Mark's Calc III - Individual question feed

Partial derivative

SpaceManSpiff — Thu, 17 Jul 2014 12:21:25 -0500

So with the function $$f(x,y)=\frac{x^{2}-y}{y^{2}-x}$$ I don't like the quotient rule because it seems like it makes way more sense to just find$$\frac{\partial}{\partial x}(x^{2}-y)(y^{2}-x)^{-1}$$$$=2x(y^{2}-x)+(-1)(y^{2}-x)^{-2}(-1)$$ $$=2x(y^{2}-x)+(y^{2}-x)^{-2}$$ But wolfram alpha is getting $$\frac{x^{2} -2xy^{2}+y}{(x-y^{2})^{2}}$$ I've tried jumping through a bunch of algebra hoops to see if my answer can be simplified to its but I can't pull it off. Is my answer somehow wrong or is Wolfram doing some weird simplification that I can't do?

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.

Quiz preparation Mystery question

SpaceManSpiff — Thu, 10 Jul 2014 12:58:28 -0500

So Mark listed a bunch of things that would be on the quiz and one of them was a mystery question. I recommend that we all use this question thread to post about different topics from this week that we think could be the mystery question. I imagine Mark didn't tell us what the topic would be precisely to start this kind of discussion. So I'll start by saying I think it will be a directional derivative question and giving a short summary of that topic and everyone else can leave answers to this post on summaries of things they think it could be. So the directional derivative is used to find the instantaneous rate of change of a given vector. I'll review this by going over a derivation for $D_{ \textbf{u}} f= \nabla f \cdot \textbf{u}$ let $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ and let $ \textbf{u}= \langle a,b \rangle $ such that $a^{2}+b^{2}=1$. That is to say $ \textbf{u}$ is a unit vector. Then $h \textbf{u}= \langle ha,hb \rangle$ and represents a step of size h in the $ \textbf{u}$ direction. So a line in the $x,y$ plane through the point $(x_{0},y_{0})$ can be parametrized as $ \textbf{p}(h)= \langle x_{0},y_{0} \rangle +\langle ah,bh \rangle$ according to the formula $ \textbf{p}(h)= \textbf{p}_{0}+ \textbf{d}h$ This simplifies to $ \textbf{p}(h)= \langle x_{0}+ah, y_{0}+bh \rangle$ So now we can say $g(x)=f(x,y)$ where $x=x_{0}+ah$ and $y=y_{0}+bh$ So $$g'(x)= \frac{df}{dt}= \frac{df}{dx} \frac{dx}{dt}+ \frac{df}{dy} \frac{dy}{dt}$$ By the multivariable chain rule. $$= \frac{df}{dx}u_{1}+\frac{df}{dy}u_{2}$$ $$=f_{x}u_{1}+f_{y}u_{2}$$ $$= \langle f_{x},f_{y} \rangle \cdot \langle u_{1},u_{2} \rangle$$ $$= \nabla f \cdot \textbf{u}$$ So what are you're guys suspicions?

How to find mixed partial derivatives

Anonymous — Tue, 08 Jul 2014 10:02:22 -0500

On the in class worksheet, question number five asks to show by example that the mixed partial derivatives of f(x,y)=$x^3y^4$ are equal. I understand how to take the partial derivatives, but what is meant by *mixed* partial derivatives?

Showing perpendicularity at every point?

Justin — Wed, 02 Jul 2014 14:33:18 -0500

This is from a few days ago, but I am curious how one would prove this. Exercise 13.2.8 reads: > Suppose that $\left|\textbf{r}(t) \right| = k$, for some constant $k$. This means that $ \textbf{r} $ describes some path on the sphere of radius $k$ with center at the origin. Show that $ \textbf{r} $ is perpendicular to $\textbf{r}'$ at every point. Hint: Use Theorem 13.2.5, part (d). Theorem 13.2.5 part (d) states: > $$\frac{d}{dt} (\textbf{r}(t) \cdot \textbf{s}(t)) = \textbf{r}'(t) \cdot \textbf{s}(t) + \textbf{r}(t) \cdot \textbf{s}'(t) $$ I'm just not sure where to begin. If someone could give me a starting point, I would really appreciate it!

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.