How far does a sprinter run in 4.5 s?

Two problems, one idea

The necklace. $1,500 at $20 a week: how many weeks? Division. With a weekly raise, the rate changes and division fails.

The forty. A 40-yard dash from a standstill. A radar gun gives speed at every instant, not position. How far after 1 second? After 4.5?

Both: rate known, total wanted. That total is an accumulation.

Start with the money

Constant rate: ${\text{total} = \text{rate} \cdot \text{time}}$.

Now the rate changes.

Picture each week as a bar, 1 week wide and as tall as that week's pay. Savings are the staircase's area. Adding up bars under a rate is the main move.

Bars estimate an accumulation; the area under the rate's graph is the exact accumulation.

Chop up the sprint

No single rate to multiply by 4.5 seconds. Chop the time into ${n}$ pieces, hold velocity constant on each, and add velocity times time: bars again, each area a distance.

Try these:

1. At ${n = 4}$, Left versus Right: which is bigger? Why, from the curve?
2. Drag ${n}$ to 50. What happens to each gap?
3. Trapezoid: over or under here? Check the hatched and dotted regions.
4. Check your second prediction against the area.

Area is accumulation

Bar area is height times width: ${\tfrac{\text{yd}}{\text{s}} \cdot \text{s} = \text{yd}}$, a distance. The exact distance is the area between the velocity curve and the ${t}$-axis. Money too: ${\tfrac{\text{dollars}}{\text{week}} \cdot \text{weeks} = \text{dollars}}$.

The rule. Split ${[a, b]}$ into ${n}$ subintervals of width ${\Delta x = \frac{b - a}{n}}$, with endpoints ${a = x_0 < x_1 < \cdots < x_n = b}$. Each term is one bar, height times width.

• Left sum: height at the start.
  $\displaystyle L_n = \sum_{i=1}^{n} \orange{f(x_{i-1})\,\Delta x}$$L_n$$=\sum _{i=1}^{n}f(x_{i-1})\Delta x$
• Right sum: height at the end.
  $\displaystyle R_n = \sum_{i=1}^{n} \orange{f(x_i)\,\Delta x}$$R_n$$=\sum _{i=1}^{n}f(x_i)\Delta x$
• Midpoint sum: height in the middle.
  $\displaystyle M_n = \sum_{i=1}^{n} \orange{f\!\left(\frac{x_{i-1} + x_i}{2}\right)\,\Delta x}$$M_n$$=\sum _{i=1}^{n}f\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x$
• Trapezoid sum: average of the two end heights.
  $\displaystyle T_n = \sum_{i=1}^{n} \orange{\frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x}$$T_n$$=\sum _{i=1}^n\frac{f(x_{i-1})+f(x_i)}{2}\Delta x$$\displaystyle = \frac{L_n + R_n}{2}$$=\frac{L_n+R_n}{2}$

From a table, widths can differ: each term uses its own.

Over or under, without the exact answer:

• ${f}$ increasing: left too low, right too high. Decreasing: reversed.
• ${f}$ concave down: trapezoid too low (slanted tops under the curve). Concave up: too high. Midpoint: opposite of trapezoid.

As ${n}$ grows, all four approach the definite integral:

Read “add up ${f(x)}$ times ${dx}$ from ${a}$ to ${b}$.” The ${\int}$ is a stretched S, for sum; ${dx}$ is a thin bar's width; ${x_i^*}$ is any point in the ${i}$th subinterval. Sprint distance: ${\int_0^{4.5} v(t)\,dt}$ yards.

Worked example, then your turn

The 40-yard dash

Unequal widths: watch them.

Free response

This one is written like an AP free-response question. The parts with a number check your answer here. Write every part out in full, with your work and units, the way you would on the exam: that's what gets scored.

Fluency

Six quick ones: widths, heights, multiply.

Mixed review

From earlier lessons: Is the Ball Ever Stopped? and Best Price.

Exit ticket