Instantaneous Rate of Change
The Instantaneous Rate of Change measures the rate of change at a single instant. This differs from the Average Rate of Change, as that measures the overall rate of change over a set interval.
The Instantaneous Rate of Change at a specific point on the graph of a function represents the slope of the tangent line to the graph at that point. This can be expressed algebraically as:
When evaluating \(I_{roc}\), it's good to make \(h\) as small as possible. For the following examples, we will set \(h = 0.01\).
Example
Determine the Instananeous Rate of Change for the function, \(f(x)=2x^3-5x^2+3x+7\), at \(x=1\).
First, we can evaluate \(f(a+h)\) by substituting \(1.01\) for \(x\):
\(f(1.01) = 2(1.01)-5(1.01)+3(1.01)+7\)
\(\textcolor{red}{f(1.01)=6.990102}\)
Next, we can evaluate \(f(a)\) by substituting \(1\) for \(x\):
\(f(1)=2(1)^3-5(1)^2+3(1)+7\)
\(\textcolor{blue}{f(1)=7}\)
Finally, we can substitute all pertinent values into the \(I_{roc}\) formula to determine its slope at these points:
\(\text{Instantaneous Rate of Change} = \cfrac{\textcolor{red}{6.990102}-\textcolor{blue}{7}}{0.01}\)
\(\text{Instantaneous Rate of Change} = \cfrac{-0.009898}{0.01}\)
\(\text{Instantaneous Rate of Change} = -0.9898\)
Therefore, we can determine that the \(I_{roc}\) of \(f(x)=2x^3-5x^2+3x+7\) at \(x = 1\) is \(\boldsymbol{-0.9898}\).
\( f(x) = 3x^2 + 2x + 1 \) at \( x = 1 \)
First, we can evaluate \(f(a+h)\) by substituting \(1.01\) for \(x\):
\(f(1.01) = 3(1.01)^2 + 2(1.01) + 1 \)
\(\textcolor{red}{f(1.01) = 6.0803}\)
Next, we can evaluate \(f(a)\) by substituting \(1\) for \(x\):
\(f(1) = 3(1)^2 + 2(1) + 1 \)
\(\textcolor{blue}{f(1) = 6}\)
Finally, we can substitute all pertinent values into the \(I_{roc}\) formula to determine its slope at these points:
\(\text{Instantaneous Rate of Change} = \cfrac{\textcolor{red}{6.0803}-\textcolor{blue}{6}}{0.01}\)
\(\text{Instantaneous Rate of Change} = \cfrac{0.0803}{0.01}\)
\(\text{Instantaneous Rate of Change} = 8.03\)
Therefore, we can determine that the Instantaneous Rate of Change of \( f(x) = 3x^2 + 2x + 1 \) at \(x = 1\) is \(\boldsymbol{8.03}\).
The function at \(x= 1\) can be represented graphically as such:
\(g(x) = x^4 + 3x^3 -2\) at \( x = -1 \)
First, we can evaluate \(f(a+h)\) by substituting \(-0.99\) for \(x\):
\(g(-0.99) = (-0.99)^4 + 3(-0.99)^3 -2\)
\(\textcolor{red}{g(-0.99) = -3.95030099}\)
Next, we can evaluate \(f(a)\) by substituting \(-1\) for \(x\):
\(g(-1) = (-1)^4 + 3(-1)^3 -2\)
\(\textcolor{blue}{g(-1) = -4}\)
Finally, we can substitute all pertinent values into the Instantaneous Rate of Change formula to determine its slope at these points:
\(\text{Instantaneous Rate of Change} = \cfrac{\textcolor{red}{-3.95030099}-\textcolor{blue}{-4}}{0.01}\)
\(\text{Instantaneous Rate of Change} = \cfrac{0.04969901}{0.01}\)
\(\text{Instantaneous Rate of Change} = 4.969901\)
Therefore, we can determine that the \(I_{roc}\) of \(f(x) = x^4 + 3x^3 -2\) at \(x = -1\) is \(\boldsymbol{4.969901}\).
The function at \(x= -1\) can be represented graphically as such:
