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unknownunknown

  • one year ago

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  1. unknownunknown
    • one year ago
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    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

  2. anonymous
    • one year ago
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    I truly wish I could help you but I'm not sure where to start with this Q. Try tagging people?

  3. unknownunknown
    • one year ago
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    @geerky42

  4. unknownunknown
    • one year ago
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    Thanks geerky won't let me reply to you there. I appreciate your help anyway.

  5. geerky42
    • one year ago
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    @freckles @amistre64 @ganeshie8

  6. geerky42
    • one year ago
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    @zepdrix Good luck, unknownunknown

  7. unknownunknown
    • one year ago
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    @triciaal

  8. geerky42
    • one year ago
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    @UsukiDoll I'm sure you would like to help @unknownunknown .

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

  10. unknownunknown
    • one year ago
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    Surely it would be -(w-w0) instead of (w-w0)

  11. geerky42
    • one year ago
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    yeah I agree.

  12. unknownunknown
    • one year ago
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    it makes a confusing concept more confusing =(

  13. unknownunknown
    • one year ago
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    Hi Usuki, thanks for coming.

  14. unknownunknown
    • one year ago
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    @Preetha @amistre64 @UnkleRhaukus @zepdrix @Luigi0210

  15. UnkleRhaukus
    • one year ago
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    \[A(x-x_0)+B(y-y_0)+C(w-w_0)=0\]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)

  16. unknownunknown
    • one year ago
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    Yep, and then it goes on to make a conclusion which I must accept for further study.

  17. UnkleRhaukus
    • one year ago
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    hmm, confusing indeed

  18. unknownunknown
    • one year ago
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    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?

  19. UnkleRhaukus
    • one year ago
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    Do you mean this? \[df(x,y, \dots) = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\cdots\]

  20. unknownunknown
    • one year ago
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    Yes. Why doesn't df/dz = df/dx + df/dy only?

  21. UnkleRhaukus
    • one year ago
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    \[\frac{df(x,y)}{dz} = \frac{\partial f}{\partial x}\frac{dx}{dz}+\frac{\partial f}{\partial y}\frac{dy}{dz}\]

  22. UnkleRhaukus
    • one year ago
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    NB these are not the same a fractions.

  23. unknownunknown
    • one year ago
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    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?

  24. unknownunknown
    • one year ago
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    By the way, defining f(x,y) = z

  25. UnkleRhaukus
    • one year ago
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    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 \[dz = 2x\cdot dx+3y^2\cdot dy\]

  26. unknownunknown
    • one year ago
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    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?

  27. UnkleRhaukus
    • one year ago
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    \[z(x,y) = x^2+y^3\] so \[\frac{\partial z}{\partial x}=2x\] this is equivalent to \[{\partial z}=2x \cdot {\partial x}\]

  28. UnkleRhaukus
    • one year ago
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    the infinitesimals have to balance

  29. unknownunknown
    • one year ago
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    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.

  30. UnkleRhaukus
    • one year ago
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    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)

  31. UnkleRhaukus
    • one year ago
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    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

  32. unknownunknown
    • one year ago
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    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?

  33. unknownunknown
    • one year ago
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    x and y *

  34. UnkleRhaukus
    • one year ago
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    i think so

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

  36. UnkleRhaukus
    • one year ago
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    for a simple straight line y = mx dy = m dx ie dy/dx = m ∆y = m ∆x ie ∆y/∆x = m

  37. UnkleRhaukus
    • one year ago
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    but then again ,geometry would perhaps be simpler than calculus for straight lines, --- but they should give the same results

  38. unknownunknown
    • one year ago
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    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.

  39. unknownunknown
    • one year ago
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    Or I see they're the components of the Natural, but how would the instantaneous change in a perpendicular direction be of use?

  40. unknownunknown
    • one year ago
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    Normal*

  41. UnkleRhaukus
    • one year ago
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    i think \(A,B,C\) the slopes in each of the directions

  42. unknownunknown
    • one year ago
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    So for any plane, are the components of the normal equivalent to the slopes of that plane?

  43. unknownunknown
    • one year ago
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    |dw:1433150445307:dw|

  44. UnkleRhaukus
    • one year ago
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    the normals point perpendicularly out from a plane |dw:1433150882524:dw|

  45. UnkleRhaukus
    • one year ago
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    such that the slope of the plane, and slope of normals are inverse (and opposite)

  46. UnkleRhaukus
    • one year ago
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    |dw:1433151130798:dw|

  47. unknownunknown
    • one year ago
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    Hmm interesting. I'm just still not completely getting how the tangent plane is calculated conceptually.

  48. UnkleRhaukus
    • one year ago
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    a tangent is the instantaneous slope at a point, the derivative at that point

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