Agent_A
  • Agent_A
Please answer all the questions (and in a neat format). Thanks! (see photo) Need all the answers at a glance.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Agent_A
  • Agent_A
1 Attachment
ganeshie8
  • ganeshie8
#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
  • Michele_Laino
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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More answers

ganeshie8
  • ganeshie8
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 \(f_x\) ?
Agent_A
  • Agent_A
The sign is positive, but decreasing....?
ganeshie8
  • ganeshie8
nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?
Agent_A
  • Agent_A
Somewhat, yes, but muddy.
ganeshie8
  • ganeshie8
First derivative at a particular point on the curve gives the slope of tangent line at that point |dw:1435034181715:dw|
ganeshie8
  • ganeshie8
Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing" |dw:1435034301743:dw|
ganeshie8
  • ganeshie8
Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing" |dw:1435034404265:dw|
ganeshie8
  • ganeshie8
That means the sign of first derivative can be used to know if a function is increasing/decreasing : \(\large f'(c)~\gt~0~~\iff~~\text{f(x) is increasing at x=c}\) \(\large f'(c)~\lt~0~~\iff~~\text{f(x) is decreasing at x=c}\)
ganeshie8
  • ganeshie8
In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.
Agent_A
  • Agent_A
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?
Agent_A
  • Agent_A
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.
ganeshie8
  • ganeshie8
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
Agent_A
  • Agent_A
OOOPS. Sorry, I was referring to another problem, when I said "zero-gradient". So that wouldn't make sense here.
Agent_A
  • Agent_A
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.
ganeshie8
  • ganeshie8
Just pick any one row and forget about everything else.
Agent_A
  • Agent_A
I'm doing two problems at once.
Agent_A
  • Agent_A
Sorry for the mess.
ganeshie8
  • ganeshie8
thats okay :)
Agent_A
  • Agent_A
Thanks for understanding.... I'm just so stressed right now.
ganeshie8
  • ganeshie8
Just answer me this, in the below highlighted row, what do you notice about the values ? |dw:1435035389370:dw|
Agent_A
  • Agent_A
Decreasing, as we go from left to right.
ganeshie8
  • ganeshie8
Yes. so \(f_x(-2,~0)~~\lt ~0 \) |dw:1435035518143:dw|
ganeshie8
  • ganeshie8
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|
Agent_A
  • Agent_A
Increasing....
ganeshie8
  • ganeshie8
good, so \(f_y(-2,~1)\gt 0\) |dw:1435035643341:dw|
Agent_A
  • Agent_A
Ohhhhhhh I get the pattern.
Agent_A
  • Agent_A
is f_xx true, then?
Agent_A
  • Agent_A
and f_yy
ganeshie8
  • ganeshie8
sry i don't seem to remember, il review quick and get back meantime try other problems maybe..
Agent_A
  • Agent_A
Okay!
Agent_A
  • Agent_A
Thanks!
ganeshie8
  • ganeshie8
@Michele_Laino
ganeshie8
  • ganeshie8
I'm thinking of working the first differences but not entirely sure
Michele_Laino
  • Michele_Laino
we can try to write the taylor series up to the second derivative of f, with for example y= const @ganeshie8
Michele_Laino
  • Michele_Laino
like this: \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)x + \frac{{{x^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]
Michele_Laino
  • Michele_Laino
oops.. \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\Delta x + \frac{{{{\left( {\Delta x} \right)}^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]
ganeshie8
  • ganeshie8
Ahh so the finite differences actually works \(f_{yy}(-2,1)~= 0\) because the first differences are constant along a column : |dw:1435036964859:dw|
ganeshie8
  • ganeshie8
|dw:1435037234174:dw|
Agent_A
  • Agent_A
If it helps, the answer is letter B.
Agent_A
  • Agent_A
For number 15. I just don't have the solutions.
Agent_A
  • Agent_A
Answers to the other numbers: 15) B 16) A 17) E 18) A 19) B
ganeshie8
  • ganeshie8
Yeah we cannot tell anything about mixed partials i guess so B makes sense to me
ganeshie8
  • ganeshie8
|dw:1435037903299:dw|
Michele_Laino
  • Michele_Laino
we have checked four conditions, so I think option D. Am I right?
Michele_Laino
  • Michele_Laino
namely: the first and fourth conditions are false; the second and the third conditions are true
Michele_Laino
  • Michele_Laino
and we can not infer anything on the fifth condition
ganeshie8
  • ganeshie8
hmm yeah since the first and fourth conditions are false we may say that only two inferences are true
ganeshie8
  • ganeshie8
#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
  • ganeshie8
|dw:1435038498718:dw|
ganeshie8
  • ganeshie8
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
Agent_A
  • Agent_A
Thanks! How were those numbers selected? Was it through the symmetric difference quotient?
ganeshie8
  • ganeshie8
Ohh no, i guess we should use symmetric difference quotient then
Agent_A
  • Agent_A
Ohhh yeah, I just am confused on how to plug the values in :P.
Agent_A
  • Agent_A
*am just
ganeshie8
  • ganeshie8
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
Agent_A
  • Agent_A
Nice! Thanks!
ganeshie8
  • ganeshie8
#17 just follows from #16
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
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 = \(\langle -2,~3 \rangle \)
Agent_A
  • Agent_A
I see!
Agent_A
  • Agent_A
Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.
Agent_A
  • Agent_A
It really was just a matter of where to look, on the chart....
Agent_A
  • Agent_A
How do we do number 18? I'm used to solving for it with the equation given.
Agent_A
  • Agent_A
Do we divide <-2, 3> by \[\sqrt{(-2)^2+(3)^2}\]?
ganeshie8
  • ganeshie8
remember, the function value doesn't change when you stay on a specific level curve ?
Agent_A
  • Agent_A
I think so.
Agent_A
  • Agent_A
When you're along one level curve, yes....
ganeshie8
  • ganeshie8
Also recall the fact that gradient vector is perpendicular to the level curve
Agent_A
  • Agent_A
Wait, never mind what I said about the root.
Agent_A
  • Agent_A
Oh yes, it is.
ganeshie8
  • ganeshie8
|dw:1435040370688:dw|
ganeshie8
  • ganeshie8
|dw:1435040574730:dw|
ganeshie8
  • ganeshie8
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
Agent_A
  • Agent_A
Ohhhh, yes
ganeshie8
  • ganeshie8
An easy way to work it if you don't happen to remember the rotation matrix is to simply use the dot product
ganeshie8
  • ganeshie8
we 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
  • Agent_A
Of course. Yes. I did not think about using the given answers.
Agent_A
  • Agent_A
Wow.
Agent_A
  • Agent_A
Wait, hmmm what is z?
Agent_A
  • Agent_A
I can't use the dot product without a z-value....
Agent_A
  • Agent_A
Ah well. I still get zero, so I guess you can use it without a z-value....
Agent_A
  • Agent_A
I used 0 for the z-value.
Agent_A
  • Agent_A
Oh yeah.... -__-
Agent_A
  • Agent_A
oops
ganeshie8
  • ganeshie8
no wait, i got confused
ganeshie8
  • ganeshie8
nope, that is correct. #18 is about directional derivative, \(u\) is a vector in \(xy\) plane, so \(z=0\)
Agent_A
  • Agent_A
:)
ganeshie8
  • ganeshie8
#19 should be easy, just work the tangent plane approximation
Agent_A
  • Agent_A
And linearization is just: \[L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)\]?
ganeshie8
  • ganeshie8
that should do
ganeshie8
  • ganeshie8
or you may use the gradient
ganeshie8
  • ganeshie8
normal vector of the required tangent plane is \(\langle f_x,~f_y,~-1 \rangle \)
ganeshie8
  • ganeshie8
so the tangent plane is \[f_x x+f_yy-z = k\] you cna work the \(k\) value by plugging in the point \((-2,~1,~33)\)
ganeshie8
  • ganeshie8
whichever way you feel easy... upto you
Agent_A
  • Agent_A
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 \[f_x = 33\]
ganeshie8
  • ganeshie8
hey still here ?
ganeshie8
  • ganeshie8
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 \(k\)
Agent_A
  • Agent_A
Hi! I'm back.
Agent_A
  • Agent_A
Ohhhhh okay. That's another method I never knew about. Thanks! That should do it! Let me try that right now.
Agent_A
  • Agent_A
This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!
ganeshie8
  • ganeshie8
I'm sure you will do great! good luck!
Agent_A
  • Agent_A
Thank You!

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