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Please answer all the questions (and in a neat format). Thanks! (see photo) Need all the answers at a glance.

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#15 For \(f_x\) we keep \(y\) fixed and see how the function changes as we change \(x\) so simply look at rows
yes! we have to apply this formula: \[\Large {f_x}\left( {{x_0},{y_0}} \right) \cong \frac{{f\left( {{x_0} + \Delta x,{y_0}} \right) - f\left( {{x_0},{y_0}} \right)}}{{\Delta x}}\]

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In each row, as \(x\) is increasing, notice that the function is decreasing. |dw:1435033219984:dw| In light of that, what can you say about the sign of \(f_x\) ?
The sign is positive, but decreasing....?
nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?
Somewhat, yes, but muddy.
First derivative at a particular point on the curve gives the slope of tangent line at that point |dw:1435034181715:dw|
Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing" |dw:1435034301743:dw|
Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing" |dw:1435034404265:dw|
That means the sign of first derivative can be used to know if a function is increasing/decreasing : \(\large f'(c)~\gt~0~~\iff~~\text{f(x) is increasing at x=c}\) \(\large f'(c)~\lt~0~~\iff~~\text{f(x) is decreasing at x=c}\)
In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.
Sorry, I don't mean to rush you, but if I may, I just want to get straight to the point, just because of the interest of time. I have a final tomorrow and a lot of these concepts are not so clear to me, and I know that, but I'm trying to save time, so I can study all of them, even though I don't have a profound understanding (just because of time, like I said). I just wanted to ask, I am familiar with the zero-gradient, but not when I have to use a contour plot, without any equation to derive it. How would this work?
Your help's been very generous, @ganeshie8 . Sorry if I sound a little blunt. I'm just rushing to study as much as I can, because I have a final in less than 15 hours.
I can understand, but there is no way to proceed further when you don't remember the relation between first derivative and increasing/decreasing functions. The question is asking exactly that. You cannot skip
OOOPS. Sorry, I was referring to another problem, when I said "zero-gradient". So that wouldn't make sense here.
Okay, I'm just totally confused now. Ughhhh. I've done this before, but the table just confuses me a lot. I don't know where to look.
Just pick any one row and forget about everything else.
I'm doing two problems at once.
Sorry for the mess.
thats okay :)
Thanks for understanding.... I'm just so stressed right now.
Just answer me this, in the below highlighted row, what do you notice about the values ? |dw:1435035389370:dw|
Decreasing, as we go from left to right.
Yes. so \(f_x(-2,~0)~~\lt ~0 \) |dw:1435035518143:dw|
Next, pick any one column and tell me what do you notice about the values as you go from bottom to top |dw:1435035560839:dw|
Increasing....
good, so \(f_y(-2,~1)\gt 0\) |dw:1435035643341:dw|
Ohhhhhhh I get the pattern.
is f_xx true, then?
and f_yy
sry i don't seem to remember, il review quick and get back meantime try other problems maybe..
Okay!
Thanks!
I'm thinking of working the first differences but not entirely sure
we can try to write the taylor series up to the second derivative of f, with for example y= const @ganeshie8
like this: \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)x + \frac{{{x^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]
oops.. \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\Delta x + \frac{{{{\left( {\Delta x} \right)}^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]
Ahh so the finite differences actually works \(f_{yy}(-2,1)~= 0\) because the first differences are constant along a column : |dw:1435036964859:dw|
|dw:1435037234174:dw|
If it helps, the answer is letter B.
For number 15. I just don't have the solutions.
Answers to the other numbers: 15) B 16) A 17) E 18) A 19) B
Yeah we cannot tell anything about mixed partials i guess so B makes sense to me
|dw:1435037903299:dw|
we have checked four conditions, so I think option D. Am I right?
namely: the first and fourth conditions are false; the second and the third conditions are true
and we can not infer anything on the fifth condition
hmm yeah since the first and fourth conditions are false we may say that only two inferences are true
#16 should be easy, just approximate the partials at (-2, 1) and take the magnitude \[\large \|\nabla f\| = \sqrt{{f_x}^2+{f_y}^2}\]
|dw:1435038498718:dw|
http://www.wolframalpha.com/input/?i=sqrt%28%28%2832.41-33%29%2F%280.1%29%29%5E2%2B%28%2833.9-33%29%2F%280.1%29%29%5E2%29
Thanks! How were those numbers selected? Was it through the symmetric difference quotient?
Ohh no, i guess we should use symmetric difference quotient then
Ohhh yeah, I just am confused on how to plug the values in :P.
*am just
Yes symmetric difference quotient also gives the same answer http://www.wolframalpha.com/input/?i=sqrt%28%28%2832.41-33.61%29%2F%282*0.1%29%29%5E2%2B%28%2833.9-32.1%29%2F%282*0.1%29%29%5E2%29
Nice! Thanks!
#17 just follows from #16
direction is \(\langle -6, ~9\rangle \) http://www.wolframalpha.com/input/?i=%3C%2832.41-33.61%29%2F%282*0.1%29%2C+%2833.9-32.1%29%2F%282*0.1%29%3E
Notice that the direction \(\langle -6, ~9\rangle \) is same as \(k\langle -6, ~9\rangle \) let \(k=\frac{1}{3}\) and we get direction = \(\langle -2,~3 \rangle \)
I see!
Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.
It really was just a matter of where to look, on the chart....
How do we do number 18? I'm used to solving for it with the equation given.
Do we divide <-2, 3> by \[\sqrt{(-2)^2+(3)^2}\]?
remember, the function value doesn't change when you stay on a specific level curve ?
I think so.
When you're along one level curve, yes....
Also recall the fact that gradient vector is perpendicular to the level curve
Wait, never mind what I said about the root.
Oh yes, it is.
|dw:1435040370688:dw|
|dw:1435040574730:dw|
just observe that rotating the gradient vector by 90 degrees gives you the tangent vector to level curve and in that direction the function remains constant
Ohhhh, yes
An easy way to work it if you don't happen to remember the rotation matrix is to simply use the dot product
we already know that gradient = \(\langle -2, 3\rangle\) dot that with each of the given options and see which one evaluates to \(0\)
Of course. Yes. I did not think about using the given answers.
Wow.
Wait, hmmm what is z?
I can't use the dot product without a z-value....
Ah well. I still get zero, so I guess you can use it without a z-value....
I used 0 for the z-value.
Oh yeah.... -__-
oops
no wait, i got confused
nope, that is correct. #18 is about directional derivative, \(u\) is a vector in \(xy\) plane, so \(z=0\)
:)
#19 should be easy, just work the tangent plane approximation
And linearization is just: \[L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)\]?
that should do
or you may use the gradient
normal vector of the required tangent plane is \(\langle f_x,~f_y,~-1 \rangle \)
so the tangent plane is \[f_x x+f_yy-z = k\] you cna work the \(k\) value by plugging in the point \((-2,~1,~33)\)
whichever way you feel easy... upto you
I tried it. I thought it was a plug-and-chug method, but again, I can't seem to find the values of \[f_x , f_y\]. I don't think \[f_x = 33\]
hey still here ?
From #16 we have \[\nabla f = \langle f_x,~f_y \rangle = \langle -6,9 \rangle \] so the tangent plane is simply \[-6x+9y-z=k\] plugin the point \((-2,1,33)\) and solve \(k\)
Hi! I'm back.
Ohhhhh okay. That's another method I never knew about. Thanks! That should do it! Let me try that right now.
This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!
I'm sure you will do great! good luck!
Thank You!

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