# Related Rates SeeSaw.

1. Jan 6, 2008

### stokes

1. The problem statement, all variables and given/known data
A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on tghe opposite end of the seesaw. The functio B(x)= 34x / 40 gives the distance that the 40 kg child must sit from the center of the seesaw when the 34 kg child sits x meters from the center. The seesaw is 9m long find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters.

2. Relevant equations
I used the quotient rule to find the derivative of the function. Which turns out to be 68 / 80.

3. The attempt at a solution

I used the quotient rule to fine the instanenous rate of change which turns out to be B'(x)=
68/80.

I dont know how to takle the average rate of change.

Last edited: Jan 6, 2008
2. Jan 6, 2008

### olgranpappy

did you have a question?

3. Jan 6, 2008

### stokes

The seesaw is 9m long find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters.

4. Jan 6, 2008

### rocomath

Let's see some work, and this should be posted in the Calculus & Beyond subforum.

5. Jan 6, 2008

### olgranpappy

"The seesaw is 9m long find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters."

that's not a question.

6. Jan 6, 2008

### stokes

I used the quotient rule to fine the instanenous rate of change which turns out to be B'(x)=
68/80.

I dont know how to takle the average rate of change.

7. Jan 6, 2008

Sorry....

8. Jan 6, 2008

### olgranpappy

... that's okay, I guess. Two things:

1. 68/80 = 34/40 (you took a derivative of B with respect to x, not time...)

2. if you want to find an average rate, I think you might have to have a time somewhere in your problem. Was one given?

9. Jan 6, 2008

### stokes

No it did not... It says find the average rate of change in distance.

10. Jan 6, 2008

### olgranpappy

11. Jan 6, 2008

### stokes

Since it is over the interval of 1.5 and 2.5. Couldnt I find B(1.5) and B(2.5) and then find the average?

12. Jan 6, 2008

### stokes

I still did not get the answer... Im stuck...

13. Jan 6, 2008

### rocomath

You're looking for the avg., not the instantaneous rate.

$$Rate_{avg}=B(x)=\frac{B(x_2)-B(x_1)}{x_2 -x_1}$$

Last edited: Jan 6, 2008
14. Jan 6, 2008

### stokes

The thing is the derivative is 34/40. Which equals to 0.85.

The prof gave us the answer which is -1.275. I havent come close to that.

I've tried that formula before too...

I used...

f(x) = f(b) - f(a) / b-a

15. Jan 6, 2008

### stokes

16. Jan 6, 2008

### olgranpappy

I"m sorry, stokes. Your statement of the problem just doesn't make sense. I don't think anyone will be able to help you with this. Are you sure you are writing down all of the information in the problem exactly as given?

17. Jan 6, 2008

### stokes

I am really sure I am writing everything down correctly.

18. Jan 6, 2008

### olgranpappy

So the statement of the problem in your text, or on your homework begins just as you wrote it:

"A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on tghe opposite end of the seesaw. The functio B(x)= 34x / 40 gives the..."

Your text used the word "tghe"?

Your text used the word "functio"?

The first sentence seems to be missing words (perhaps an "and" in between "34 Kg" and "is")

Etc Etc Etc.

If you can't even take the time to write down the correct statement of the problem, then you should not expect to get much help here.

19. Jan 6, 2008

### stokes

A child weighs 34 Kg is seated on a seesaw. While a child who weighs 40 kg is situated on the opposite end of the seesaw. The function B(x)= 34x / 40 gives the distance that the 40 kg child must sit from the center of the seesaw when the 34 kg child sits x meters from the center. The seesaw is 9m long find the average rate of change in distance as the lighter childs distance changes from 1.5m to 2.5meters.

Im sorry I just reread my mistake.

20. Jan 6, 2008

### rocomath

Who cares, grow up.

Sorry no one is unable to help you at the moment stokes, just be patient and I'm sure someone may be able to.