Question

[CALCULUS III—DIRECTIONAL DERIVATIVES] Given f(x, y), find the directional derivative of f(x, y) at point (x, y) along direction v = i - j. figure inside

Answer 1

Given:\[z=x\ sin(y/x)\]
Find the direction derivative of z at (1, pi/2) along direction v = i - j.

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx

Answer 3

First find the unit vector \(\vec{u}\)

Answer 4

attachment: definition of a directional derivative

\[D_uf(x,y)=\lim_{h\rightarrow 0}\frac{f(x+ah,\ y+bh)-f(x,y)}{h}\]

Answer 5

Then find the gradient, \(\nabla f(x,y)\)

Answer 6

That's not a unit vector.

Answer 7

\[v=<-1,1>\]\[m_v=\sqrt{2}\]\[u_v=<-\sqrt{2}/2,\sqrt{2}/2>\]

Answer 8

What's the gradient?

Answer 9

I do not recall what a gradient is, but following the procedure described in the definition of a directional derivative, I am going to proceed to evaluate the lim by solving f(x+ah, y+bh) - f(x,y)

Answer 10

That is not going to be an easy limit.

Answer 11

And no, that limit will be the directional derivative, it will not be the gradient.

Answer 12

http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx

Answer 13

\[
\nabla z(x,y) = \left<\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right>
\]

Answer 14

It's not javascript. It's MathJax using latex. The latex is \nabla

Answer 15

\[\nabla z(x,y)=<\frac{-y\ cos(y/x)+x\ sin(y/x)}{x},cos(y/x)>\]

Answer 16

So they want \[
D_{\vec{u}}\;z(x,y) = \nabla z(x,y) \cdot \vec{u}
\]

Answer 17

Let me set up before I proceed to evaluation

Answer 18

Would not the dot product of those two vectors provide a scalar

Answer 19

Result of a dot product should be scalar function.

Answer 20

\[D_u^{\rightarrow}z(x,y)=<\frac{-y\ cos(y/x)+x\ sin(y/x)}{x},\ cos(y/x)>\]\[⋅<\sqrt2/2,- \sqrt2/2>\]

Answer 21

are you also checking these evaluations with me or solely providing procedural guidance?

Answer 22

Ummm, so far it looks good. I can try to check

Answer 23

http://www.wolframalpha.com/input/?i=%5Bgrad+x+sin%28y%2Fx%29+%5D+dot+%5B2%2Fsqrt%282%29%3B+2%2Fsqrt%282%29%5D

Answer 24

\[D_u^{\rightarrow}z(x,y)=\sqrt2\frac{(-y\ cos(y/x)+x\ sin(y/x)-x\ cos(y/x))}{2x}\]

Answer 25

Ok I missed a sign apparently on the x cos (y/x)

Your input was 2/sqrt2 instead of sqrt(2) / 2

adjusted: http://www.wolframalpha.com/input/?i= [grad+x+sin%28y%2Fx%29+]+dot+[sqrt%282%29%2F2%3B+sqrt%282%29%2F2]

Answer 26

The unit vector I set up was +, - but it should be -, +

Answer 27

ok

Answer 28

I see my mistakes. Thank you for your guidance, you were helpful.

Answer 29

final answer (see attachment)