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Posted on 2014-10-19
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A particle p moves on the curve y = x pow(2) + 1 in such a way that dx / dt = 4.  If s is the distance of P from the point ( 0, 6 ), find ds / dt :  ( a ) in general as a function of x and ( b ) when P is at ( 2, 5).

Please explain how to get started with this problem.  

This is University calculus I  related rate problem.
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Question by:naseeam
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8 Comments
 
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Accepted Solution

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phoffric earned 1600 total points
ID: 40390942
To get started with this problem, draw a line from a point (x,y) on the curve to (0,6). Then, s = f(x,y), and you are given that y = g(x).
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by:naseeam
ID: 40391176
What do they mean by particle p moves on the curve y = x pow(2) + 1 in such a way that dx / dt = 4?
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LVL 84

Expert Comment

by:ozo
ID: 40391186
Solving
dx / dt = 4
we get
x = 4t +C
Given that, can you find y(t)?
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LVL 32

Assisted Solution

by:phoffric
phoffric earned 1600 total points
ID: 40391194
>> What do they mean by particle p moves on the curve y = x^2 + 1 in such a way that dx / dt = 4?
You have a parabola and a particle is moving on it in such a way that its speed in the x-direction is always 4 length-units/second.
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Author Comment

by:naseeam
ID: 40391244
dx / dt = 4
 we get
 x = 4t +C
 Given that, can you find y(t)?

I believe this is integration.  Students haven't learned integration in this class yet.
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LVL 32

Assisted Solution

by:phoffric
phoffric earned 1600 total points
ID: 40391269
You don't need integration to compute the derivative ds/dt.

I already mentioned that you can determine s = f(x,y).
And you know that the particle p moves on the curve y = x^2 + 1
So, y = g(x).

So, implicitly, s = h(x), right?
And dx/dt = 4 from your OP.

So, now you can take derivative of s --> ds/dt using what rule?
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LVL 84

Assisted Solution

by:ozo
ozo earned 400 total points
ID: 40391276
If you have x(t) and y(x), can you get y(t)?
From y(t) and x(t) can you get s(t)?
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LVL 1

Author Closing Comment

by:naseeam
ID: 40391903
Great answers!  Not too much help, not too little help.  Perfect balance!
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