Directional Derivative

  • Thread starter Punkyc7
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  • #1
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find the directional derivative of z=2x^2-y^3 at (1,1)

is it just <4,-3>
 

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  • #2
Dick
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That's the gradient. It's not a directional derivative. You can use the gradient to find directional derivatives, but it's not one by itself.
 
  • #3
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so then how would you go about finding it because where not comparing it with another point.... is it just 1
 
  • #4
Dick
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so then how would you go about finding it because where not comparing it with another point.... is it just 1

The directional derivative at (1,1) is the derivative of f(x,y) in some direction. You need to specify the direction to find the directional derivative. Suppose I told you the direction is <u,v>. What's the directional derivative in that direction?
 
  • #5
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u-1, v-1 and you would dot that with our gradient
 
  • #6
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we would have to make those unit vector though
 
  • #7
Dick
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u-1, v-1 and you would dot that with our gradient

Ok, if (u,v) is a point and you want the directional derivative in the direction which is the difference between (1,1) and (u,v), then sure, it's <u-1,v-1>.<4,-3>. If you are just given the direction <u,v>, I'd say it's <u,v>.<4,-3>. Since they didn't give you a direction I'm not sure what they are asking.
 
  • #8
Dick
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we would have to make those unit vector though

If that's your definition of directional derivative, then absolutely, normalize them. The point is that all you can say is that it is the dot product of the gradient with the direction.
 

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