Distance and accumulated change [5.1]

Graphically and analytically we have developed the idea of the derivative as the instantaneous rate of change of a function.

distance $\to$ velocity

If $d(t)$ is the distance a car travel, as a function of time, the derivative of $d$, which we write $d'(t)$, is the car's instantaneous velocity at time $t$.: $d'(t)\equiv v(t)$.

velocity $\to$ distance

Now, we'll try to go the other way: Let's say we know the velocity of a car, $v(t)$. We should like to find the distance travelled (which depends on the time spent travelling).

Some examples will try to make a graphical connection as well, with this reverse process, which we'll call "accumulated change".

Steady velocity

Suppose car A is moving with a steady velocity of 30 miles per hour. After 20 minutes spent travelling at this speed, how much distnace has the car "accumulated"? That is:. How far does car A travel in this time?

• Distance = Velocity (rate) $\times$ Time.
• 20 minutes = $\frac 13$ hour.
• Distance = 30 $\frac{\text{miles}}{\text{hour}} \times \frac 13$ hour=10 miles
• Graphical interpretation: Area under $v(t)$ graph $$A=\frac 13 \times 30=10$$
  The units of the area: $$A=\text{hours} \times\frac{\text{miles}}{\text{hours}}=\text{miles}.$$

What if the velocity is not constant??

Suppose the velocity of car B, in meters / sec, is given by $v(t)=5t$ where $t$ is the time in seconds.

• At $t=0$ speed = 0: the car is not yet moving.
• At $t=1$ second, $v(1)=5*1=5$ m / sec.
• At $t=2$ seconds, $v(2)=10$ m/sec.
• etc.

How far has the car gone in 10 seconds

• First approach: Find the average velocity of the car: $v_{ave}=$___?
• For a constantly increasing speed, $v_{ave}=\frac{v_f-v_i}{2}=\frac{50-0}{2}=25$ m/sec.
• Distance = average Velocity $\times$ Time = 25 * 10 = 250 meters.
  
  
• Second approach: Area under $v(t)$ graph? [Will this give the same answer??]
• Area = $\frac 12$ base $\times$ height=$\frac 12 10 \times 50=250$ m!

  BTW: what are the accelerations of the two cars? (Acceleration is the derivative of velocity, or the second derivative of distance.))

  • 0 meters per second${}^2$
  • 5 meters per second${}^2$

A general approach for finding areas under curves

What if we'd like to find the area underneath a more complicated curve? Suppose $f$ is a nonnegative function and $a\lt b$ are numbers. That is, we'd like to find the shaded area shown below, which is

1. under the curve $y=f(x)$,
2. above the $x$-axis,
3. and between $x=a$ and $x=b$

Well, when we started working with derivatives, we started by making approximations, and then came up with a way to make those approximations more and more exact. Along similar lines, we'll start by

Approximating the area:

We can obtain a lower/under estimate of this area by

1. drawing four rectangles of equal width and varying heights, each rectangle as high as possible while keeping the rectangles completely inside the shaded area,
2. finding the areas of each of these rectangles, and
3. finding their sum.

We can obtain an upper/over estimate of this area by

1. drawing four rectangles of equal width and varying heights, as low as possible while still covering all portions of the desired area,
2. finding the areas of each of these rectangles, and
3. finding their sum.

The exact answer for the shaded area should be somewhere between these approximations.

What could we do to get better upper or better lower estimates?

Distance and accumulated change (.ppt)

We will do this handout (.pdf)