# CLEAR Calculus

## Lab 13: Example (distance-time-velocity)

Notation

In order to get our approximations accurate to with in about a tenth of a mile, we had to break the two-hour time interval up into twelve 10-minute subintervals. In order to get much more accuracy, we'll need a large number of subintervals and a more efficient way to

1. Express these computations

2. Compute the results

In order to do this we'll use summation notation which is also called sigma notation. We have divided our 2-hour time interval into $$n = 12$$ subintervals of $$\Delta t= 1/6$$ hour each. This generates 13 time points where we need to compute the velocity: \begin{align*} & t_0 = 0,\ t_1 = \frac{1}{6},\ t_2 = \frac{1}{3},\ t_3 = \frac{1}{2},\ t_4 = \frac{2}{3},\ t_5 = \frac{5}{6},\ t_6= 1,\\ & t_7 = \frac{7}{6},\ t_8 = \frac{4}{3},\ t_9 = \frac{3}{2},\ t_{10} = \frac{5}{3},\ t_{11} =\frac{11}{6},\text{ and } t_{12}= 2. \end{align*}

We then express the sums for our approximations as \begin{align*} \text{Underestimate } & = \sum_{i=0}^{11} v(t_i) \frac{1}{6} = 0.67169 \text{ miles and } \\ \text{Overestimate } & = \sum_{i=1}^{12}v(t_i) \frac{1}{6} = 0.77929 \text{ miles}. \end{align*}

• Here we use the capital Greek letter sigma ($$\Sigma$$) to denote a sum.

• The variable $$i$$ is called an index. It ranges through the values that we'll use on the subscripts for the times listed above. Some textbooks may use a $$k$$ or $$n$$ to denote the index.

• The values below and above the $$\Sigma$$ give the starting and ending values of the index to be used in each sum. So the underestimate will run through values of $$i$$ from 0 to 11 while the overestimate will run through values of $$i$$ from 1 to 12.

• The expression to the right of the $$\Sigma$$ shows what we will add up. In this case, it is several distances, each computed as a product of a velocity, $$v(t_i)$$ mph, and a time, 1/6 hour.

In general we could perform this exact same analysis on any time interval $$[a,b]$$. That is $$a < t < b$$. We could choose any number of subintervals $$n$$. This would mean each subinterval of time would be $$\Delta t = \frac{b - a}{n}$$ hours long. The endpoints of these subintervals would then be $\small t_0 = a,\ t_1 = a + \Delta t,\ t_2 = a + 2\Delta t,\ t_3 = a + \Delta t, \dots, t_{n-1} = a + (n-1)\Delta t,\ t_n = a + n\Delta t = b.$

The left hand sum is generated by evaluating the velocity function at the earliest time in each subinterval: $\sum_{i=0}^{n-1} v(t_i) \Delta t.$

The right hand sum is generated by evaluating the velocity function at the latest time in each subinterval: $\sum_{i=1}^n v(t_i) \Delta t.$

Calculators

Summation notation gives us a way to quickly write down the long sums we'll need to compute, but we now need a way to automate the actual multiplying and adding.

TI calculators use the sum and seq commands to handle this very easily. The basic command is

sum(seq(expression to add, variable, start value, end value, increment))

In our case,

• we want to add values given by the expression $$\sin\sqrt{9 - t^2}$$,

• the variable is $$t$$,

• the start value is 0 for the underestimate and $$\Delta t$$ for the overestimate,

• the end value is $$2 - \Delta t$$ for the underestimate and 2 for the overestimate,

(but for reasons we'll discuss later, we'll make these both slightly larger: $$2 - \Delta t/2$$ and $$2 + \Delta t/2$$),

• the increment is $$\Delta t$$.

So for our example with $$\Delta t = 1/6$$, we would enter the following to get the underestimate:

sum(seq(sin(√(9-t^2))*1/6, t, 0, 2-1/12, 1/6))

We would make the following small changes to get the overestimate:

sum(seq(sin(√(9-t^2))*1/6, t, 1/6, 2 + 1/12, 1/6))

These commands return the values 0.67169 and 0.77929, respectively.

Now we can easily generate much more accurate approximations. For example, breaking the two hours into 120 one-minute intervals can be accomplished with

sum(seq(sin(√(9-t^2))*1/60, t, 0, 2-1/120, 1/60))

which returns the underestimate of 0.71884 miles and

sum(seq(sin(√(9-t^2))*1/60, t, 1/60, 2+1/120, 1/60))

which returns the overestimate of 0.72960 miles.

These approximations give us an error bound of 0.01076 miles. Notice that this is 10 times smaller than the error bound that we got from using twelve 10-minute intervals. This is to be expected since we already found that our error bound would be 0.64563$$\Delta t$$.

Now suppose that we want to compute the distance traveled accurate to 5 decimal places given in the orginal problem, 0.72421 miles. We would need the error bound to be 0.000005. So we can set $0.64563 \Delta t = 0.000005 \text{ and solve } \Delta t = 0.0000077437$ or $n = \frac{2}{0.0000077437} = 258,252 \text{ subintervals*}$

*I am not responsible for your calculator melting if you actually try this.