http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 11 Jul 2014 08:51:04 -0500http://calc3.askbot.com/question/86/rate-of-change-of-potential/In section 14.5, question #6, it asks: Suppose the electric potential at $(x,y)$ is $ln\sqrt {(x^2+y^2)}$. Find the rate of change of the potential at $(3,4)$ toward the origin and also in a direction at a right angle to the direction toward the origin. I know how to calculate $D_uf$ but I am lost on what $u$ would be in this problem.Fri, 11 Jul 2014 07:14:49 -0500http://calc3.askbot.com/question/86/rate-of-change-of-potential/http://calc3.askbot.com/question/86/rate-of-change-of-potential/?answer=88#post-id-88This question confused me as well. The only thing I could think was that you could find a $\vec u$ that was perpendicular to $\langle 3,4 \rangle$ so you could do $\langle 4,-3\rangle$ or $\langle -4,3\rangle$. But when I used those I didn't come up with the same answer as the book. So hopefully someone will be able to clarify this for both of us. Thanks Christina! I need to go back and restudy how to do my derivatives apparently :(Fri, 11 Jul 2014 08:38:32 -0500http://calc3.askbot.com/question/86/rate-of-change-of-potential/?answer=88#post-id-88http://calc3.askbot.com/question/86/rate-of-change-of-potential/?answer=90#post-id-90To find the unit vector in this case remember you are starting at the point (3,4) and moving toward the origin. If you went 3 in the x direction and up 4 in the y direction from the origin to get to (3,4), to go back you go backward 3 in the x direction and down 4 in the y direction. Your direction vector would then be $\langle-3,-4\rangle$. Then you just normalize that by dividing by $\sqrt{(-3)^2+(-4^2)}$ which is 5. Using this unit vector will give you the correct answer. As far as the second part, think about what happens when you move perpendicular. No change, so the answer is zero. ADDED: in response to Tiffany....Not sure you are taking your derivatives correctly.Remember that the derivative of the natural log function is 1\argument of natural log multiplied by the derivative of the argument. (I am not going to show all those steps here) That would make your gradient vector $$ \nabla f = \langle \frac{x}{{(x^2+y^2)}}, \frac{y}{\{(x^2+y^2)}}\rangle$$ And at the point, we get $$ \nabla f(3,4) = \langle \frac{3}{\{(3^2+4^2)}}, \frac{4}{\{(3^2+4^2)}}\rangle$$ $$ = \langle \frac {3}{25} , \frac {4}{25}\rangle$$ Now if you calculate (don't forget to normalize the directional vector): $$ D_\vec u f = \langle \frac {3}{25} , \frac {4}{25}\rangle \cdot \langle \frac{-3}{5},\frac{-4}{5} \rangle$$ $$=\frac{-1}{5}$$ The 0 is from considering the rate of change perpendicular to the gradient, which would be zero. Hope this helps and have a great weekend!Fri, 11 Jul 2014 08:51:04 -0500http://calc3.askbot.com/question/86/rate-of-change-of-potential/?answer=90#post-id-90