Calculate directional deriv of f(x,y) = cos^-1 (xy)

[jwpaine]

Calculate the directional derivative of f(x,y) = cos^-1 (xy) in the direction v = i - 2j, at the point P = (0.5, 0.5)

What am I doing wrong here?

f(x,y) = cos^-1 (xy)

First computing the gradient at P = (0.5, 0.5)

gradF = <d/dx f(x,y) , d/dy f(x,y)> = < -y / sqrt(1 + x^2*y^2) , -x / sqrt(1 + x^2 * y^2) >
gradF_P = gradF_(0.5, 0.5) = < -0.5 / sqrt(1 + (0.5)^2*(0.5)^2) , -0.5 / sqrt(1 + (0.5)^2*(0.5)^2)> = <-2/sqrt(17) , -2/sqrt(17) >
DvF(P) = GradF_P * v = <-2/sqrt(17) , -2/sqrt(17) > * <1, -2> = -2/sqrt(17) + 4/sqrt(17) > = 2sqrt(17) / 17

which is NOT correct.

What am I doing wrong, here?

 

[galactus]

Re: Calculate directional deriv

Hey JW:

Since the vector v is not a unit vector we have to normalize it.

\(\displaystyle u=\frac{V}{||v||}=\frac{1}{\sqrt{5}}\cdot (i-2j)=\frac{1}{\sqrt{5}}i-\frac{2}{\sqrt{5}}j\)

From which we get \(\displaystyle u_{1}=\frac{1}{\sqrt{5}}, \;\ u_{2}=\frac{-2}{\sqrt{5}}\)

The partials of f are \(\displaystyle f_{x}(x,y)=\frac{-y}{\sqrt{1-x^{2}y^{2}}}\)

\(\displaystyle f_{y}(x,y)=\frac{-x}{\sqrt{1-x^{2}y^{2}}}\)

We have \(\displaystyle f_{x}(\frac{1}{2},\frac{1}{2})=\frac{-2}{\sqrt{15}}\)

\(\displaystyle f_{y}(\frac{1}{2},\frac{1}{2})=\frac{-2}{\sqrt{15}}}\)

So, from \(\displaystyle D_{u}f(\frac{1}{2},\frac{1}{2})=f_{x}(\frac{1}{2},\frac{1}{2})u_{1}+f_{y}(\frac{1}{2},\frac{1}{2})u_{2}=\boxed{\frac{2\sqrt{3}}{15}}\)

Check my work. It's late.

 

[jwpaine]

Re: Calculate directional deriv

Thanks a lot for getting back to me.

Sorry about not using LaTeX.

I'll review this, first thing tomorrow. Exhausted.

Cheers,
JWP