IB

# Rates of Change

Rates of Change Instantaneous Rate of Change As we have seen, the slope of the tangent at point P is the limit of the slope of the secant between points...

#### Definition of Average Rate of Change

The expression

$\frac{f\left(a+h\right)-f\left(a\right)}{h}$

is called the difference quotient for  f  at  a  and represents the average rate of change of  y = f(x)  from   to  a + h .

Geometrically, the rate of change of  f  from  a  to  a+h   the slope of the secant line through the point  P(a, f(a))  and  Q(a+h, f(a+h)).

If  f(t) is the position function of a particle that is moving on a straight line, then in the time interval from  t = a to t = a+h, the change in position is f(a+h) − f(a), and the average velocity of the particle over the time interval is

$\begin{array}{l}average\text{\hspace{0.17em}}velocity\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{displacement}{time}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{f\left(a+h\right)-f\left(a\right)}{h}\end{array}$

Example 1. The displacement of a particle moving in a straight line is given by the equation of motion f(t) = t3 − 4t + 3. Find the average velocity of the particle over the interval 0 ≤ t ≤ 4.

Solution :

$\begin{array}{l}average\text{\hspace{0.17em}}velocity\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{f\left(4\right)-f\left(0\right)}{4-0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\left(64-16+3\right)-3}{4-0}=12\end{array}$

#### Exercises – Rate of Change

Multiple Choice Questions

1.  The traffic flow at a particular intersection is modeled by the function f defined by f(t) = 25+6cos(x/3) for 0 ≤ t ≤ 120. What is the average rate of change of the traffic flow over the time interval 30 ≤ t ≤ 40.

(A) 0.743

(B) 0.851

(c) 0.935

(d) 1.176

2. The rate of change of the altitude of a hot air balloon rising from the ground is given by y(t) = t3−3t2 + 3t for 0 ≤ t ≤ 10. What is the average rate of change in altitude of the balloon over the time interval 0 ≤ t ≤ 10.

(A) 56

(B) 73

(c) 85

(d) 94

#### Free Response Questions

 t(sec) 0 10 20 30 40 50 60 70 80 90 f(t)(ft/sec) 0 28 43 67 82 85 74 58 42 35

3. The table above shows the velocity of a car moving on a straight road. The car’s velocity v is measured in feet per second.

(a) Find the average velocity of the car from t = 60 to t = 90.

(b) The instantaneous rate of change of f  with respect to x at x = a can be approximated by finding the average rate of change of f near x = a. Approximate the instantaneous rate of change of f at x = 40 using two points, x = 30 and x = 50

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