Rate of change of potential

Anonymous — Fri, 11 Jul 2014 07:14:49 -0500

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.

Find the direction

Anonymous — Fri, 11 Jul 2014 07:23:36 -0500

I am stumped on how to begin question #16 in section 14.5: Find the directions in which the directional derivative of $f(x,y)=x^2+sin(xy)$ at the point $(1,0)$ has the value 1. I'm sure $<0,1>$ would work but I don't see how to find all other directions.

Finding the cosine of the angle between two curves?

Tiffany — Mon, 30 Jun 2014 14:42:31 -0500

So, I've been trying to work on the homework 13.2, and I'm stuck on both questions 6 and 7. But we'll start with question 6. "6. Find the cosine of the angle between the curves $\langle0,t^2,t\rangle$ and $\langle\cos(\frac{\pi t}{2}),\sin(\frac{\pi t}{2}),t\rangle$ where they intersect." Well I can tell that they intersect at t=1 And from the book on page 336 it states... " $ \cos(\theta) $ = $\mathbf {\frac{\vec r' * \vec s'}{|\vec r'||\vec s'|}}$ = $ \mathbf {\frac {\vec r'}{|\vec r'|} \cdot \frac{\vec s'}{|\vec s'|}} $ " So letting $\vec r$ = $\langle0,t^2,t\rangle$ and $\vec s$ = $\langle \cos(\frac{\pi t}{2}), \sin(\frac{\pi t}{2}),t\rangle$ I end up with $\vec r'$ = $\langle 0,2t,1\rangle$ and $\vec s' $ = $\langle (\frac{-1}{2} \pi \sin(\frac{\pi t}{2})), (\frac 12 \pi \cos(\frac{\pi t}{2})), 1\rangle $ using this to come up with | $ \vec r'$| I get |$\vec r'$| = $\sqrt{0^2 + (2t)^2 + 1^2} $ = $\sqrt{4t^2 +1} $ and |$\vec s'$|= $\sqrt{(\frac{-1}{2}\pi \sin(\frac {\pi t}{2}))^2 + (\frac 12 \pi \cos(\frac{\pi t}{2}))^2 + 1^2) } $ Now when we plug this into the $\cos(\theta)$ equation, do we substitute t=1 for the t in all of the equations to find the answer, or how exactly are we supposed to approach this problem? I end up with this ridiculous answer with a whole bunch of randomness in it that doesn't match up with the back of the book, so I'm beyond confused on this problem, and feel like there is a much easier way than how i'm trying to figure it out. HELLLPPPP! :) *Comment*: When you take the norm of $\vec{s}'$, note that you get a $\sin^2+\cos^2$. That yields a significant simplificatoin that might help.

Mark's Calc III - Individual question feed

Rate of change of potential

Find the direction

Finding the cosine of the angle between two curves?