A moving point along a graph

[mathkid]

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!

[MarkFL]

You want to use:



So, compute and use the given value of . Then plug in the given values for x.

[simons545]

Go Parametrics!
Don't forget that reciprocals can be useful too!

[mathkid]

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

[MarkFL]

Your expression for dy/dx isn't quite right. You want x to be squared within the parentheses.

[mathkid]

Oh yes. OK I fixed that. Now what?

[MarkFL]

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.

[mathkid]

Show me Mark. I will then frame it and hang it on my f-----! Wall as a master key

[mathkid]

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

[mathkid]

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

[MarkFL]

Yes, those numbers are right.

As a follow-up, can you find the point(s) at which dy/dt is changing the most rapidly?

[mathkid]

Not sure. How?

Do I divide nim by den?

[MarkFL]

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...