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 October 11th, 2012, 09:06 AM #1 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 A moving point along a graph A point is moving along the graph of y = 1/(1+x^(2)) such that dx/dt is 2 inches per second. Find dy/dt for (a) x = -2, (b) x = 0, (c) x = 1 and (d) x = 3 If someone could show me how to set this problem up and solve for one of these above x's, I could take notes and to the rest. Thanks!
 October 11th, 2012, 09:28 AM #2 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: A moving point along a graph You want to use: $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$ So, compute $\frac{dy}{dx}$ and use the given value of $\frac{dx}{dt}$. Then plug in the given values for x.
 October 11th, 2012, 11:13 AM #3 Newbie   Joined: Oct 2012 Posts: 7 Thanks: 0 Re: A moving point along a graph Go Parametrics! Don't forget that reciprocals can be useful too!
 October 11th, 2012, 02:22 PM #4 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: A moving point along a graph Mark The implicit differentiation lingo is happening very slow for me. I know first to take the derV. In doing so I get -2x/(1+x)^(2) = dy/dt but I don't know how to proceed from here
 October 11th, 2012, 03:07 PM #5 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: A moving point along a graph Your expression for dy/dx isn't quite right. You want x to be squared within the parentheses.
 October 11th, 2012, 03:19 PM #6 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: A moving point along a graph Oh yes. OK I fixed that. Now what?
 October 11th, 2012, 03:23 PM #7 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: A moving point along a graph Multiply both sides by dx/dt, then by the chain rule, the left side is dy/dt, which is what you want. You are given values for x and dx/dt, so plug and chug.
 October 11th, 2012, 03:32 PM #8 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: A moving point along a graph Show me Mark. I will then frame it and hang it on my f-----! Wall as a master key
 October 11th, 2012, 03:38 PM #9 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: A moving point along a graph At x =10 I'm getting... -2x(DX/DT)/(1+x^(2)) ^(2) = -2(10)(2)/(1+10^(2))^(2) = -40/10201 inches per sec
 October 11th, 2012, 05:00 PM #10 Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: A moving point along a graph Is this right Mark? dy/dx =(-2x DX/dt)/(1+x^(2))^(2) for x = -2. 8/25 in per sec for x = 0. 0 in/sec for x = 6. -24/1369 in/sec for x = 10. -40/10201 in/sec

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a point is moving along the graph of the function y=1/(1 x^2)

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