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# Quick Review – p.254 - PowerPoint PPT Presentation

Quick Review – p.254. 1. 2. 3. 4. Quick Review – p.254. 5. 6. 7. 8. Quick Review – p.254. 9. 10. Rectangular Approximation method (RAM). Section 5.1a. Solve the following:. A train moves along a track at a steady rate of 75mph from

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### Rectangular Approximation method (RAM)

Section 5.1a

A train moves along a track at a steady rate of 75mph from

7:00am to 9:00am. What is the total distance traveled by the

train?

Distance = Rate x Time

Distance = (75 mph)(2 hr) = 150 miles

Now, how does this result relate to the graph of the train’s

velocity over time???

velocity

(mph)

It’s the area under

the curve!!!

75

time (hr)

7

9

(mph)

It’s the area under

the curve!!!

75

time (hr)

7

9

Would this same principle apply if the velocity varied over time?

velocity

Yes, but how could we

find the area under this

irregularly-shaped curve?

Area = Distance

time

Yes, but how could we

find the area under this

irregularly-shaped curve?

time

The area could be found by slicing the region into small strips…

…and then summing the area of the strips (which are essentially

rectangles).

The narrower the strips, the more

accurate our estimate of area!!!

the x-axis with velocity v(t) = t for time t > 0. Where is the

particle at t = 3?

2

Graph the velocity…

v

and partition the time interval

[0, 3] into subintervals of length

t (here, we’ll use 12 sub-

intervals of length 3/12).

9

Zoom in on one of these sub-

intervals and find the area:

v

t

0

1

2

3

We’ll approximate the area

using the following rectangle:

t

0

0.5

0.75

1

the x-axis with velocity v(t) = t for time t > 0. Where is the

particle at t = 3?

2

Subinterval

Midpoint

Height =

Area =

the x-axis with velocity v(t) = t for time t > 0. Where is the

particle at t = 3?

2

Continue this pattern, sum the area for each

of the 12 subintervals:

 We conclude that the particle has moved

approximately 9 units in 3 seconds!!!

(If it starts at x = 0, then it is very

close to x = 9 when t = 3…)

What we just used is called the Midpoint Rectangular

Approximation Method (MRAM). (we evaluated

the height of each rectangle at the midpoint of each

subinterval)

Evaluating the function at the left-hand endpoint yields

the LRAM approximation, and right-hand endpoints

gives the RRAM approximation for area.

but only six subintervals:

3

3

3

LRAM

MRAM

RRAM

A = 6.875

LRAM: A =

but only six subintervals:

3

3

3

LRAM

MRAM

RRAM

A = 6.875

A = 8.9375

MRAM: A =

but only six subintervals:

3

3

3

LRAM

MRAM

RRAM

A = 6.875

A = 8.9375

A = 11.375

RRAM: A =

of n subintervals…

n LRAM MRAM RRAM

n

n

n

6 6.875 8.9375 11.375

12 7.90625 8.984375 10.15625

24 8.4453125 8.99609375 9.5703125

48 8.720703125 8.999023438 9.283203125

100 8.86545 8.999775 9.13545

1000 8.9865045 8.99999775 9.0135045

For large values of n, we can see that all three sums

approach the same number (in this case, 9).

given function on the interval [0, 3].

First, look at a graph…

Next, use a calculator to evaluate RAM values…

n LRAM MRAM RRAM

n

n

n

5 5.15480 5.89668 5.91685

10 5.52574 5.80685 5.90677

25 5.69079 5.78150 5.84320

50 5.73615 5.77788 5.81235

100 5.75701 5.77697 5.79511

1000 5.77476 5.77667 5.77857

All three converge!!!

(at n = 1000 they agree for the first 3 digits!!!)

given function on the interval [0, 3].

First, look at a graph…

Next, use a calculator to evaluate RAM values…

n LRAM MRAM RRAM

n

n

n

5 5.15480 5.89668 5.91685

10 5.52574 5.80685 5.90677

25 5.69079 5.78150 5.84320

50 5.73615 5.77788 5.81235

100 5.75701 5.77697 5.79511

1000 5.77476 5.77667 5.77857

Note: the exact area is –7cos(3) + 6sin(3) – 2,

which is 5.77666752456 to twelve digits…