🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →

HomeClass Notes1,200,000CA670,000

MAT135H1 Lecture Notes - Lecture 27: Linearization, Thx

69 views2 pages

Verified Note

emeraldshark622

9 Mar 2019

School

Department

Course

Professor

Thaddeus Janisse

For unlimited access to Class Notes, a Class+ subscription is required.

MAT135H1 verified notes

MAT135H1 Lecture Notes - Lecture 26: Inverse Function, Lego Mindstorms

MAT135H1 Lecture Notes - Lecture 27: Linearization, Thx

MAT135H1 Lecture Notes - Lecture 28: Maxima And Minima, Thx

Document Summary

Where: y are related but still we find ttx rules think of treat y x y as a and can"t we function of function and x and the derivative . In this type of situation necessary y , for solve. 3. 9 linear the tangent approximation line approximation : is differentiable at suppose tangent f line approximation hast a fix ) to flash - a) a fix ) For values of x near a the is : We thinking of is a called the fixed . as local linearization of so that tea) and fca ) error. El x ) in the approximation is defined by : are. 11 x ) ekx linearization of ek " near. I t t tho ) ( x - o ) Suppose the tangent and local linearity . is line differentiable approximation. E ix ) fix ) flat at x -

Related Documents

MAT135H1 Lecture 15: Lecture Note

emeraldshark622

MAT135H1 Lecture Notes - Lecture 17: Thx

emeraldshark622

MAT135H1 Lecture Notes - Lecture 12: Ath, Constant Function

emeraldshark622

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.