Question

Calculus Integration
Help please!

Answer 1

One sec

Answer 2

The answer that I got to #1 part A was 20 mph or 0.333 m/min.
For part B I got 6:10.
For part c I got 0.333 m/min.

Answer 3

@mathmale

Answer 4

Happy to see you back.
Generally, I like to see the actual work done by others before commenting on their results.
Since you use the word "integration," I assume you're approximating integration by using the trapezoidal rule (or a version thereof) to calculate the area under the curve formed by the trapezoids in question and then dividing by the equivalent of "40 minutes" in "hours."

Are any or all of these assumptions correct?

Answer 5

Yes I am using the trapezoid rule, and the review is that applications of integrals
would you like me to post what led me to the answers that I provided?

Answer 6

Only if you're really unsure of your results. What average speed did you arrive at in Part A?

Answer 7

I got 20 mph

Answer 8

as the average speed

Answer 9

I've done some quick calcs using the Trap Rule and have obtained an ave. speed of 21 2/3 mph.

Did you use the Trap Rule in Part A?

Answer 10

First I found the average of all those values then I tried again using the trap rule, they both were about 20 so I figured that was the true value

Answer 11

turns out that I made a simple arithmetic mistake, so you'll need to disregard my 21 2/3.

My gut feeling is that you really need to find the "area under the curve" between 6:00 and 6:40 a.m., and then divide the result by 40 minutes, to obtain the average.

Don't accept what I say here unless it makes sense to you.

Note that the Trap Rule is numeric approximation, not calculus.

Answer 12

The average value of a function is (area under the curve between a and b) / width of interval (b-a).

Answer 13

I'd definitely use the Trap Rule myself, but there are other methods as well. You could use the midpoint rule, for example.

Answer 14

\[\int\limits_{0}^{40}v(t)dx\]

Answer 15

Is that correct?

Answer 16

Yes, that's what we're trying to approximate...the actual area under the "speed" curve.
But you don't have a function of time here; all you have are a bunch of evenly spaced speed values. That dictates using 1) Trap Rule, or 2) Midpoint Rule or (3) Simpson's Rule. All are used for approximating areas under a curve between limits a and b.

Answer 17

\[\frac{ 17+20+22+21+17 }{ 5 }\]?

Answer 18

= 19.4

Answer 19

Looking for an appropriate way to phrase this: your result is a simple average, not a technique for approximating the area under a curve. No matter which method you use, yours, or the Trap Rule, or the Midpoint Rule, or Simpson's Method, you're going to get a slightly different result.

My preference would be that you choose and use one of the latter 3 rules, but if you'd like to stick with your 19.4, then estimate at what time or times the speed of the boat is 19.4 mph.

Answer 20

okay ill use the trap rule

Answer 21

Yes, you'd be safer doing that. Note that the problem statement clearly specifies that you "use trapezoids."

Answer 22

Using that I get 20 mph

Answer 23

so the average speed is reached at 6:10

Answer 24

I haven't actually tried graphing this situation myself, so can't vouch for your result. Did you mean to say that you've obtained just one result, and are stopping there, or is there another time at which the speed is = to or close to the average speed?

Answer 25

Also at 6:30

Answer 26

Yes, I did expect there'd be 2 diff times at which the actual speed was near to the ave. speed.

Answer 27

How would you attack Part C?

Answer 28

for c after 40 minutes wouldn't it have traveled 20 miles

Answer 29

oh wait it would be less than that

Answer 30

Again, I'd much appreciate your sharing your work and the reasoning behind it.

Answer 31

distance = (average speed) * (elapsed time)

Answer 32

40 minutes = (2/3) hour

Answer 33

\[d=20*2/3?\]

Answer 34

Assuming that you believe your average speed is 20 mph, then yes, (20 mph) * (2/3 hr) is the approx. distance.

Answer 35

okay so 13.333

Answer 36

Think: how could you make this response a bit more sensible? What does 13.3 measure? Does an approximation warrant 3 decimal place accuracy?

Answer 37

The distance traveled by the bottle is 13.3 miles.

Answer 38

Better. Or, "approx. 13 miles."

Answer 39

I there a formula or something for average acceleration, or is that the derivative of the velocity function?

Answer 40

Acceleration is definitely the derivative of the velocity function (which we don't have).
If velocity increases over a given time interval, the acc'n is pos.; if decreases, neg.
I'm not so confident about answering this question.

Answer 41

okay

Answer 42

The data given you definitely does show changes in the velocity.

You could focus on the CHANGES in the velocity, dividing them by time. I'd have to spend more time thinking about this before deciding what units time should have in this case: minutes? Hours? seconds?

Answer 43

With that said, would you mind focusing on the next couple of problems and letting me know if you have any questions? If at all possible, please graph this new situation and share your graph with me.

Answer 44

\[\frac{ \Delta v }{ \Delta t }\] is this a way of solving it?

Answer 45

Yes, that's an approx. approach; a more exact approach would be the 2nd derivative, but we don't have any specific function to differentiate.

Answer 46

Oh okay

Answer 47

Thank you for the help sir I have to go help paint at my church soon so I have to go

Answer 48

What a kind and thoughtful thing to do. Have a great day. See you again.