Instantaneous Rate of Change:
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1 Instantaneous Rate of Cange: Last section we discovered tat te average rate of cange in F(x) can also be interpreted as te slope of a scant line. Te average rate of cange involves te cange in F(x) over a designated interval [x 1, x ] or between te two points (x 1, F(x 1 )) and (x, F(x )). Te secant line passes troug tese two points. A logical extrapolation would indicate tat te instantaneous rate of cange in F(x) at a point x would be te same as te slope of a tangent line toucing te grap of F(x) at tat same point (x, F(x)). In te case of average rate of cange you can find te slope of te secant line easily because you ave two points to work wit. But in te case of te instantaneous rate of cange tere is only one point and terefore it is not possible to directly calculate te slope of te tangent line. Is it possible to find tis value tat will be bot te slope of te tangent line and te instantaneous rate of cange? We will now define a process tat will elp us do just tat. Finding Instantaneous Rate of Cange and te Slope of te Tangent Line. We will again consider te following profit function from te last section: P(q) = -q + 4q 5 were q represents quantity of items sold. Profit and slope of tangent line P Profit Tangent line x Figure 1 We would like te find te instantaneous rate of cange in profit for q = 4. We will find te instantaneous rate of cange numerically. Tis metod proceeds as follows: Since we can not find te slope of te tangent line toucing te grap at q = 4 by direct calculation we will try to guess its value as precisely as possible. In order to do tis we must find slopes of secant lines tat are very close to te tangent line at te point
2 (4, 78). First we will find te slope of secant line troug te points ( 4, 78) and (41, 83). Ten we will repeat te process for points ( 4, 78) and ( 4.5, ). We ten will repeat again for te points ( 4, 78), (4.5, ) and so on. In order to be very precise we will also need to consider points on te left side of q = 4 as well suc as ( 39.99, ) and ( 4, 78). Te process is better illustrated in te table below. using points from te rigt side of q = 4 x1 x P(x1) P(x) [P(x) - P(x1)] / [x - x1] using points from te left side of q = 4 x1 x P(x1) P(x) [P(x) - P(x1)] / [x - x1] Figure As we look at te slopes of te many secant lines in te cart above we notice tat as te secant lines pass troug points tat are closer and closer togeter teir slopes seem to be closer and closer to te same value bot on te rigt and left side of te point (4, 78). Terefore we will make an estimate (or a good guess) tat te slope of te tangent line is m = 4. Te closer te points involved in calculations te better te estimate. In te process above we found te slope of te tangent line numerically. We will also learn to find it algebraically. In order to develop te algebraic approac we need to go back to idea of a scant line. If we ave a function f(x) a secant line intersect te grap of tis function in two places. Terefore it sares two points wit f(x). We will call tose two points (x, f(x)) and (x +, f(x + )). In order to find te slope of tis second line we need to find te cange in te y-values divided by te cange in te x-values. Tis will give us te secant line f ( x+ ) f( x) f( x+ ) f( x) slope m = =. Tis is a generalization of te slope of ( x+ ) x any secant line. If we now tink back to te numerical process above, you will remember tat we found te slopes of several secant lines. Eac secant line was closer to te tangent line of interest tan te secant line before it and te slope values were getting closer to te slope value of te tangent line. For eac succeeding secant line te two points used were closer
![599) and ( 4, 78). Te process is better illustrated in te table below. using points from te rigt side of q = 4 x1 x P(x1) P(x) [P(x) - P(x1)] / [x - x1] 4 41 78 838 38 4 4.5 78 7919.5 39 4 4.](/docs-images/47/20936890/images/page_2.jpg "5 78 7811.995 39.9 4 4.5 78 781. 39.99 4 4.5 78 78.1 39.999 using points from te left side of q = 4 x1 x P(x1) P(x) [P(x) - P(x1)] / [x - x1] 4 39 78 7558 4 4 39.9 78 7775.98 4. 4 39.99 78 7797.6 4.")
3 togeter tan before. One way to express tis cange in points is to use te notation Limit. Tis notation means tat te distance between te x values is getting smaller and smaller. Te distance is approacing te value zero. Terefore te limit of te slope values of te secant lines will approac te slope value of te tangent. Te symbolic notation tat describes tis process is given as follows. dy f ( x + ) f ( x) y dx x x ( dy y is notation for te slope of te tangent line and is notation for te slope of te dx x secant line. Tis formula is also called te Derivative (to be discussed in a later section) Now we will look at an example ow to find te slope of a tangent line toucing te grap of f(x) at some value x = a. Example 1 Find te slope of te tangent line toucing te grap ( ) x 3 f x = + at x = 5. Solution: Before we start te problem we will review notation: Review te following evaluations: f(x) = x + 3 f(4) = (4) + 3 = 19 f(7) = (7) + 3 = 5 f(m) = m + 3 f(x + ) = (x + ) + 3 Now we will now proceed to solve te example problem. dy f ( x + ) f ( x) Te slope of te tangent line is given to be so we need to dx make te appropriate substitution as follows: dy [( x + ) + 3] [ x + 3] x + x x 3 ( x + ) limit limit Limit dx = = = = lim it( x + ) = x = (5) = 1. Terefore te slope of te tangent line toucing te grap at x = 5 is m = 1. See te grap below.
4 F(x) and tangent line x f(x) T Figure 3 In order to verify tat tis is te correct Tangent line we will find te equation of te tangent line tat touces te grap of f( x) = x + 3 at x = 5. Solution: In order to find te equation of any line you need te following: 1) a point ) a slope 3) te point slope form of te equation y y1 = m( x x1) Since we are interested in te tangent line tat touces te grap of f( x) = x + 3 at 4) x = 5, we need to find tat te y value f(5) = 8. So now we ave te point (5, 8). Next we find te slope to te tangent line at x = 5. Since te derivative of dy = x is te general slope formula for any tangent line toucing f (x) we found dx dy above ave tat te slope of te tangent line at x = 5 is 1 dx =. Finally we use te point and slope information to create te equation y y1 = m( x x1) to get y 8 = 1( x 5). Now simplify to get y = 1x wic is te tangent line equation pictured in Figure 3 above. Example Find te instantaneous rate of cange in Revenue R = te quantity of items sold. q + 4q at q = 3 were q is a) Using te table in Figure as a guide, find te instantaneous rate of cange numerically from te left side. Calculate te slope of secants lines troug te following points were 1) q 1 =3 and q = 9.5 3) q 1 =3 and q = 9.99 ) q 1 =3 and q = 9.9 4) q 1 =3 and q = 9.999
5 b) Now find te instantaneous rate of cange numerically from te rigt side. Calculate te slope of secants lines troug te following points were 1) q 1 =3 and q = 3.1 3) q 1 =3 and q = 3.1 ) q 1 =3 and q = 3.1 4) q 1 =3 and q = 3.1 c) Use your calculations from a) and b) to estimate te slope of te tangent line at q = 3 wic will also te instantaneous rate of cange in revenue at q = 3. Now we will do te problem again algebraically: To find te instantaneous rate of cange at q = 3, we need to find dr R( x + ) f ( x) ( ( q + ) + 4( q + )) ( q + 4 q) dq o q 4q 4q 4 q 4 q ( 4q 4) = -4q + 4 = -4(3) + 4 = Te Marginal Concept Revisited: In te section on average rate of cange we defined te term marginal as te average rate cange in te dependent variable over a one unit cange in te independent variable. Wen considering instantaneous rate of cange you are considering a very small cange in te independent variable. However wen you are considering a cange in quantity of items te smallest cange will be a unit of one. ( you would not make a fraction of an item to sell) Terefore te instantaneous rate of cange wen quantity is involved will be equivalent to te definition of marginal.
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