Question

use partial derivatives to approximate the value of \[f(6.1, 1.9)\] where \[f(x,y) = (x^{2}-y^{3}-1)^{\frac{1}{3}}\] at \[(6,2)\]

Answer 1

I wonder how much longer I should stare at this... ^_^

Answer 2

Well, @Ray10 ...
ready when you are:

Answer 3

oh sorry!!!
My laptop froze and I didn't realize!! :O @terenzreignz

Answer 4

I am ready :D

Answer 5

Well I'm not.
You took a while.
Just let me patch things up with her
http://openstudy.com/study#/updates/52319b0de4b03eb771a27ef9
(don't worry, it's just algebra)
and I'll get back to you.

I'll do my best to hurry it along, but remember, her learning is a priority ^_^

Answer 6

yes sure thing TJ! :D
of course! I understand :) I'll be waiting

Answer 7

you sent me the link to my own question :P

Answer 8

Whoops... well if you're curious, check this http://openstudy.com/study#/updates/5231a34de4b012524e87f0c4

Answer 9

you teach very well :)

Answer 10

Thank you ^_^

Answer 11

You know what?
While you're at it, find \(\Large f_x\) and \(\Large f_y\)
They'll be needed.

Answer 12

Okay, I'm basically done.
Have you found these partials?

Answer 13

so far this is what I have got:
\[f(x,y) = (6,2) => f(x,y)=3\]
I found fx and fy to be
\[fx=\frac{ 2x }{( \sqrt[3]{3(x^{2}-y^{3}-1)} })^{2}\] and
\[fy=\frac{ -y^{2} }{(\sqrt[3]{x^{2}-y^{3}-1} )^{2}}\]
\[fx=(x=6), = \frac{ 4 }{ 9 }\]
so far correct ?

Answer 14

small typo at fx, the bracket is meant to be on the denominator :P

Answer 15

I hate fractional exponents -.-

Answer 16

oh me too!! -.-

Answer 17

By the way... f_x for subscripts.
\[\Large f(x,y) = (x^{2}-y^{3}-1)^{\frac{1}{3}}\]

Answer 18

\[\Large f_x(x,y)=\frac13(x^2-y^3-1)^{-\frac23}\cdot2x\] right?

Answer 19

yes that is correct :)

Answer 20

Find \(\large f_y\)
And for heaven's sake, don't use radicals.

Answer 21

what are radicals? :S

Answer 22

These nasty things:
\[\Huge \color{red}{\sqrt{\qquad}}\]

Answer 23

\[fy(x,y)=1/3(x^{2}−y^{3}−1)^{−2/3}⋅-y^{2}\]

Answer 24

Fraction
\frac{numerator}{denominator}
subscripts
f_x etc

Answer 25

why is your formula's nice and big to read, and mine are small? :S

Answer 26

increase size...
\large
\Large
\LARGE
\huge
\Huge

before the whole thing.

Answer 27

I use \Large by default

Answer 28

Okay, good.
Enter: the total differential:
\[\Large dz = f_x(x_o,y_o)\partial x + f_y(x_o,y_o)\partial y\]

Answer 29

\[\Large \frac{ 1 }{ 3 }(x^{2}-y^{3}-1)^\frac{ 2 }{ 3 } \times -y^{2}\] ^_^

Answer 30

Here, of course, the partial derivatives are evaluated at the point (6,2)
What are \(\large f_x(6,2)\) and \(\large f_y(6,2)\)?

Answer 31

When I use that total differential, I see to arrive at:
\[\Large f(6.1,1.9)=f(6,2)+(0.1)\frac{ 4 }{ 9 }+(-0.1)\frac{ 4 }{ 9 }\]

Answer 32

You get 4/9 at both partial derivatives?

Answer 33

For \[\Large fx\] I get \[\Large \frac{ 4 }{ 9 }\]
and for \[\Large fy\] I get \[\Large -\frac{ 4 }{ 9 }\]
is that correct?

Answer 34

abusing the size option? :P

Answer 35

haha sorry, it looks neater :P I'll dial it down :P

Answer 36

You might want to redo your \(\Large f_y\)

<again and again... I told you that to make subscripts, you just have f_{subscript}>

Answer 37

\[\large f_{y}= \frac{ 1 }{ 3 }(x^{2}-y^{3}-1)^\frac{ -2 }{ 3 } \times -y^{2}\]
\[\large f_{y}= \frac{ 1 }{ 3 }(6^{2}-2^{3}-1)^\frac{ -2 }{ 3 } \times -2^{2}\]
\[\large f_{y} = \frac{ 4 }{ 27 }\] is what I get

Answer 38

I wonder how the negative sign... just disappeared... strange :3
\[\large f_{y}= \frac{ 1 }{ 3 }(6^{2}-2^{3}-1)^\frac{ -2 }{ 3 } \times \boxed-2^{2}\]

Answer 39

Carelessness is your worst enemy. ^_^

Answer 40

ah I keep forgetting to type it in -.- ah it's a task to type each line out :P

Answer 41

I wonder how the negative sign... just disappeared... strange :3
\[\large f_{y}= \frac{ 1 }{ 3 }(6^{2}-2^{3}-1)^\frac{ -2 }{ 3 } \times \boxed-2^{2}\]

Answer 42

\[\large \frac{-4}{27}\] correct? :D

Answer 43

I then still don't get the correct final answer :(

Answer 44

What is it?

Answer 45

I mean, the correct final answer

Answer 46

the answer should be around 6.39 :/
I get 3....

Answer 47

Impossible.

Answer 48

to work out the correct answer, it's just:
\[\large (6.1^{2}+1.9^{2})^\frac{ 1 }{ 2 }\] ??

Answer 49

<smh>
The correct answer is
\[\Large f(6.1,1.9)=(6.1^2-1.9^3-1)^{\frac13}\]
How on earth did you get\[\large (6.1^{2}+1.9^{2})^\frac{ 1 }{ 2 }\]???

Answer 50

ohhhhhh there was a misunderstanding in my lecture notes!! let me show you :P

Answer 51

Yes... show me~ good idea :)

Answer 52

I thought the bottom line is a standard equation to work out the actual value -.-

Answer 53

this is yet another good reason for not doing work soo late into the night :P

Answer 54

You do realise that this is an entirely different function, right? -.-

Answer 55

oh yeah of course! That is just the lecture notes. I was applying it to another question and was following the method :S

Answer 56

Oh... too methodical :P
I haven't taken notes (in maths) since eighth grade :P

Answer 57

Anyway, do the total differential again. You were close.

Answer 58

\[\Large f(6.1,1.9)=f(6,2)+(0.1)\frac{ 4 }{ 9 }+(-0.1)\frac{ -4 }{ 27 } = 3.06\]
\[\Large :D\]

Answer 59

Aren't you glad this all worked out in the end? :)

Answer 60

I am more than glad, you seem to always give that extra external insight which helps the poster, work out the problem :D

Answer 61

By the way, allow me to nitpick...\[\Large f(6.1,1.9)\color{red}\approx f(6,2)+(0.1)\frac{ 4 }{ 9 }+(-0.1)\frac{ -4 }{ 27 } \color{red}\approx 3.06\]

Answer 62

Reminding you that this is JUST an approximation :P

Answer 63

Oh of course!! :P Thank you for reminding!
And thank you very much foryour help!! :D

Answer 64

No problem ^_^

Answer 65

Here's that insight.
If something changes in x and y, something will change in z = f(x,y)
\[\Large f(x+\Delta x , y+\Delta y) =f(x,y)+\color{red}{\Delta z}\]

Answer 66

Now derivatives are all about differences, and \(\large \Delta z\) is approximated by \[\Large dz = f_x\partial x + f_y \partial y\]

Answer 67

Which in turn is approximated by \[\Large dz \approx f_x\Delta x + f_y \Delta y \]

Answer 68

And that's where THIS approximation comes from:
\[\Large f(x+\Delta x , y+\Delta y) \color{green}\approx f(x,y)+\color{red}{f_x(x,y)\Delta x + f_y(x,y) \Delta y }\]

Answer 69

ahhh there we go!! Now I have a clearer understanding :D thank you TJ!!
sorry for the late reply!! My computer crashed, so I had to restart it and fix it up :/ @terenzreignz