Two variables, x and y, are related by the equation

$\displaystyle y=\frac{3}{4}(\frac{x}{12}-1)^6.$

Given that both x and y vary with time, find the value of y when the rate of change of y is 12 times the rate of change of x.

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- Feb 28th 2010, 12:44 AMPunchConnected rate of change
Two variables, x and y, are related by the equation

$\displaystyle y=\frac{3}{4}(\frac{x}{12}-1)^6.$

Given that both x and y vary with time, find the value of y when the rate of change of y is 12 times the rate of change of x. - Feb 28th 2010, 04:38 AMskeeter
- Feb 28th 2010, 04:49 AMtom@ballooncalculus
And, just in case a picture helps fill out the chain rule...

http://www.ballooncalculus.org/asy/diffChain/rates.png

... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Starting then with...

http://www.ballooncalculus.org/asy/d...ratesRatio.png

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Feb 28th 2010, 06:02 AMskeeter
$\displaystyle y = \frac{3}{4}\left(\frac{x}{12} - 1\right)^6$

$\displaystyle \frac{dy}{dt} = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12} \cdot \frac{dx}{dt}$

$\displaystyle 12 \cdot \frac{dx}{dt} = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12} \cdot \frac{dx}{dt}$

assuming $\displaystyle \frac{dx}{dt} \ne 0$ ...

$\displaystyle 12 = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12}$

$\displaystyle 12 = \frac{3}{8}\left(\frac{x}{12} - 1\right)^5$

$\displaystyle 32 = \left(\frac{x}{12} - 1\right)^5$

$\displaystyle 2^5 = \left(\frac{x}{12} - 1\right)^5$

$\displaystyle x = 36$

calculate the value of y from the original equation. - Feb 28th 2010, 04:13 PMPunch
- Feb 28th 2010, 04:21 PMskeeter
if $\displaystyle \frac{dy}{dt} = 12 \cdot \frac{dx}{dt}$ , then $\displaystyle \frac{dy}{dx} = 12$ because ...

$\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx}$ - Mar 1st 2010, 02:39 AMPunch
I understood this part but... what I was trying to clarify was... From the sentence "find the value of y when the rate of change of y is 12 times the rate of change of x.", What would I be able to say from the sentence?