Rate of change from one point towards another

  • Thread starter Punkyc7
  • Start date
  • #1
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Suppose z=e^(xy+x-y). How fast is z changing when we move from the origin towards the point (2,1)?



Would it just be 1/sqrt((2^2+1))=1/sqrt(5), i am asking this because does the z=e^(xy+x-y) get used at all?
 

Answers and Replies

  • #2
1,255
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The function is what entirely determines the answer.
1/sqrt(5) is how much (x,y) is changing----they're asking for how fast z is changing.

Anytime you're looking for a rate of change, you should be thinking about 'derivatives'. In this case, because they're asking about a particular direction of motion, you should be thinking about 'directional derivatives'.
 
  • #3
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So e^(xy+x-y)
<y+1(e^(xy+x-y)), x-1(e^(xy+x-y))>

would you use the starting point or the ending point?
 
  • #4
1,255
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Great start Punkyc.

So e^(xy+x-y)
<y+1(e^(xy+x-y)), x-1(e^(xy+x-y))>
You have the gradient there, good; thats the first part of finding the http://en.wikipedia.org/wiki/Directional_derivative" [Broken].
The key is taking the dot product of that, with a unit vector in the direction of interest.

would you use the starting point or the ending point?
Look at the wording of the question carefully. You're not actually moving from the origin to (1,2), but you are at the origin moving in that direction.
 
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  • #5
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So it woud be <1,-1> dot (1/sqrt(5)<1,2>

so we have -1/(sqrt(5))
 
  • #6
1,255
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Yup! I think that's it.
Keep in mind: even though this is almost the exact answer you guessed in the first place, that's just a coincidence, and the process is important.
 
  • #7
420
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Thanks for your help
 

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