http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 10 Jul 2014 12:58:28 -0500http://calc3.askbot.com/question/82/quiz-preparation-mystery-question/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?SpaceManSpiffThu, 10 Jul 2014 12:58:28 -0500http://calc3.askbot.com/question/82/http://calc3.askbot.com/question/56/showing-perpendicularity-at-every-point/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! JustinWed, 02 Jul 2014 14:33:18 -0500http://calc3.askbot.com/question/56/