Question

@tkhunny
I have another question that continued off the previous question

Answer 1

Let's see it. Don't forget to show me how far you have gotten.

Answer 2

Continuing off 32t as the equation:

Use left and right sums with /_\ (delta) T to find upper and lower bounds on the distance that the object travels in four seconds.

Left sum = ft
Right sum = ft
Find the precise distance using the area under the curve.
Precise distance = ft

Answer 3

You'll have to specify \(\Delta t\). Since we're going only to t = 4, you should keep \(\Delta t\) small enough to be meaningful.

Answer 4

Oh, thats right. I forgot that delta T is = 1. Sorry

Answer 5

Perfect. We need to build little rectangles - four of them.
We have v(t) = 32t

The left sum requires:
v(0) =
v(1) =
v(2) =
v(3) =

The right sum requires:
v(1) =
v(2) =
v(3) =
v(4) =

Fill in the blanks!

Answer 6

The left sum requires:
v(0) = 0
v(1) = 32
v(2) = 64
v(3) = 96

The right sum requires:
v(1) = 32
v(2) = 64
v(3) = 96
v(4) = 128

Answer 7

Okay, then we have a tiny thought to get into our heads. Our task is to produce lower and upper bounds. We have to make sure we know what we are doing.

In this case, since v(t) is monotonic increasing (never flattens out - never turns around), it will ALWAYS be the case that a left sum is a lower bound and a right sum is an upper bound. Do you have enough of a handle on what we are doing to see that?

Answer 8

I'm not sure I understand?

Answer 9

We are chopping up v(t) into four pieces. \(\Delta t\) has defined the base of the four rectangles we need. Our task is to decide on the height of each piece.

Any clearer?

Answer 10

Ah Ok. Much clearer.

Answer 11

Well, then. Let's consider the first rectangle. Its base is [0,1] - width 1.

Left

If we use the left-most value in the interval, f(0) = 0, we get a really small area, 1 * 0 = 0.

Right

If we use the right-most value in the interval, f(1) = 32, we get an area, 1 * 32 = 32.

That's the first piece of each of the two sums we need . Find the area of the other pieces.

Answer 12

1*64 = 64 and 1 * 96 = 96

Answer 13

There should be four pieces for each sum.

Answer 14

0 + 43 + 64 + 96 = 203

Answer 15

43? s/b 32

Answer 16

32 im sorry
0 + 32 + 64 + 96 = 192

Answer 17

Okay, that's the left-hand sum. Because of the nature of the function, this is also a lower bound!

Let's see the other one.

Answer 18

32 + 64 + 96 + 128 = 320

Answer 19

Perfect. We're off to part number 3.

Evaluate: \(\int\limits_{0}^{4}32t\;dt\)

It had BETTER land somewhere between 192 and 320!

Answer 20

I'm not sure about this part. Also, can you explain that strange equation you posted?

Answer 21

i believe that means integral, but don't know beyond that

Answer 22

It's just an integral. You must find the anti-derivative of 32t and then evaluate that anti derivative at t = 4 and t = 0.

Answer 23

16t^2?

Answer 24

It's one of those funny "Fundamental Theorem" things.

If F(x) is an antiderivative of f(x), then \(\int\limits_{b}^{a}f(x)\;dx = F(a) - F(b)\), for suitable a and b.

You have \(x(t) = 16t^{2}\). That is exactly correct. Now, you need \(x(4) - x(0)\).

Answer 25

256

Answer 26

Did we just establish that \(192 \le \int\limits_{0}^{4}32t\;dt = 256 \le 320\)? Yes, I believe we did! I think we're done. Plenty to ponder with this exercise.

Past my bedtime! :-)

Answer 27

Thank you!