Mr. Simonds’ MTH 251 class
Timmy was done with his bath and decided to stay in the tub whilst the water drained. The water did not drain at a constant rate, in part because Timmy couldn’t keep his toes off the stopper lever as the water was draining. For simplicities sake, let’s say that the volume of water (gal) remaining in the tub t minutes after the drainage began was given by the formula V  t   128  2 t 2 . Let’s determine the average rate of change in V (including unit) over the time it took the tub to drain and write a sentence interpreting that rate. Definition
The average rate of change in the continuous function f over the interval  a, b  is:
f b  f  a 
.
ba
Average Rates of Change |1
Mr. Simonds’ MTH 251 class
Let’s find a formula for the average rate of change in V  t   128  2 t 2 over the interval t0 , t1  . Remember … It would behoove you to copy the notes using the same format and showing the same steps as illustrated by Mr. Simonds; your grade will in part depend upon your ability to follow this format while working your homework and taking your tests. You will get the opportunity to be creative in this class, but the manner in which you present algebra is not one of those places. 2| Average Rates of Change
Mr. Simonds’ MTH 251 class
Let’s find the difference quotient for the function V  t   128  2 t 2 . Definition The difference quotient for the function f is f  x  h  f  x
h
. Please note that application of this formula needs to be adapted to alternative function names and/or independent variables. Average Rates of Change |3
Mr. Simonds’ MTH 251 class
Let’s show that, in general, the difference quotient for the function y  V  t  finds the average rate of change in V over the interval t , t  h  . We determined that the average rate of change for Timmy’s tub function over the interval t0 , t1  is given by the formula 2  t0  t1  and that the difference quotient for the function simplifies to the formula 2  2t  h  . Let’s use each of these formulas to find (without unit) the average rates of change in V over the intervals  0.5,1.5 and 3,3.5 . 4| Average Rates of Change
Mr. Simonds’ MTH 251 class
It appears that the drainage rate is increasing over time; this is supported by the values of V shown in Table 1. Let’s use the average rate of change formula to deduce the actual rate of change in V exactly one minute in the drainage process and then use the difference quotient to deduce the actual rate of change exactly three minutes into the drainage process. Let’s do all of this without units. Table 1: V  t   128  2 t 2 t 0 1 2 3 4 5 6 7 8 V(t) 128 126 120 110 96 78 56 30 0 Average Rates of Change |5
Mr. Simonds’ MTH 251 class
Figure 1 shows the function V  t   128  2 t 2 along with the secant line connecting the fixed point 1,V 1  to the variable point 1  h,V 1  h   . Let’s state each of the expressions requested below the figure in terms of V and h . This is the point 1,V 1 .
This is the point 1  h,V 1  h  .
run
Figure 1: A secant line to V  t   128  2 t 2
The run between the two points is: x2  x1  The rise between the two points is: y2  y1  rise

The slope of the secant line is: y2  y1
 x2  x1
6| Average Rates of Change



Mr. Simonds’ MTH 251 class
4
and use that to find the slope of the t 8
secant line between the points where t   7 and t   4 . Let’s find the difference quotient for the function z  t  
t  8
y0
Figure 2:
y
4
t 8
Average Rates of Change |7
Mr. Simonds’ MTH 251 class
 t 
 where t is the amount of time (s)  8 
v  4  v  2
The vertical velocity of a yo‐yo (in in/s) is given by v  t    sin 
that has passed since some yo‐yo started to yo‐yo. Find and interpret 4s  2s
. The amount of time (minutes) it takes Vimel to walk four miles is given by w  t   50  10 e  t where t is the amount of time (months) that has passed since Vimel started to walk to work. Find and interpret w  2  w 0
2 mo  0 mo
. 8| Average Rates of Change