Calculus III Exam 3 Review

• Partial Derivatives
• The Tangent Plane to a Surface in $\R^3$
• 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

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

## 弱项分析

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

## 公式整理

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

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

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

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

## 例题分析

### Differentials

#### Tangent Plane Sample Problem

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

$z=x^2+y^2 \rArr x^2+y^2-z=0$

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

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

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

$\vec {n}\cdot(\vec {r} - \vec {r_0}) = 0 \rArr <2,2,-1>\cdot=0$

End - of - File