Calculus III Exam 3 Review

这学期的微积分已经学了很多知识啦,这学期的第三次考试应该会是最难的一次,所以 Kiyosh 决定好好地复习一下。
从学过的知识和考过的考点来看,这次考试会包含 Exploration 8 - 14 中的内容,主题如下:

  • Partial Derivatives
  • The Tangent Plane to a Surface in R3\R^3
    • Implicit Differentiation
  • The Directional Derivative
  • Extreme Value Problems for Funtions of Two Variables
    • Second Derivative Test
    • Extreme Value Theorem
    • Extreme Value Problems with Constraints: Lagrange Multipliers
  • Double Integrals and Iterated Integrals
    • Double Integrals Over Non-Rectangular Regions

教授已经发布了 Exam 3 Review Problems,可以从题目下手,再针对弱项复习。
课堂中教授提供的复习大纲:

  • Differentials
  • Find Linearization of A Function f(x,y)f(x,y) at Some Point (x0,y0)(x_0, y_0)
  • Max/Min (Optimization) for f(x,y)f(x,y)
    • Second Derivative Test
    • Extreme Values Over a Closed Bounded Region
  • Lagrangian
  • Directional Derivative

所以在实际考试中并没有积分部分。

弱项分析

在做 Review Problems 时,发现自己对于微分部分的知识掌握有问题,其他部分的知识主要是公式的忘记,基本的道理还是懂的。这一次复习应该着重复习以下知识点:

  • Differentials and Linear Approximation
    • Linearization Formula (Tangent Planes to Surfaces)
  • Optimization Problems
    • Second Derivative Test
    • Lagrangian

根据以上内容各个击破!

公式整理

  • Linearization of ff: L(x,y)=f(r0)(rr0)+f(r0)L(x,y)=\vec {\nabla}f(\vec {r_0}) \cdot (\vec {r} - \vec {r_0})+f(\vec {r_0})

  • Directional Dericative: Du^f(r0)=f(r0)u^D_{\hat{u}}f(\vec{r_0})=\vec{\nabla}f(\vec{r_0})\cdot\hat{u}

  • Second Derivative Test Discriminant: D=fxx(a,b)fxy(a,b)fyx(a,b)fyy(a,b)=fxx(a,b)fyy(a,b)fxy2(a,b)D = \begin{vmatrix} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b) \end{vmatrix} = f_{xx}(a,b)f_{yy}(a,b) - f^2_{xy}(a,b)

    • If D>0D > 0
      • fxx(a,b)<0f_{xx}(a,b) < 0, then f(a,b)f(a,b) is a local maximum.
      • fxx(a,b)>0f_{xx}(a,b) > 0, then f(a,b)f(a,b) is a local minimum.
    • If D<0D < 0, then f(a,b)f(a,b) is a saddle point.
    • If D=0D = 0, the test fails.
  • Lagrangian: L={fx(x,y)=λgx(x,y)fy(x,y)=λgy(x,y)g(x,y)=k\vec{\nabla}\mathscr{L} = \begin{cases} f_x(x,y) = \lambda g_x(x,y) \\ f_y(x,y) = \lambda g_y(x,y) \\ g(x,y) = k \end{cases}

例题分析

Differentials

Tangent Plane Sample Problem

Find the equation of tangent plane to z=f(x,y)=x2+y2z=f(x,y)= x^2+y^2 at point (1,1,2)(1,1,2) on surface.
Solution
Need a normal vector to the surface z=f(x,y)z=f(x,y) at point (1,1,2)(1,1,2)

z=x2+y2x2+y2z=0z=x^2+y^2 \rArr x^2+y^2-z=0

Note: z=x2+y2z=x^2+y^2 is a level set (contour) of F(x,y,z)=x2+y2zF(x,y,z)=x^2+y^2-z (when the value of the function is zero).

F Surface z=x2+y2 at (1,1,2)\vec{\nabla}F \perp {\rm ~Surface~} z=x^2+y^2 {\rm ~at~} (1,1,2)

Use F(1,1,2)=n\vec{\nabla}F(1,1,2)=\vec{n}, we have F=<2x,2y,1>\vec{\nabla}F=<2x,2y,-1>
n=F(1,1,2)=<2,2,1>\vec {n}=\vec{\nabla}F(1,1,2)=<2,2,-1> points to the tangent plane
and we know r0=<1,1,2>\vec{r_0}=<1,1,2>
Now we have

n(rr0)=0<2,2,1><x1,y1,z2>=0\vec {n}\cdot(\vec {r} - \vec {r_0}) = 0 \rArr <2,2,-1>\cdot<x-1,y-1,z-2>=0

后记

很厉害的事情发生了就。因为 Kiyoshi 写 Blog 写得太慢了,而且在考试前两天才开始写,所以已经考完了但是都还没有写完这一篇 hhhh
所以就这样了!

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文章作者: Kiyoshi
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