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Quiz preparation Mystery question

asked 2014-07-10 12:58:28 -0600

SpaceManSpiff gravatar image

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?

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answered 2014-07-10 21:09:16 -0600

Dylan gravatar image

updated 2014-07-10 21:34:14 -0600

I think that it's possible we could see a component-wise proof on the gradient, such as the one we saw on the worksheet today. (Show that $ \nabla \left[ \sin(f(x,y))\right] = \cos(f(x,y)) \nabla \left[f(x,y)\right]$.)

There's really a lot of possibilities with that; abstract stuff and cases with specific functions. I think that I want to be fully comfortable with doing any component-wise proof with the gradient for tomorrow.

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answered 2014-07-10 21:32:19 -0600

SpaceManSpiff gravatar image

How did you get through that one by the way?

I ended up getting to it like this.

$$ \nabla (\cos(f(x,y)))= \langle \frac{ \partial }{\partial x} \sin(f(x,y)), \frac{\partial}{\partial y} \sin(f(x,y)) \rangle$$ $$= \langle f_{x} \cos(x,y), f_{y} \cos(x,y) \rangle$$ $$=\cos(f(x,y)) \langle f_{x},f_{y} \rangle$$ $$=\cos(f(x,y) \nabla f(x,y)$$

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Asked: 2014-07-10 12:58:28 -0600

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Last updated: Jul 10 '14