect to time, which is simply the velocity. -
Instantaneous velocity is the velocity at which an object is travelling at exactly the instant that is specified.
If I travel north at exactly 10m/s for exactly ten seconds, then turn west and travel exactly 5m/s for another ten seconds exactly, my average velocity is roughly 5.59m/s in a (roughly) north-by-northwest direction. However, my instantaneous velocity is my velocity at any given point: at exactly five seconds into my trip, my instantaneous velocity is 10m/s north; at exactly fifteen seconds in, it's 5m/s west.
Questions
Derivatives
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Tangent Line to a Curve
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Normal Line to a Tangent
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Slope of a Curve at a Point
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Average Velocity
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Instantaneous Velocity
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Limit Definition of Derivative
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First Principles Example 1: x²
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First Principles Example 2: x³
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First Principles Example 3: square root of x
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Standard Notation and Terminology
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Differentiable vs. Non-differentiable Functions
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Rate of Change of a Function
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Average Rate of Change Over an Interval
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Instantaneous Rate of Change at a Point