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unknownunknown
 one year ago
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unknownunknown
 one year ago
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unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Two 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/1802scmultivariablecalculusfall2010/2.partialderivatives/partafunctionsoftwovariablestangentapproximationandoptimization/session27approximationformula/MIT18_02SC_MNotes_ta2_3.pdf (equation 3). How is equation 3 derive from the prior equation? How is (ww0) transferred like that? Thanks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I truly wish I could help you but I'm not sure where to start with this Q. Try tagging people?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Thanks geerky won't let me reply to you there. I appreciate your help anyway.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1@freckles @amistre64 @ganeshie8

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1@zepdrix Good luck, unknownunknown

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1@UsukiDoll I'm sure you would like to help @unknownunknown .

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1geerky42, without looking at the math, just simple algebra. Wouldn't you agree equation 3 can't follow from the previous one?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Surely it would be (ww0) instead of (ww0)

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1it makes a confusing concept more confusing =(

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Hi Usuki, thanks for coming.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1@Preetha @amistre64 @UnkleRhaukus @zepdrix @Luigi0210

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[A(xx_0)+B(yy_0)+C(ww_0)=0\]where \(C\neq0\): \[\frac AC(xx_0)+\frac BC(yy_0)+(ww_0)=0\\ \frac AC(xx_0)+\frac BC(yy_0)=(ww_0)\] with \(a=A/C\), \(b=B/C\) \[a(xx_0)+b(yy_0)=(ww_0)\\ w+w_0 = a(xx_0)+b(yy_0)\] which is not the same as (3)

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Yep, and then it goes on to make a conclusion which I must accept for further study.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2hmm, confusing indeed

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Just 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
 one year ago
Best ResponseYou've already chosen the best response.2Do you mean this? \[df(x,y, \dots) = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\cdots\]

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Why doesn't df/dz = df/dx + df/dy only?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{df(x,y)}{dz} = \frac{\partial f}{\partial x}\frac{dx}{dz}+\frac{\partial f}{\partial y}\frac{dy}{dz}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2NB these are not the same a fractions.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Why 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
 one year ago
Best ResponseYou've already chosen the best response.1By the way, defining f(x,y) = z

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.1Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.2\[z(x,y) = x^2+y^3\] so \[\frac{\partial z}{\partial x}=2x\] this is equivalent to \[{\partial z}=2x \cdot {\partial x}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2the infinitesimals have to balance

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Ohh yes, thank you. So here for example: http://ocw.mit.edu/courses/mathematics/1802scmultivariablecalculusfall2010/2.partialderivatives/partafunctionsoftwovariablestangentapproximationandoptimization/session27approximationformula/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
 one year ago
Best ResponseYou've already chosen the best response.2When 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
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1Makes sense, thanks very much! Now if only that equation (3) made sense.. I wish they'd fix the errors there.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2for a simple straight line y = mx dy = m dx ie dy/dx = m ∆y = m ∆x ie ∆y/∆x = m

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2but then again ,geometry would perhaps be simpler than calculus for straight lines,  but they should give the same results

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1What I don't get here, is for A(xx0) + B(yy0) + C(zz0), these A,B,C are points right? And yet they're treated as a slope, which we'd need two points for.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Or I see they're the components of the Natural, but how would the instantaneous change in a perpendicular direction be of use?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2i think \(A,B,C\) the slopes in each of the directions

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1So for any plane, are the components of the normal equivalent to the slopes of that plane?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433150445307:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2the normals point perpendicularly out from a plane dw:1433150882524:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2such that the slope of the plane, and slope of normals are inverse (and opposite)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433151130798:dw

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.1Hmm interesting. I'm just still not completely getting how the tangent plane is calculated conceptually.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2a tangent is the instantaneous slope at a point, the derivative at that point
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