Related Rates Seesaw Problem: Finding Average Rate of Change

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Homework Help Overview

The problem involves a seesaw scenario where two children of different weights are positioned at varying distances from the center. The original poster seeks to find the average rate of change in distance for one child as their position changes from 1.5 meters to 2.5 meters, using a function that relates their distances based on their weights.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the quotient rule to derive a function but questions their understanding of the average rate of change. Some participants suggest verifying the use of given values in calculations, while others question the completeness of the problem statement.

Discussion Status

The discussion is ongoing, with participants offering guidance on checking calculations and suggesting that the original poster consult their professor for further clarification. There is no explicit consensus on the approach to take, and multiple interpretations of the problem are being explored.

Contextual Notes

There is mention of a previous post in another forum regarding the same problem, indicating potential confusion or missing information in the current formulation. The original poster expresses uncertainty about how to calculate the average rate of change.

stokes
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1. Homework Statement
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.


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

The prof gave us the answer which is -1.275. I haven't come close to that.

I used...

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


I don't know how to tackle the average rate of change.
 
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you have forgotten to use the given x value.

i.e.

34*x/40 = (34 * 1.5) / 40 = 1.275
 
Hi stokes,

Was the problem formulated as you have given? Are you sure something's not missing?
 
stokes, it would be best to take help from your professor.

And please let us know how he solved it.
 

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