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MATH 234 Lecture Notes - Lecture 20: Fxx, Maxima And Minima, Talking Lifestyle 1278

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26 Jul 2018

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Lecture #20: find the second order taylor expansion: = 6 + 0 + (cid:886)(cid:884) x2 = 6 + 2x2. Critical point is minimum because second derivative is positive. 2 variables f(x) is a function of one variable. This formula is the second order taylor series of a function of two variables. Example: let f(x, y) = e2x sin (3y) Find second order taylor expansion at point (0, 0) 2 small real numbers f(0 + x , 0 + y) = f( x, y) . 0 + 0 x + 3 y + (cid:883)(cid:884) (0 ( x)2 + 12 x y + 0 ( y)2) We get a quadratic form at the point (a, b) = q ( x, y) = (cid:2869)(cid:2870) (fxx (a, b) ( x)2) Consider x & y as symbols, as variables. Example: let f(x, y) = x3 + y3 3xy f(cid:4666)(cid:882),(cid:882)(cid:4667)=(cid:882) fx(cid:4666)x,y(cid:4667)=(cid:884)e(cid:2870)xsin(cid:4666)(cid:885)y(cid:4667) @ (cid:4666)(cid:882),(cid:882)(cid:4667)=(cid:882) fy(cid:4666)x,y(cid:4667)=(cid:885)e(cid:2870)xcos(cid:4666)(cid:885)y(cid:4667) @ (cid:4666)(cid:882),(cid:882)(cid:4667)=(cid:885) fxx(cid:4666)x,y(cid:4667)=(cid:886)e(cid:2870)xsin(cid:4666)(cid:884)y(cid:4667) @ (cid:4666)(cid:882),(cid:882)(cid:4667)=(cid:882) fxy(cid:4666)x,y(cid:4667)=6e(cid:2870)xcos(cid:4666)(cid:885)y(cid:4667) @ (cid:4666)(cid:882),(cid:882)(cid:4667)=6 fyy(cid:4666)x,y(cid:4667)= 9e(cid:2870)xsin(cid:4666)(cid:885)y(cid:4667) @ (cid:4666)(cid:882),(cid:882)(cid:4667)=(cid:882)

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Related Questions

answers the Qs? 5 stars for corect and complete answers

This exercise is intended to give you some insight into why the second derivative test works. A good pan of what I'm asking you to do has been adapted form the Discovery Project on page 821 of our text. In 11.4, we wrote down the tangent plane or linear approximation of a function f(x, y) at a point. To simplify this exercise, we're going to assume that the point is located at the origin, x=0, y=0. Then At a critical point, the two partial derivatives are 0, so the linear approximation just simplifies f(x, y) f(0,0), which is the equation of a horizontal plane. This just doesn't give enough information to determine whether we have a max, min, or saddle. So we look at the quadratic approximation at (0,0), which is defined as follows: At a critical point, this will reduce to It can be proved that if the function f(x, y) has continuous second partials, then it will have the same type of critical point as its quadratic approximation. Let f(x, y) = e-x2-2y2 Find the first partial derivatives Find the first partial derivatives fx(x, y) and fy(x, y): Evaluate fx(0,0) and fy(0, 0). Find the second partial derivatives fxx(x, y), fxy(x, y) and fyy(x, y): Evaluate fxx, fxy and fyy at the critical point. In order to recognize the shape of this graph from its equation, it will be helpful to rewrite the equation in a different form so that: Complete the square for the quadratic approximation and state the values of a, b, c, and d. What docs the graph of the quadratic approximation look like? A: paraboloid opening up. B: paraboloid opening down, C: saddle. Judging from your knowledge of the shape of the graph in exercise 16, does the quadratic approximation have a max, min or saddle? Enter MAX, MIN, or SADDLE. Recall that in the second derivative test, at a critical point (a. b). For f(x, y), state the values of D and fxx. Does the second derivative test tell you that the critical point of f is a local maximum, local minimum, or saddle? After you've completed the lab, we'll take time out to see how the quadratic approximation relates to the second derivative test.