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 October 21st, 2012, 06:49 PM #1 Newbie   Joined: Oct 2012 Posts: 11 Thanks: 0 Change of rate I tried the dy/dt formula and steps, but I can't seem to get any of the answer choices. Question: a point's moving on a graph of 4x^(3)+6x^(3)=xy. When at Point ((1/10),(1/10)), the x coord is decreasing by 4 u/s... what's the speed of Y coordinate? Choices: 2,1,-2,-1 or 0 units/sec If you know a similar prob or anything, that would be great for help.
 October 21st, 2012, 07:20 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Change of rate I suspect you have mistyped the given curve. You want to differentiate with respect to time t, then solve for $\frac{dy}{dt}$ and plug in the given values for $x,\,y,\,\frac{dx}{dt}$. If you can correct the equation of the curve, I will be glad to help.
October 22nd, 2012, 01:55 PM   #3
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Re: Change of rate

Quote:
 Originally Posted by allylee I tried the dy/dt formula and steps, but I can't seem to get any of the answer choices Question:a points moving on a graph of 4x^(3)+6x^(3)=xy.
MarkFL suspects you have copied this incorrectly because it would be more simply written 10x^3= xy.
However, if x= 1/10, then you would, in fact, have 10(1/1000)= 1/100= (1/100)(1/100) so that may well be what is intended.

Quote:
 When at Point ((1/10),(1/10)), the x coord is decreasing by 4 u/s...whats the speed of Y coordinate Choices: 2,1,-2,-1 or 0 units/sec If you know a similar prob or anything, that would be great for help
For any y as a function of x, by the chain rule, $\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}$. So, whatever your formula is, differentiate with respect to x to find dy/dx.

 March 8th, 2013, 06:35 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 If the intended equation is 4x³ + 6y³ = xy, one gets -1 units/sec as the answer.

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