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## Derivatives and Rates of Change

The problem of finding the tangent line to a curve and the problem of finding the velocity of an object both involve finding the same type of limit.

*This special type of limit is called a derivative and we will *
see that it can be interpreted as a rate of change in any of
the natural or social sciences or engineering.

## Tangents

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## Tangents

*If a curve C has equation y = f(x) and we want to find the *
*tangent line to C at the point P(a, f(a)), then we consider a *
*nearby point Q(x, f(x)), where x* *≠ a, and compute the slope *
*of the secant line PQ:*

*Then we let Q approach P along the curve C by letting *
*x approach a.*

## Tangents

*If m*_{PQ}*approaches a number m, then we define the tangent *
*t to be the line through P with slope m. (This amounts to *

saying that the tangent line is the limiting position of the
*secant line PQ as Q approaches P. See Figure 1.)*

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## Tangents

## Example 1

*Find an equation of the tangent line to the parabola y = x*^{2}
*at the point P(1, 1).*

Solution:

*Here we have a = 1 and f(x) = x*^{2}, so the slope is

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*Example 1 – Solution*

=

= 1 + 1

= 2

Using the point-slope form of the equation of a line, we find that an equation of the tangent line at (1, 1) is

*y – 1 = 2(x – 1) or y = 2x – 1*

cont’d

## Tangents

We sometimes refer to the slope of the tangent line to a
**curve at a point as the slope of the curve at the point.**

The idea is that if we zoom in far enough toward the point, the curve looks almost like a straight line.

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## Tangents

*Figure 2 illustrates this procedure for the curve y = x*^{2} in
Example 1.

*Zooming in toward the point (1, 1) on the parabola y = x*^{2}

**Figure 2**

## Tangents

The more we zoom in, the more the parabola looks like a line.

In other words, the curve becomes almost indistinguishable from its tangent line.

There is another expression for the slope of a tangent line that is sometimes easier to use.

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## Tangents

*If h = x – a, then x = a + h and so the slope of the secant *
*line PQ is*

*(See Figure 3 where the case h > 0 is illustrated and Q is to *
*the right of P. If it happened that h < 0, however, Q would *
*be to the left of P.)*

**Figure 3**

## Tangents

*Notice that as x approaches a, h approaches 0 (because *
*h = x – a) and so the expression for the slope of the *

tangent line in Definition 1 becomes

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## Velocities

In general, suppose an object moves along a straight line
*according to an equation of motion s = f(t), where s is the *
displacement (directed distance) of the object from the
*origin at time t.*

*The function f that describes the motion is called the *
**position function of the object. **

*In the time interval from t = a to t = a + h the change in *

## Velocities

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See Figure 5.

## Velocities

**Figure 5**

## Velocities

The average velocity over this time interval is

which is the same as the
slope of the secant line
*PQ in Figure 6.*

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## Velocities

Now suppose we compute the average velocities over
*shorter and shorter time intervals [a, a + h]. *

*In other words, we let h approach 0. As in the example of *
**the falling ball, we define the velocity (or instantaneous **
* velocity) v(a) at time t = a to be the limit of these average *
velocities:

*This means that the velocity at time t = a is equal to the *
*slope of the tangent line at P.*

Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground.

(a) What is the velocity of the ball after 5 seconds?

(b) How fast is the ball traveling when it hits the ground?

Solution:

*We will need to find the velocity both when t = 5 and when *

## Example 3

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*Example 3 – Solution*

*Using the equation of motion s = f(t) = 4.9t*^{2}, we have

cont’d

*Example 3 – Solution*

*(a) The velocity after 5 seconds is v(5) = (9.8)(5) *

cont’d

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*Example 3 – Solution*

(b) Since the observation deck is 450 m above the ground,
*the ball will hit the ground at the time t when *

*s(t) = 450, that is,*

*4.9t*^{2} = 450
This gives

*t*^{2} = *and t =* ≈ 9.6 s

cont’d

*Example 3 – Solution*

The velocity of the ball as it hits the ground is therefore

cont’d

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## Derivatives

## Derivatives

We have seen that the same type of limit arises in finding the slope of a tangent line (Equation 2) or the velocity of an object (Equation 3).

In fact, limits of the form

arise whenever we calculate a rate of change in any of the sciences or engineering, such as a rate of reaction in

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## Derivatives

*If we write x = a + h, then we have h = x – a and h*

*approaches 0 if and only if x approaches a. Therefore an *
equivalent way of stating the definition of the derivative, as
we saw in finding tangent lines, is

## Example 4

*Find the derivative of the function f(x) = x*^{2} *– 8x + 9 at the *
*number a.*

Solution:

From Definition 4 we have

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*Example 4 – Solution*

_{cont’d}

## Derivatives

*We defined the tangent line to the curve y = f(x) at the point *
*P(a, f(a)) to be the line that passes through P and has *

*slope m given by Equation 1 or 2. *

Since, by Definition 4, this is the same as the derivative
*f′(a), we can now say the following.*

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## Derivatives

If we use the point-slope form of the equation of a line, we can write an equation of the tangent line to the curve

*y = f(x) at the point (a, f(a)):*

*y – f(a) = f′(a)(x – a)*

## Rates of Change

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## Rates of Change

*Suppose y is a quantity that depends on another quantity x. *

*Thus y is a function of x and we write y = f(x). *

*If x changes from x*_{1} *to x*_{2}*, then the change in x (also called *
**the increment of x) is**

*∆x = x*_{2} *– x*_{1}

*and the corresponding change in y is*

*∆y = f(x*_{2}*) – f(x*_{1})

## Rates of Change

The difference quotient

**is called the average rate of **
**change of y with respect to x ***over the interval [x*_{1}*, x*_{2}] and

can be interpreted as the slope
*of the secant line PQ*

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## Rates of Change

By analogy with velocity, we consider the average rate of
*change over smaller and smaller intervals by letting x*_{2}

*approach x*_{1} and therefore letting *Δx approach 0. *

The limit of these average rates of change is called the

**(instantaneous) rate of change of y with respect to x at ***x = x*_{1}, which (as in the case of velocity) is interpreted as
*the slope of the tangent to the curve y = f(x) at P(x*_{1}*, f(x*_{1})):

*We recognize this limit as being the derivative f′(x*_{1}).

## Rates of Change

*We know that one interpretation of the derivative f′(a) is as *
*the slope of the tangent line to the curve y = f(x) when *

*x = a. We now have a second interpretation:*

The connection with the first interpretation is that if we
*sketch the curve y = f(x), then the instantaneous rate of *

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## Rates of Change

This means that when the derivative is large (and therefore
*the curve is steep, as at the point P in Figure 9), the *
*y-values change rapidly. *

**Figure 9**

*The y-values are changing rapidly *
*at P and slowly at Q.*

## Rates of Change

When the derivative is small, the curve is relatively flat (as
*at point Q) and the y-values change slowly.*

*In particular, if s = f(t) is the position function of a particle *
*that moves along a straight line, then f′(a) is the rate of *
*change of the displacement s with respect to the time t. *

*In other words, f′(a) is the velocity of the particle at time*
*t = a. *

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## Example 6

A manufacturer produces bolts of a fabric with a fixed width.

*The cost of producing x yards of this fabric is *
*C = f(x) dollars. *

*(a) What is the meaning of the derivative f′(x)? What are its *
units?

(b) In practical terms, what does it mean to say that
*f*′(1000) = 9?

*(c) Which do you think is greater, f′(50) or f′(500)? *

*What about f*′(5000)?

*Example 6(a) – Solution*

*The derivative f′(x) is the instantaneous rate of change of C *
*with respect to x; that is, f′(x) means the rate of change of *
the production cost with respect to the number of yards
produced.

Because

*the units for f′(x) are the same as the units for the *
difference quotient ∆C/∆x.

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*Example 6(b) – Solution*

*The statement that f*′(1000) = 9 means that, after 1000

yards of fabric have been manufactured, the rate at which the production cost is increasing is $9/yard.

*(When x = 1000, C is increasing 9 times as fast as x.)*
Since ∆x = 1 is small compared with x = 1000, we could
use the approximation

and say that the cost of manufacturing the 1000th yard (or the 1001st) is about $9.

cont’d

*Example 6(c) – Solution*

The rate at which the production cost is increasing

*(per yard) is probably lower when x = 500 than when x = 50 *
(the cost of making the 500th yard is less than the cost of
the 50th yard) because of economies of scale. (The

manufacturer makes more efficient use of the fixed costs of production.)

So

*f′(50) > f′(500)*

cont’d

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*Example 6(c) – Solution*

But, as production expands, the resulting large-scale operation might become inefficient and there might be overtime costs.

Thus it is possible that the rate of increase of costs will eventually start to rise.

So it may happen that

*f′(5000) > f′(500)*

cont’d