anonymous
  • anonymous
Multivariable Calculus question It's part b)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
Do I need like... an unit vector of some sort?
anonymous
  • anonymous
Isn't it the case that \( \nabla f(p) \cdot \mathbf v = D_\mathbf v f(p) \)?

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anonymous
  • anonymous
All I know is that: |dw:1378076224262:dw|
anonymous
  • anonymous
I have never seen that formula before.
anonymous
  • anonymous
Well remember that \[ \nabla f(p) = \]And in this case: \[ \mathbf v = \]
anonymous
  • anonymous
Except we're doing 2D vectors not 3D ones...
anonymous
  • anonymous
Right. Yeah whoops :P .
anonymous
  • anonymous
Disregard the z then :P .
anonymous
  • anonymous
They managed to get a system of equations somehow. Confused about that.
anonymous
  • anonymous
How about this: \[ \mathbf v(\theta) = <\cos(\theta), \sin(\theta)> \]Then solve for any value of \(\theta\) where\[ \nabla f\cdot \mathbf v (\theta) = 1 \]
anonymous
  • anonymous
My way sort of avoids systems of equations by using polar coordinates... If you wanna figure out how they got a system, then let me think for a second.
anonymous
  • anonymous
Well that works too but how do you isolate the dot product term then?
anonymous
  • anonymous
Well here is their work. It makes no sense to me though.
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anonymous
  • anonymous
First of all, can you tell me what you got for a)?
anonymous
  • anonymous
<2,1>
anonymous
  • anonymous
That is correct.
anonymous
  • anonymous
Okay, so what would \( \nabla f\cdot \mathbf v \) be if \(\mathbf v = \)?
anonymous
  • anonymous
Well woudn't that just be fx and fy evaluated at (1,0) ?
anonymous
  • anonymous
It would just be \( <2,1> \cdot \)
anonymous
  • anonymous
\(=2x+y\)
anonymous
  • anonymous
Thus \[ \nabla f\cdot \mathbf v = 1 \iff 2x+y = 1 \]
anonymous
  • anonymous
I see! :) .
anonymous
  • anonymous
What about the second one? That's more confusing.
anonymous
  • anonymous
Nvm, I think I see it. Because to find an unit vector, you need a magnitude/norm. That's the second one.
anonymous
  • anonymous
LIkewise: \[ \|\mathbf v\| = \sqrt{\mathbf v\cdot \mathbf v} = \sqrt{x^2+y^2} \]So \[ \|\mathbf v\| = 1 \iff \sqrt{x^2+y^2} = 1 \]They squared both sides: \[ \implies x^2+y^2 = 1^2 \iff x^2+y^2 = 1 \]
anonymous
  • anonymous
I see it :) . Thanks a lot!!
anonymous
  • anonymous
Ya. I think that makes much more sense now :) .

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