Difficult partial derivative of a log

[dorikin]

Hello!

I am trying to solve the partial derivative 'P' http://www.flickr.com/photos/61865210@N07/5757168138/ ,

which is part of a larger equation:
http://www.flickr.com/photos/61865210@N07/5757300018/
(Sorry, can't seem to display to pictures, using insert image)

Someone told me that solving P isn't possible but it's from a book with answers to an example (no worked solution).

I've tried to solve 'q' but I don't get the same answer as the book. Can someone tell me if it is possible or not? and if it is, check my working out.

If you need more context about the equation I have attached some pages of the book (PFM.pdf) that I got the equation from. The equation is on page 109, eq.4.54 and I'm trying to do example 4.4 on page 111.

Also, this book got the equations from: Real-Time Obstacle Avoidance Using Harmonic Potential Functions.pdf

Thank you for your help in advance!

ken

Homework Statement

x=0.5
y=0.86397
c=1
d=4

Homework Equations

P=
http://www.flickr.com/photos/61865210@N07/5757168138/

where R=sqrt((x-c)^2+(y-d)^2)
and n is the normal unit vector, ( -sqrt(3)/2 , 1/2)

The Attempt at a Solution

P=d/dn f(x,y)= n . grad f

f(x,y)=ln(R)=ln( [ (x-c)^2+(y-d)^2 ]^0.5 )

Using chain rule:

grad ln(R) = 1/R * 0.5*[ (x-c)^2+(y-d)^2 ]^-0.5 *(2x-2c) i + 1/R * 0.5*[ (x-c)^2+(y-d)^2 ]^-0.5 *(2y-2d) j

cancel down and sub in variables:
grad ln(R) = -0.3149 i - 1.975 j

P=d/dn f(x,y)= n . grad f

P=-sqrt(3)/2*-0.3149 +0.5*-1.975
P= -0.7148

 

[hunt_mat]

In order to find a normal derivative you will need to use a surface of some kind, as the normal in question will be the normal to that you that the normal derivative from.