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

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4#15 For \(f_x\) we keep \(y\) fixed and see how the function changes as we change \(x\) so simply look at rows

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we have to apply this formula: \[\Large {f_x}\left( {{x_0},{y_0}} \right) \cong \frac{{f\left( {{x_0} + \Delta x,{y_0}} \right)  f\left( {{x_0},{y_0}} \right)}}{{\Delta x}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4In 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 \(f_x\) ?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1The sign is positive, but decreasing....?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Somewhat, yes, but muddy.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4First derivative at a particular point on the curve gives the slope of tangent line at that point dw:1435034181715:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing" dw:1435034301743:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing" dw:1435034404265:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That means the sign of first derivative can be used to know if a function is increasing/decreasing : \(\large f'(c)~\gt~0~~\iff~~\text{f(x) is increasing at x=c}\) \(\large f'(c)~\lt~0~~\iff~~\text{f(x) is decreasing at x=c}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, 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 zerogradient, but not when I have to use a contour plot, without any equation to derive it. How would this work?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Your 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I 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

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1OOOPS. Sorry, I was referring to another problem, when I said "zerogradient". So that wouldn't make sense here.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Okay, 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Just pick any one row and forget about everything else.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1I'm doing two problems at once.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Thanks for understanding.... I'm just so stressed right now.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Just answer me this, in the below highlighted row, what do you notice about the values ? dw:1435035389370:dw

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Decreasing, as we go from left to right.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes. so \(f_x(2,~0)~~\lt ~0 \) dw:1435035518143:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Next, pick any one column and tell me what do you notice about the values as you go from bottom to top dw:1435035560839:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4good, so \(f_y(2,~1)\gt 0\) dw:1435035643341:dw

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhhhhh I get the pattern.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4sry i don't seem to remember, il review quick and get back meantime try other problems maybe..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I'm thinking of working the first differences but not entirely sure

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can try to write the taylor series up to the second derivative of f, with for example y= const @ganeshie8

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1like this: \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)x + \frac{{{x^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\Delta x + \frac{{{{\left( {\Delta x} \right)}^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh so the finite differences actually works \(f_{yy}(2,1)~= 0\) because the first differences are constant along a column : dw:1435036964859:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435037234174:dw

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1If it helps, the answer is letter B.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1For number 15. I just don't have the solutions.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Answers to the other numbers: 15) B 16) A 17) E 18) A 19) B

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yeah we cannot tell anything about mixed partials i guess so B makes sense to me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435037903299:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have checked four conditions, so I think option D. Am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely: the first and fourth conditions are false; the second and the third conditions are true

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and we can not infer anything on the fifth condition

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4hmm yeah since the first and fourth conditions are false we may say that only two inferences are true

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4#16 should be easy, just approximate the partials at (2, 1) and take the magnitude \[\large \\nabla f\ = \sqrt{{f_x}^2+{f_y}^2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435038498718:dw

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Thanks! How were those numbers selected? Was it through the symmetric difference quotient?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ohh no, i guess we should use symmetric difference quotient then

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Ohhh yeah, I just am confused on how to plug the values in :P.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes symmetric difference quotient also gives the same answer http://www.wolframalpha.com/input/?i=sqrt%28%28%2832.4133.61%29%2F%282*0.1%29%29%5E2%2B%28%2833.932.1%29%2F%282*0.1%29%29%5E2%29

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4#17 just follows from #16

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4direction is \(\langle 6, ~9\rangle \) http://www.wolframalpha.com/input/?i=%3C%2832.4133.61%29%2F%282*0.1%29%2C+%2833.932.1%29%2F%282*0.1%29%3E

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Notice that the direction \(\langle 6, ~9\rangle \) is same as \(k\langle 6, ~9\rangle \) let \(k=\frac{1}{3}\) and we get direction = \(\langle 2,~3 \rangle \)

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1It really was just a matter of where to look, on the chart....

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1How do we do number 18? I'm used to solving for it with the equation given.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Do we divide <2, 3> by \[\sqrt{(2)^2+(3)^2}\]?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4remember, the function value doesn't change when you stay on a specific level curve ?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1When you're along one level curve, yes....

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Also recall the fact that gradient vector is perpendicular to the level curve

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Wait, never mind what I said about the root.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435040370688:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435040574730:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4just 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4An easy way to work it if you don't happen to remember the rotation matrix is to simply use the dot product

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4we already know that gradient = \(\langle 2, 3\rangle\) dot that with each of the given options and see which one evaluates to \(0\)

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Of course. Yes. I did not think about using the given answers.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1I can't use the dot product without a zvalue....

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Ah well. I still get zero, so I guess you can use it without a zvalue....

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1I used 0 for the zvalue.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4no wait, i got confused

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nope, that is correct. #18 is about directional derivative, \(u\) is a vector in \(xy\) plane, so \(z=0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4#19 should be easy, just work the tangent plane approximation

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1And linearization is just: \[L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(xx_0)+f_y(x_0, y_0)(yy_0)\]?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4or you may use the gradient

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4normal vector of the required tangent plane is \(\langle f_x,~f_y,~1 \rangle \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4so the tangent plane is \[f_x x+f_yyz = k\] you cna work the \(k\) value by plugging in the point \((2,~1,~33)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4whichever way you feel easy... upto you

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1I tried it. I thought it was a plugandchug method, but again, I can't seem to find the values of \[f_x , f_y\]. I don't think \[f_x = 33\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4From #16 we have \[\nabla f = \langle f_x,~f_y \rangle = \langle 6,9 \rangle \] so the tangent plane is simply \[6x+9yz=k\] plugin the point \((2,1,33)\) and solve \(k\)

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhhh okay. That's another method I never knew about. Thanks! That should do it! Let me try that right now.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I'm sure you will do great! good luck!
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