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: 11,566 Thanks: 108 Math Focus: The calculus  Re: A moving point along a graph
You want to use: So, compute and use the given value of . Then plug in the given values for x. 
October 11th, 2012, 11:13 AM  #3 
Joined: Oct 2012 Posts: 6 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: 11,566 Thanks: 108 Math Focus: The calculus  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: 11,566 Thanks: 108 Math Focus: The calculus  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 
October 11th, 2012, 06:46 PM  #11 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,566 Thanks: 108 Math Focus: The calculus  Re: A moving point along a graph
Yes, those numbers are right. As a followup, can you find the point(s) at which dy/dt is changing the most rapidly? 
October 11th, 2012, 06:50 PM  #12 
Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: A moving point along a graph
Not sure. How? Do I divide nim by den? 
October 11th, 2012, 07:17 PM  #13 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,566 Thanks: 108 Math Focus: The calculus  Re: A moving point along a graph
Hint: you want to find where: Do you see why? Hey, I know you are loaded down with homework, so only if you get done with your assigned work and you feel like fooling with this... 

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