Related Rates Seesaw Problem: Finding 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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