6.A Accumulating Change

Activities

Growing Bacteria

A bacteria population is growing at a rate given by \[f(t)=3+0.1t^2 \quad \text{ millions of bacteria per hour}\]

A plot of this growth rate is shown below.

We have already estimated the change in population from \(t=0\) to \(t=3\) using intervals of width 1. Find a better estimate of the increase in population by

  1. Using intervals of width 1/2 and evaluating \(f(t)\) at the right endpoint of each interval.
  2. Using intervals of width 1/2 and evaluating \(f(t)\) at the left endpoint of each interval.

You can do this by hand, or use Desmos to help out. Decide whether your value is an over-estimate or an under-estimate.

The Change in World Population

The rate of change of the world’s population, in millions of people per year, is in the table below.

Year 1950 1960 1970 1980 1990 2000
Rate of Change 37 41 78 77 86 79

Using this information, estimate the total change in the world’s population between 1950 and 2000 by using

  1. A left endpoint estimate
  2. A right endpoint estimate

Speed Tests

The velocity of a new electric car is measured every two seconds. The data is shown below.

Time (seconds) 0 2 4 6 8
Velocity (m/s) 0 8 16.7 27.3 36.2

Estimate the total distance traveled by this car in the first eight seconds by using

  1. A left endpoint estimate
  2. A right endpoint estimate

Estimating Area with RStudio

Here is some RStudio code that will estimate the area under the curve \(f(x)=2+4x^2-x^3\) on the interval \(0 \leq x \leq 4\) using a right endpoint estimation on intervals of size \(0.5\).

f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.5
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29
  1. Update this code to estimate this area using
    1. A right endpoint estimation on intervals of size \(0.1\)
    2. A right endpoint estimation on intervals of size \(0.01\)
    3. A left endpoint estimation on intervals of size \(0.5\)
    4. A left endpoint estimation on intervals of size \(0.01\)
    5. A left endpoint estimation on intervals of size \(0.00001\)
  2. Make a plot of \(f(x)\) on \([0,4]\). Is is clear whether each of the values about is an over-estimate? an under-estimate? or neither?

Solutions

Growing Bacteria

  1. Right endpoint estimate using intervals of width 1/2.
f = makeFun(3+0.1*t^2~t)
# right endpoint approximation
f(0.5)*0.5 + f(1)*0.5 + f(1.5)*0.5 + f(2)*0.5 + f(2.5)*0.5 + f(3)*0.5 
## [1] 10.1375
  1. Left endpoint estimate using intervals of width 1/2.
f = makeFun(3+0.1*t^2~t)
# left endpoint approximation
f(0)*0.5 + f(0.5)*0.5 + f(1)*0.5 + f(1.5)*0.5 + f(2)*0.5 + f(2.5)*0.5 
## [1] 9.6875

The Change in World Population

  1. Left endpoint estimate
37*10 + 41*10 + 78*10 + 77*10 + 86*10
## [1] 3190
  1. Right endpoint estimate
41*10 + 78*10 + 77*10 + 86*10 + 79*10
## [1] 3610

Speed Tests

  1. Left endpoint estimate
0*2 + 8*2 + 16.7*2 + 27.3*2
## [1] 104
  1. Right endpoint estimate
8*2 + 16.7*2 + 27.3*2 + 36.2*2
## [1] 176.4

Estimating Area with RStudio

  1. We estimate the area below \(f(x)=2+4x^2-x^3\) on the interval \(0 \leq x \leq 4\) using
    1. A right endpoint estimation on intervals of size \(0.1\)
f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.1
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29.32
  1. A right endpoint estimation on intervals of size \(0.01\)
f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.01
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29.3332
  1. A left endpoint estimation on intervals of size \(0.5\)
f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.5
points = seq(from=a, to=b-base, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29
  1. A left endpoint estimation on intervals of size \(0.01\)
f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.01
points = seq(from=a, to=b-base, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29.3332
  1. A left endpoint estimation on intervals of size \(0.00001\)
f=makeFun(2+4*x^2-x^3~x)
a=0
b=4
base = 0.00001
points = seq(from=a, to=b-base, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 29.33333
  1. Make a plot of \(f(x)\) on \([0,4]\). Is is clear whether each of the values about is an over-estimate? an under-estimate? or neither?
f=makeFun(2+4*x^2-x^3~x)
slice_plot(f(x)~x, domain(x=0:4)) + ylab("f(x)")

It is not clear whether any of these values is an over-estimate or an under-estimate. There is a maximum at \(x=2.75\).

  • A left endpoint estimate will be an under-estimate on \([0,2.75]\) and an over-estimate on \([2.75,4]\).
  • A right endpoint estimate will be an over-estimate on \([0,2.75]\) and an under-estimate on \([2.75,4]\).

So we can’t easily tell whether it’s an under-estimate or an over-estimate.