Directional Derivative

  • Thread starter jegues
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Homework Statement



Find the rate of change of the function,

[tex]f(x,y,z) = cos(\pi x y) + xln(z^{2}+1),[/tex]

with respect to length [tex]s[/tex] along the curve,

[tex]y=-3x, z=x^{2}-y^{2}+9,[/tex]

directed so that x increases, at the point (-1,3,1).

Homework Equations





The Attempt at a Solution



See figure attached.

I wrote parametric equations for the curve, and threw them into a vector.

Since at the point (-1,3,1) t=-1, I evaluated the vector accordingly.

The question says "so that x increase" so I reversed the direction of the vector by multiplying it by negative 1.

After this I found a unit vector that points in the same direction as the vector described previously.

Then I found the gradient of f and evaluated it at the point (-1,3,1).

After this I dotted the two to get the rate of change of f in the desired direction.

The solution lists the answer as,

[tex]\frac{ln(2) - 16}{\sqrt{266}}[/tex]

Where did I go wrong?
 

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Answers and Replies

  • #2
vela
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You want the vector tangent to the curve, which is the velocity. You have to differentiate with respect to t to get the velocity vector before setting t=-1.
 
  • #3
1,097
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You want the vector tangent to the curve, which is the velocity. You have to differentiate with respect to t to get the velocity vector before setting t=-1.

Thank you vela, that fixed everything.
 

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