- unknownunknown

Question to follow..

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- unknownunknown

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

- anonymous

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

- unknownunknown

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- unknownunknown

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

- geerky42

- geerky42

@zepdrix
Good luck, unknownunknown

- unknownunknown

- geerky42

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

- unknownunknown

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

- unknownunknown

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

- geerky42

yeah I agree.

- unknownunknown

it makes a confusing concept more confusing =(

- unknownunknown

Hi Usuki, thanks for coming.

- unknownunknown

- UnkleRhaukus

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

- unknownunknown

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

- UnkleRhaukus

hmm, confusing indeed

- unknownunknown

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?

- UnkleRhaukus

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

- unknownunknown

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

- UnkleRhaukus

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

- UnkleRhaukus

NB these are not the same a fractions.

- unknownunknown

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?

- unknownunknown

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

- UnkleRhaukus

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

- unknownunknown

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?

- UnkleRhaukus

\[z(x,y) = x^2+y^3\]
so \[\frac{\partial z}{\partial x}=2x\]
this is equivalent to
\[{\partial z}=2x \cdot {\partial x}\]

- UnkleRhaukus

the infinitesimals have to balance

- unknownunknown

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.

- UnkleRhaukus

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)

- UnkleRhaukus

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

- unknownunknown

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?

- unknownunknown

x and y *

- UnkleRhaukus

i think so

- unknownunknown

Makes sense, thanks very much!
Now if only that equation (3) made sense.. I wish they'd fix the errors there.

- UnkleRhaukus

for a simple straight line
y = mx
dy = m dx ie dy/dx = m
∆y = m ∆x ie ∆y/∆x = m

- UnkleRhaukus

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

- unknownunknown

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.

- unknownunknown

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

- unknownunknown

Normal*

- UnkleRhaukus

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

- unknownunknown

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

- unknownunknown

|dw:1433150445307:dw|

- UnkleRhaukus

the normals point perpendicularly out from a plane
|dw:1433150882524:dw|

- UnkleRhaukus

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

- UnkleRhaukus

|dw:1433151130798:dw|

- unknownunknown

Hmm interesting. I'm just still not completely getting how the tangent plane is calculated conceptually.

- UnkleRhaukus

a tangent is the instantaneous slope at a point, the derivative at that point

Looking for something else?

Not the answer you are looking for? Search for more explanations.