## A moving point along a graph

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### 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!
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mathkid

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### 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.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### Re: A moving point along a graph

Go Parametrics!
Don't forget that reciprocals can be useful too!
simons545
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### 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
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mathkid

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### 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.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### Re: A moving point along a graph

Oh yes. OK I fixed that. Now what?
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mathkid

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### 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.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### 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
“If you can't explain it simply, you don't understand it well enough”

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mathkid

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### 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
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mathkid

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### 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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mathkid

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### Re: A moving point along a graph

Yes, those numbers are right.

As a follow-up, can you find the point(s) at which dy/dt is changing the most rapidly?
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### Re: A moving point along a graph

Not sure. How?

Do I divide nim by den?
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mathkid

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### Re: A moving point along a graph

Hint: you want to find where:

$\frac{d^2y}{dx^2}=0$

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...
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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