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### Course: Multivariable calculus>Unit 3

A worked example for finding the quadratic approximation of a two-variable function. Created by Grant Sanderson.

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• Is a cubic approximation any better than a quadratic approximation? I know it would be even more monstrous with the third partial derivative and the third partial derivative would most likely be 0 but is it any better?
• Yes, the cubic approximation would be better than a quadratic approximation. I assume you've completed single-variable calculus. What Grant is doing in this video is basically a "two-term Taylor expansion" for multivariable functions (but the multivariable version has much more than the two terms of it's single-variable cousin).
A cubic approximation would be a "three-term Taylor approximation" basically, and as you probably know, the more terms you add in the Taylor approximation, the more accurate the approximation is. A quartic approximation would be better than a cubic approximation, and a quintic approximation would be even better than a quartic approximation, etc...
At a certain point, adding terms would affect the approximated value so minutely that it would be unnecessary and too cumbersome to do by hand for most practical situations.
• Hello. Can You explain please, why did you put 1/2 in front of "fxx" and "fyy" in the formula, I did not understand that part? Thank you n advance.
• It was shown in the previous video where Grant derive the formula for QA.
• how to find the value of cos 61.8 by approximation formula?
• This is not a multivariable calculus problem. You might want to review Sal's approximation formula video for single variable calculus: https://www.khanacademy.org/math/differential-calculus/derivative-applications/differentiation-application/v/approximating-incremental-cost-with-derivative

The linear approximation formula is f(x) ~= f(x_0) + f'(x_0) * (x - x_0)
In the case of cos(61.8 degrees), or cos(1.0786 radians) --> x = 1.0786; x_0 = pi/3 = 1.0472 :
cos(1.0786) ~= cos(pi/3) - sin(pi/3) * (1.0786 - pi/3)
cos(1.0786) ~= 1/2 - √3/2 * (0.0314) = 0.4728

In reality, cos(1.0786) = 0.4726.

Realistically, there would be no easy way to convert 61.8 degrees into radians and to convert √3/2 into decimal form without a calculator, and if you have a calculator, you might as well directly plug in cos(61.8) into it.

If you wanted an even more accurate quadratic approximation, you would simply add
f''(x_0)/2 * (x-x_0)^2 to the first result (0.4728) so
cos(1.0786) ~= 0.4728 + (-cos(pi/3)/2 * (pi/3 - 1.0786)^2) = 0.4728 - 0.0002 = 0.4726
which is basically equal to cos(1.0786), to the nearest tenthousandths place.
• How does this situation or quadratic approximation equation change in the case that we're dealing with a function where ꝺ^2Q/ꝺxꝺy ≠ ꝺ^2Q/ꝺyꝺx ?
• The Hessian matrix captures such an eventuality, and you could adapt the maths from there - although you might not be able to do anything with it.
Researching Clairaut's theorem (aka: Schwarz's theorem, or Young's theorem) states that where there is continuity of the second derivatives in the neighbourhood of your input point, the mixed second derivatives are the same irrespective of the order of differentiation.
https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Schwarz's_theorem
Hmmm...: for some reason, this text editor doesn't seem to like apostrophes in URLs, but the link still gets you there.
(1 vote)
• At , did you forget to put 1/2 before fxx and fyy?
• Hi everyone - I think I missed the nature of the mistake that was made at . Can someone explain why he needed to have 1/2 as a coefficient in front of the f terms? Thanks for any insight.
(1 vote)
• The reason for the 'half' is explained in the previous video...
The quadratic function, Q(x, y) that we want to use to approximate the original function needs the same 2nd derivatives as the original. We're assuming it has a term of the form:
a.(x - x_0)^2
When we take the 2nd derivative of Q with respect to x, all other terms vanish, and we're left with 2a.
Therefore, taking the second derivative of the original function evaluated at (x_0, y_0), we're looking for '2a'...
'a' (what we need for the coefficient of the '(x - x0)' term of 'Q') is obviously half of '2a'.
The same logic applies to the (y-y_0) term.