How to Approximate Instantaneous Rate of Change in Math

You can approximate the instantaneous rate of change at a point $P$ using average rate of change. To do this, select two points that are very close to each other and near $P$ , so that the slope of the line between them is nearly the slope of the tangent at $P$ .

Approximation of instantaneous rate of change

Approximation is performed because it is difficult to position the tangent accurately. It can often be too slack or too steep, and then the numbers become wrong. When the tangent is too slack, you get a lower value than the answer, and when the tangent is too steep you get a value too high for the answer you are looking for. There is another, more accurate way to calculate the instantaneous growth compared to trying to work with the tangent line. By using the formula for average rate of change and choosing

x_{2}

very close to the point where you will find the growth,

x_{1}

, you’ll get an approximate value for the instantaneous rate of change. Here is an example.

Example 1

Look at $f (x) = 0.5 x^{2} - 0.5 x + 2$ . You want to find the instantaneous rate of change when $x = 4$ by approximation

You use the average rate of change formula:

average rate of change

\begin{array}{l} = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

\begin{array}{l} average rate of change & = \frac{change in y}{change in x} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} \end{array}

where

x_{1} = 4

. You’ll now choose

x_{2} = 4.01

(or another value that is very close to

x_{1}

). You then have to find

y_{1} = f (x_{1})

and

y_{2} = f (x_{2})

, and those are

\begin{array}{l} y_{1} & = f (4) \\ = 0.5 {(4)}^{2} - 0.5 (4) + 2 \\ = 8 \\ y_{2} & = f (4.01) \\ = 0.5 {(4.01)}^{2} - 0.5 (4.01) + 2 \\ = 8.035 05 \end{array}

You enter the values into the formula and get:

\begin{array}{l} approximation & \approx \frac{change in y}{change in x} \\ \approx \frac{8 - 8.035 05}{4.01 - 4} \\ \approx 3.505 \end{array}

The exact instantaneous rate of change is $3.5$ . You can see by how close the answer here is, that approximation is a pretty good approach.

You can get an even better result if you choose an $x_{2}$ that is even closer to $x_{1}$ . For example, if you select $x_{2} = 4.0001$ , then the calculation works out like this:

\begin{array}{l} y_{2} & = f (4.0001) \\ = 0.5 {(4.0001)}^{2} - 0.5 (4.0001) + 2 \\ = 8.000 350 005 \end{array}

You then enter the values into the formula and get:

\begin{array}{l} approximation & = \frac{change in y}{change in x} \\ = \frac{8 - 8.000 350 005}{4.0001 - 4} \\ = 3.500 05 \end{array}

So as you can see, the closer the $x$ -values are to each other, the better the approximation will be. As we saw before, the exact value for instantaneous growth for $x = 4$ is $3.5$ , so this method works well if you use two $x$ values that are very close to each other.

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How to Interpret and Find Instantaneous Rate of Change in Math

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