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anonymous

  • 5 years ago

the surface z=(x,y) has this line tangent to it: x=2+2t, y=1, z=-2+3t at the point (2,1,-2). Find f_x(2,1)

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  1. anonymous
    • 5 years ago
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    What is f_x(2,1)

  2. anonymous
    • 5 years ago
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    Is it the derivative?

  3. nowhereman
    • 5 years ago
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    Do you mean the surface is \( (x,y, f(x,y))\) and \(f_x = \frac{∂f}{∂x}\)?

  4. anonymous
    • 5 years ago
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    i meant f sub x , yes I believe it has to do with partial derivatives

  5. nowhereman
    • 5 years ago
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    Then it is \(\frac{3}{2}\) because the tangent line is in x-direction.

  6. anonymous
    • 5 years ago
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    are you just taking coefficient in front of the t variable?

  7. nowhereman
    • 5 years ago
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    Yes.

  8. anonymous
    • 5 years ago
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    Okay, I'm still kind of lost as to the problem. I believe the first deriv the tangent line is 0 ? or something of that nature. I'm confused to what the answer is actually telling me.

  9. anonymous
    • 5 years ago
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    at the point\[F_x (2,1)\] the tangent line is 3,2?

  10. nowhereman
    • 5 years ago
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    You see, the function is \(f: ℝ×ℝ→ℝ\) and the tangent lines of the graph at a point \((x,y)\) are of the form \[(x + d_xt,y + d_yt,f(x,y) + f_x(x,y)d_xt + f_y(x,y)d_yt)\] where \((d_x, d_y)\) is the direction of the argument. Here you can see, that \(d_x = 2\) and \(d_y = 0\) so you have \[(x + 2t, y, f(x,y) + 2f_x(x,y)t)\] Then you only have to plug in the values for of the base coordinates and the function at that point.

  11. amistre64
    • 5 years ago
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    it gives you a point and a tangent line; determine the equation of the surface

  12. amistre64
    • 5 years ago
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    nowheres looks more technical :)

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