Rate of change of potential

answered 2014-07-11 08:51:04 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

updated 2014-07-11 13:00:12 -0600

To 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!