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

- Agent_A

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- Agent_A

##### 1 Attachment

- 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

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

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

The sign is positive, but decreasing....?

- ganeshie8

nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?

- Agent_A

Somewhat, yes, but muddy.

- ganeshie8

First derivative at a particular point on the curve gives the slope of tangent line at that point
|dw:1435034181715:dw|

- ganeshie8

Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing"
|dw:1435034301743:dw|

- ganeshie8

Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing"
|dw:1435034404265:dw|

- 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

In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.

- 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

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

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

OOOPS. Sorry, I was referring to another problem, when I said "zero-gradient". So that wouldn't make sense here.

- 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

Just pick any one row and forget about everything else.

- Agent_A

I'm doing two problems at once.

- Agent_A

Sorry for the mess.

- ganeshie8

thats okay :)

- Agent_A

Thanks for understanding.... I'm just so stressed right now.

- ganeshie8

Just answer me this,
in the below highlighted row, what do you notice about the values ?
|dw:1435035389370:dw|

- Agent_A

Decreasing, as we go from left to right.

- ganeshie8

Yes. so \(f_x(-2,~0)~~\lt ~0 \)
|dw:1435035518143:dw|

- 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

Increasing....

- ganeshie8

good, so \(f_y(-2,~1)\gt 0\)
|dw:1435035643341:dw|

- Agent_A

Ohhhhhhh I get the pattern.

- Agent_A

is f_xx true, then?

- Agent_A

and f_yy

- ganeshie8

sry i don't seem to remember, il review quick and get back
meantime try other problems maybe..

- Agent_A

Okay!

- Agent_A

Thanks!

- ganeshie8

@Michele_Laino

- ganeshie8

I'm thinking of working the first differences but not entirely sure

- 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

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

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

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

|dw:1435037234174:dw|

- Agent_A

If it helps, the answer is letter B.

- Agent_A

For number 15. I just don't have the solutions.

- Agent_A

Answers to the other numbers:
15) B
16) A
17) E
18) A
19) B

- ganeshie8

Yeah we cannot tell anything about mixed partials i guess
so B makes sense to me

- ganeshie8

|dw:1435037903299:dw|

- Michele_Laino

we have checked four conditions, so I think option D. Am I right?

- Michele_Laino

namely:
the first and fourth conditions are false;
the second and the third conditions are true

- Michele_Laino

and we can not infer anything on the fifth condition

- ganeshie8

hmm yeah since the first and fourth conditions are false
we may say that only two inferences are true

- 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

|dw:1435038498718:dw|

- 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

Thanks! How were those numbers selected? Was it through the symmetric difference quotient?

- ganeshie8

Ohh no, i guess we should use symmetric difference quotient then

- Agent_A

Ohhh yeah, I just am confused on how to plug the values in :P.

- Agent_A

*am just

- 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

Nice! Thanks!

- ganeshie8

#17 just follows from #16

- 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

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

I see!

- Agent_A

Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.

- Agent_A

It really was just a matter of where to look, on the chart....

- Agent_A

How do we do number 18? I'm used to solving for it with the equation given.

- Agent_A

Do we divide <-2, 3> by \[\sqrt{(-2)^2+(3)^2}\]?

- ganeshie8

remember, the function value doesn't change when you stay on a specific level curve ?

- Agent_A

I think so.

- Agent_A

When you're along one level curve, yes....

- ganeshie8

Also recall the fact that gradient vector is perpendicular to the level curve

- Agent_A

Wait, never mind what I said about the root.

- Agent_A

Oh yes, it is.

- ganeshie8

|dw:1435040370688:dw|

- ganeshie8

|dw:1435040574730:dw|

- 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

Ohhhh, yes

- 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

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

Of course. Yes. I did not think about using the given answers.

- Agent_A

Wow.

- Agent_A

Wait, hmmm what is z?

- Agent_A

I can't use the dot product without a z-value....

- Agent_A

Ah well. I still get zero, so I guess you can use it without a z-value....

- Agent_A

I used 0 for the z-value.

- Agent_A

Oh yeah.... -__-

- Agent_A

oops

- ganeshie8

no wait, i got confused

- ganeshie8

nope, that is correct.
#18 is about directional derivative, \(u\) is a vector in \(xy\) plane, so \(z=0\)

- Agent_A

:)

- ganeshie8

#19 should be easy, just work the tangent plane approximation

- 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

that should do

- ganeshie8

or you may use the gradient

- ganeshie8

normal vector of the required tangent plane is \(\langle f_x,~f_y,~-1 \rangle \)

- 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

whichever way you feel easy... upto you

- 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

hey still here ?

- 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

Hi! I'm back.

- 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

This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!

- ganeshie8

I'm sure you will do great! good luck!

- Agent_A

Thank You!

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