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5.1 Estimating with Finite SumsPowerPoint Presentation

5.1 Estimating with Finite Sums

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Go to problem #9 in 5.1 and read the instructions. We will do it to help you see that these problems are simpler than they look:

Consider an object moving at a constant rate of 3 ft/sec.

What is the distance travelled by that object over 4 seconds?

Since rate . time = distance:

If we draw a graph of the velocity, how is that distance represented geometrically?

The distance that the object travels is equal to the area under the line.

After 4 seconds, the object has gone 12 feet.

velocity

time

Velocity is the first derivative of position.

Mr. Murphy’s fall from the 196 foot platform can

be expressed with the equation:

…where t is in seconds and s is measured in feet. Given the above statement, the equation for his velocity is:

…which is expressed in what units?

feet/second

How can we determine from this graph how far Mr. Murphy fell?

t (seconds)

Remember that whenever you are lost…

…units can lead you home.

v(t) (feet/second)

How can we determine from this graph how far Mr. Murphy fell?

So what will the final units here be?

t (seconds)

feet

How can we convert the units we have on the graph into feet?

v(t) (feet/second)

By multiplying:

v(t)

(feet/second)

t

(seconds)

×

(3.5, 112)

112

(feet/second)

×

3.5

(seconds)

How can we determine from this graph how far Mr. Murphy fell?

112 × 3.5 =

t (seconds)

392

feet

How can this number be represented geometrically on this graph?

…392 is the area of this rectangle.

v(t) (feet/second)

But Mr Murphy only fell 196 feet, not 392.

How can we determine from this graph how far Mr. Murphy fell?

But Mr Murphy only fell 196 feet, not 392.

t (seconds)

What are we missing here?

196 is ½ of 392

v(t) (feet/second)

…which is also the area of this triangle.

How can we determine from this graph how far Mr. Murphy fell?

So the lesson here is:

t (seconds)

The distance Mr Murphy fell can be found by finding

The area of the region between the curve (or the line) and the x-axis

v(t) (feet/second)

over a particular interval

[0, 3.5] seconds

If the velocity is not constant, fell?

The distance traveled is still equal to the area under the curve.

But how would we handle something like this:

A =

A =

A =

A = 1

Example:

And since in this case, each rectangle has a base of 1…

What would be a good way of at least approximating the area under this curve?

Try drawing a few rectangles side by side under the curve…

Because of where each rectangles touch the curve, this is called the Left-hand Rectangular Approximation Method (text books call it LRAM).

If the velocity is not constant, fell?

The distance traveled is still equal to the area under the curve.

But how would we handle something like this:

A =

A =

A =

A = 1

Example:

And since in this case, each rectangle has a base of 1…

How do we use this info to approximate the area?

Each rectangle has a base of

1

The heights are determined by

V

Approximate area:

Approximate area: fell?

If the velocity is not constant,

The distance traveled is still equal to the area under the curve.

But how would we handle something like this:

Example:

We could also use a Right-hand Rectangular Approximation Method (RRAM).

Approximate area: fell?

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

In this example there are four subintervals.

How can we increase the accuracy?

Increase the number of subintervals

Approximate area: fell?

The exact answer for this

problem is .

With 8 subintervals

(using midpoint method):

width of each rectangle (or subinterval)

Circumscribed fell? rectangles are all above the curve:

Inscribed rectangles are all below the curve:

We will be learning how to find the exact area under a curve if we have the equation for the curve. Rectangular approximation methods are still useful for finding the area under a curve if we do not have the equation…

But what if we did and the graph of the equation were simple…

Go to problem #9 in 5.1 and read the instructions. We will do it to help you see that these problems are simpler than they look:

The size of each subinterval is:

2

If we used the left-hand method, we would wind up with...

You get the idea. And so the answer is…

The size of each subinterval

The left hand height of the first rectangle

…and the next one…

…and the next one…

Go to problem #9 in 5.1 and read the instructions. We will do it to help you see that these problems are simpler than they look:

The size of each subinterval is:

2

If we used the left-hand method, we would wind up with...

44.8 (mg/L) sec

Now use the right-hand method. Be careful not to do more work than is necessary…

Answer?

44.8 (mg/L) sec

How is it the same?

Go to problem #9 in 5.1 and read the instructions. We will do it to help you see that these problems are simpler than they look:

Now use the right-hand method. Be careful not to do more work than is necessary…

Answer?

44.8 (mg/L) sec

How is it the same?

Both the left most and right most rectangle heights are 0.

Notice more importantly that in going from left hand to right, you don’t need to start from scratch. Just subtract the left-most value and add the right-most value.

Now use the right-hand method. Be careful not to do more work than is necessary…

Answer?

44.8 (mg/L) sec

How is it the same?

Lastly, there are less structured approaches to this. For instance, you could also have looked at each square on the grid (each has an area of 2) and simply added up the ones under the curve.

In the case of ones that are cut off by the curve (like the first one on the left) you can cut the area in half. Try it to see if your approximation will be better than the other two methods...

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