Question

Question to follow..

Answer 1

Two questions.

1) When calculating the instantaneous change in f(x,y), why can we not sum the partial derivatives as such (excuse notation for d as there is no partial derivative d I can see in the latex): \[df/dz = df/dx + df/dy \] ?

Why does the above not work?

2) I am having trouble with conceptual understanding of derivation of the tangent plane. Could someone please check if this is correct: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-27-approximation-formula/MIT18_02SC_MNotes_ta2_3.pdf (equation 3). How is equation 3 derive from the prior equation? How is (w-w0) transferred like that?

Thanks

Answer 2

I truly wish I could help you but I'm not sure where to start with this Q. Try tagging people?

Answer 3

@geerky42

Answer 4

Thanks geerky won't let me reply to you there. I appreciate your help anyway.

Answer 5

@freckles @amistre64 @ganeshie8

Answer 6

@zepdrix

Good luck, unknownunknown

Answer 7

@triciaal

Answer 8

@UsukiDoll I'm sure you would like to help @unknownunknown .

Answer 9

geerky42, without looking at the math, just simple algebra. Wouldn't you agree equation 3 can't follow from the previous one?

Answer 10

Surely it would be -(w-w0) instead of (w-w0)

Answer 11

yeah I agree.

Answer 12

it makes a confusing concept more confusing =(

Answer 13

Hi Usuki, thanks for coming.

Answer 14

@Preetha @amistre64 @UnkleRhaukus @zepdrix @Luigi0210

Answer 15

\[A(x-x_0)+B(y-y_0)+C(w-w_0)=0\]where \(C\neq0\):

\[\frac AC(x-x_0)+\frac BC(y-y_0)+(w-w_0)=0\\
\frac AC(x-x_0)+\frac BC(y-y_0)=-(w-w_0)\]
with \(a=A/C\), \(b=B/C\)
\[a(x-x_0)+b(y-y_0)=-(w-w_0)\\
-w+w_0 = a(x-x_0)+b(y-y_0)\]
which is not the same as (3)

Answer 16

Yep, and then it goes on to make a conclusion which I must accept for further study.

Answer 17

hmm, confusing indeed

Answer 18

Just out of curiosity, do you have any idea why we can't simply add the partial derivatives together, to get the partial derivative of the overall function? Wouldn't the change in the xz plane and the yz plane by definition be the overall change?

Answer 19

Do you mean this?
\[df(x,y, \dots) = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\cdots\]

Answer 20

Yes. Why doesn't df/dz = df/dx + df/dy only?

Answer 21

\[\frac{df(x,y)}{dz} = \frac{\partial f}{\partial x}\frac{dx}{dz}+\frac{\partial f}{\partial y}\frac{dy}{dz}\]

Answer 22

NB these are not the same a fractions.

Answer 23

Why is it necessary there to differentiate \[ \frac{ dx }{ dz }\] in addition to \[\frac{ \partial f }{ \partial x }\]

Is it applying the chain rule? My understanding is the partial derivative is already calculating the change in x, and also for y, isn't that sufficient without dx/dz and dy/dz?

Answer 24

By the way, defining f(x,y) = z

Answer 25

Maybe an example will help

\[z(x,y) = x^2+y^3\]

\[\frac\partial{\partial x} z = 2x\\
\frac\partial{\partial y}z =3y^2\]

so

\[dz = 2x\cdot dx+3y^2\cdot dy\]

Answer 26

Hmm I see. So why is it not \[\frac{ \partial }{ \partial x } z = 2x dx\] already included there in the partial differentiation, since we are differentiating x right?

Answer 27

\[z(x,y) = x^2+y^3\]

so \[\frac{\partial z}{\partial x}=2x\]
this is equivalent to
\[{\partial z}=2x \cdot {\partial x}\]

Answer 28

the infinitesimals have to balance

Answer 29

Ohh yes, thank you. So here for example: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-27-approximation-formula/MIT18_02SC_MNotes_ta4.pdf

The first equation (6), my understanding is it's difficult to get an exact value when differentiating, but why is that? Whatever dx is, can we not just multiply it like you did now, to get dw? Why must the change in x be different to dx here?

Sorry for being so precise, I just want to make sure I understand this correctly before moving on.

Answer 30

When the difference is finite (not infinitesimal)\[\Delta f(x,y, \dots) \approx \frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y+\cdots\]

because the \(\Delta \) terms are too large (grainy)

Answer 31

the equality in only met in the instantaneous case; when the difference are smooth enough to be approximated by a straight line, i.e infinitesimals

Answer 32

So do we only use the approximation formula because we are unsure if the function is smooth? If the function is smooth, could we take the limit of x here and use equality instead of approximation?

Answer 33

x and y *

Answer 34

i think so

Answer 35

Makes sense, thanks very much!

Now if only that equation (3) made sense.. I wish they'd fix the errors there.

Answer 36

for a simple straight line
y = mx

dy = m dx ie dy/dx = m

∆y = m ∆x ie ∆y/∆x = m

Answer 37

but then again ,geometry would perhaps be simpler than calculus for straight lines, --- but they should give the same results

Answer 38

What I don't get here, is for A(x-x0) + B(y-y0) + C(z-z0), these A,B,C are points right? And yet they're treated as a slope, which we'd need two points for.

Answer 39

Or I see they're the components of the Natural, but how would the instantaneous change in a perpendicular direction be of use?

Answer 40

Normal*

Answer 41

i think \(A,B,C\) the slopes in each of the directions

Answer 42

So for any plane, are the components of the normal equivalent to the slopes of that plane?

Answer 43

the normals point perpendicularly out from a plane

Answer 44

such that the slope of the plane, and slope of normals are inverse (and opposite)