14.1 Functions of Several Variables #
 In the real world, most things don’t depend on a single variable
 Temperature may depend on the .$(x,y)$ (latitude/longitude) position
 Volume of a cylinder depends on radius .$r$ and height .$h$: .$V=\pi r^2 h$
 Formal Definition: A function .$f$ of two variables is a rule that assigns to each ordered pair of real numbers .$(x,y)$ in a set .$D$ a unique real number denoted by .$f(x,y)$. The set .$D$ is the domain of .$f$ and its range is the set of values that .$f$ takes on, that is, .$\{f(x,y)\ \ (x,y) \in D \}$.
 E.x. for .$f(x,y) = \frac{\sqrt{x+y+1}}{x1}$, the domain is .$D= \{(x,y)\ \ x + y + 1 \geq 0, x \neq 1\}$ which can be graphed with a solid line following .$y=x1$ with a dotted line at .$x=1$
 E.x. for .$f(x,y) = x\ln(y^2x)$, the domain is .$D=\{(x,y)\ \ x<y^2\}$. This can be graphed with a dotted line following the curve .$x = y^2$
 For the equation .$z = f(x,y)$, .$x,y$ are the independent variables and .$z$ is the dependent variable – similar to single variable equations
 We can visualize functions of two variables (i.e .$f(x,y)$) by graphing them in 3D as .$(x,y,f(x,y))$
 We can then write level curves for the function by setting .$f(x,y) = k$ for some .$k$onstant in the range of .$f$. This will result in a graph similar to a
Topographic map
Level Curve: The level curves of a function .$f$ of two variables are the curves with equations .$f(x,y) = k$, where .$k$ is a constant (in the range of .$f$).
 We can then write level curves for the function by setting .$f(x,y) = k$ for some .$k$onstant in the range of .$f$. This will result in a graph similar to a
Topographic map
14.2 Limits and Continuity #
Limits #
Vector Limit: Let .$f$ be a function of two variables whose domain .$D$ includes points arbitrarily close to .$(a, b)$. Then we say that the limit of .$f(x, y)$ as .$(x, y)$ approaches .$a, b.$ is .$L$ and we write $$\lim_{(x,y)\to(a,b)} f(x,y) = L$$ if for every number .$\varepsilon < 0$ there is a corresponding number .$\delta < 0$ such that if .$(x, y) \in D$ and .$0 < \sqrt{(xa)^2 + (yb)^2} < \delta$ then .$\vert f(x,y)  L \vert < \varepsilon$
 Notice that .$\vert f(x,y)  L \vert$ is the distance between the numbers .$f(x, y)$ and .$L$, and .$\sqrt{(xa)^2 + (yb)^2}$ is the distance between the point .$(x, y)$ and the point .$(a, b)$.
 If .$f(x,y) \to L_1$ as .$(x,y) \to (a,b)$ along a path .$C_1$ and .$f(x,y) \to L_2$ as .$(x,y) \to (a,b)$ along a path .$C_2$, where .$L_1 \neq L_2$, then .$\lim_{(x,y)\to(a,b)} f(x,y)$, does
not exist.
 We can test this by setting .$x$ and .$y$ to various different values (e.x. .$x=0, y=0, x=y, …,$ etc)
Continuity #
A function .$f$ of two variables is called continuous at .$(a,b)$ if $$\lim_{(x,y)\to (a,b)} f(x,y) = f(a,b)$$ We say .$f$ is continuous on .$D$ if .$f$ is continuous at every point .$(a,b)$ in .$D$.
 That is, we need the limit to exist ands for .$f(a,b)$ to be defined
 Continuous functions: .$x,y, c, \text{ trig (on domain)}$
 Arithmetic, composition, exponent all preserve continuity (on domain!)
 Dividing doesn’t necessarily preserve continuity
14.3 Partial Derivatives #
If .$f$ is a function of two variables, its partial derivatives are the functions .$f_x$ and .$f_y$ defined by $$f_x(x,y) = \frac{\delta f}{\delta x} = \lim_{h\to0} \frac{f(x+h,y)f(x,y)}{h}$$ $$f_y(x,y) = \frac{\delta f}{\delta y} = \lim_{h\to0} \frac{f(x,y+h)f(x,y)}{h}$$
 Notice we use .$\delta$ instead of .$d$ for partial derivatives
 These can be written at a single point .$(a,b)$ with respect to .$x$ and .$y$ by treating the remaining variables as a constant $$f_x(a,b) = g'(a); \ \ \ g(x) = f(x,b)$$ $$f_y(a,b) = h'(b); \ \ \ h(y) = f(a,y)$$
 Which can be extrapolated for 3 (or more) variables: $$f_z(a,b,c) = k'(c); \ \ \ k(z) = f(a,b,z)$$
Higher Derivatives #
 Just like regular derivatives, we can do many partial derivatives
 For example, the following are second partial derivatives of .$z=f(x,y)$: $$(f_x)_x = f_{xx} = \frac{\delta }{\delta x}\bigg(\frac{\delta f}{\delta x}\bigg) = \frac{\delta^2f}{\delta x^2}$$ $$(f_x)_y = f_{xy} = \frac{\delta }{\delta y}\bigg(\frac{\delta f}{\delta x}\bigg) = \frac{\delta^2f}{\delta y \delta x}$$ $$(f_y)_x = f_{yx} = \frac{\delta }{\delta x}\bigg(\frac{\delta f}{\delta y}\bigg) = \frac{\delta^2f}{\delta x \delta y}$$ $$(f_y)_y = f_{yy} = \frac{\delta }{\delta y}\bigg(\frac{\delta f}{\delta y}\bigg) = \frac{\delta^2f}{\delta y^2}$$
Clairautâ€™s Theorem Suppose .$f$ is defined on a disk .$D$ that contains the point .$(a,b)$. If the functions .$f_{xy}$ and .$f_{yx}$ are both continuous on .$D$, then $$f_{xy}(a,b) = f_{yx}(a,b)$$
Partial Differential Equations #
 In the sciences, we typically want to find how a system changes with respect to multiple variables
 Partial derivatives occur in partial differential equations, e.x Laplace’s Equation:
$$ \frac{\delta^2 u}{\delta x^2} + \frac{\delta^2 u}{\delta y^2} = 0$$
 Solutions to Laplace’s are always harmonic functions, such as .$u=e^x \sin(y)$
14.4 Tangent Planes and Linear Approximations #
Tangent Planes #
 Tangent planes are to surfaces as tangent lines are to curves
 Tangent planes contain both the partial derivative lines w.r.t .$x$ and .$y$
 All we need to know to write a tangent plane is a point .$(a,b)$ and direction of the two partial derivatives .$\langle 1,0, f_x(a,b)\rangle, \langle 0,1, f_y(a,b)\rangle$
 Direction vector can be found with .$\langle 1,0, f_x(a,b)\rangle \times \langle 0,1, f_y(a,b)\rangle = \langle f_x, f_y, 1 \rangle$ which we can dot with .$\langle xx_0, yy_0, zf(x_0,y_0) \rangle$
Suppose .$f$ has continuous partial derivatives. An equation of the tangent plane to the surface .$z = f(x, y)$ at the point .$P(x_0,y_0,z_0)$ is $$zf(x_0, y_0) = f_x(x_0,y_0)(xx_0) + f_y(x_0, y_0)(yy_0)$$
 Direction vector can be found with .$\langle 1,0, f_x(a,b)\rangle \times \langle 0,1, f_y(a,b)\rangle = \langle f_x, f_y, 1 \rangle$ which we can dot with .$\langle xx_0, yy_0, zf(x_0,y_0) \rangle$
Linear Approximations #
 As we get very close to the surface, then the tangent plane (at point .$(a,b)$) looks more and more like the surface
 Thus, we can use it to for making approximations when we are near near .$(a,b)$
If .$z=f(x,y)$, then .$f$ is differentiable at .$(a,b)$ if .$\Delta z$ can be expressed in the form $$\Delta z = f_x(a,b)\Delta x + f_y(a,b)\Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y$$ where the error terms, .$\varepsilon_1$ and .$\varepsilon_2 \to 0$ as .$(\Delta x, \Delta y) \to (0,0)$ and the other terms are the linearization of the function.
 Thus, we can use it to for making approximations when we are near near .$(a,b)$
 Rewriting this, we can use the given .$f(x,y)$ to write the linear approximation which is an estimate for point/state at .$f(a,b)$ $$f(x,y) \approx f(a,b) + f_x(a,b) (xa) + f_y(a,b)(yb)$$
 We can similarly define the 3D linear approximation, increment of .$w$, and differential .$dw$ as: $$f(x,y,z) \approx f(a,b,c) + f_x(a,b,c) (xa) + f_y(a,b,c)(yb) + f_z(a,b,c) (zc)$$ $$\Delta w = f(x+\Delta x, y + \Delta y, z + \Delta z)  f(x,y,z)$$ $$dw = \frac{\delta w}{\delta x}dx + \frac{\delta w}{\delta y}dy + \frac{\delta w}{\delta z}dz$$
Differentials #
 With one variable functions, i.e. .$y=f(x)$, we defined the differential .$dx$ to be independent so we had to write .$dy$ as .$dy = f'(x)\ dx$
 Given a differentiable function of two variables, i.e. .$z=f(x,y)$, we know both .$dx$ and .$dy$ are independent so we write: $$dz = f_x(x,y)\ dx + f_y(x,y)\ dy = \frac{\delta z}{\delta x}dx + \frac{\delta z}{\delta y}dy$$
 If the partial derivatives .$f_x$ and .$f_y$ exist near .$(a,b)$ and are continuous at .$(a,b)$, then .$f$ is differentiable at .$(a,b)$.
14.5 Chain Rule #
Chain Rule #
Suppose that .$z=f(g_1 (x_1,_{\dots}, x_m), g_2 (x_1,_{\dots}, x_m), g_n (x_1,_{\dots}, x_m))$ is a differentiable function of the .$n$ variables .$g_1,_{\dots}, g_n$ and is .$g_j$ is a differentiable function of the .$m$ variables .$x_1,_{\dots}, x_m$. Then .$z$ is a function of .$x_1,_{\dots}, x_m$ and $$ \frac{\delta f}{\delta x_i} = \frac{\delta f}{\delta g_1} \frac{\delta g_1}{\delta x_i} + \frac{\delta f}{\delta g_2} \frac{\delta g_2}{\delta x_i} + \dots + \frac{\delta f}{\delta g_m} \frac{\delta g_m}{\delta x_i}$$ for each .$i= 1,2,\dots,m$
Implicit Differentiation #

If .$F(x,y) = 0$ defines .$y$ implicitly as a function of .$x$ (that is, .$y = f(x)$, where .$F(x,f(x))=0$ for all .$x$ in the domain of .$f$), then $$ \frac{dy}{dx} =  \frac{\frac{\delta F}{\delta x}}{ \frac{\delta F}{\delta y}} =  \frac{F_x}{F_y}$$

If .$F(x,y,z) = 0$ defines .$z$ implicitly as a function of .$x,y$, then
$$ \frac{dz}{dx} =  \frac{\frac{\delta F}{\delta x}}{ \frac{\delta F}{\delta z}} =  \frac{F_x}{F_z};\ \ \frac{dz}{dy} =  \frac{\frac{\delta F}{\delta y}}{ \frac{\delta F}{\delta z}} =  \frac{F_y}{F_z}$$
14.6 Directional Derivatives and the Gradient Vector #
Directional Derivatives #
Direction Derivative For function .$f$ at .$(x_0, y_0)$ in the direction of unit vector .$\hat u = \langle a, b \rangle$ is $$D_{\hat u} f(x_0, y_0) = \lim_{h\to 0} \frac{f(x_0 + ha, y_0 + hb)  f(x_0, y_0)}{h} = f_x (x,y) a + f_y (x,y) b$$ if the limit exists (for the former) and if .$f$ is a differentiable function of .$x$ and .$y$ (for the latter)
 That is, .$\hat u = \hat i = \langle 1, 0 \rangle$ for .$D_\hat{i} = f_x$ and .$\hat u = \hat j = \langle 0, 1 \rangle$ for .$D_\hat{j} = f_y$
 Differentiability is important because it means that as you approach the surface very closely, it looks more and more like the tangent plane.
 Directional derivatives can be thought of as the slope of the tangent line at a given point
 This definition can be extrapolated to (three variables/higher dimensions) trivially, shown below with gradient vectors
Gradient Vector #
Gradient: If .$f$ is a function of variable .$x,y,z$, then the gradient of .$f$ is the vector function .$\nabla f$ defined by $$\nabla f(x,y,z) = \big\langle f_x(x,y,z) + f_y (x,y,z) + f_z (x,y,z) \big\rangle = \frac{\delta f}{\delta x}\hat i + \frac{\delta f}{\delta y}\hat j + \frac{\delta f}{\delta z}\hat k$$
 We can now use the gradient to rewrite our directional derivative equation as $$D_{\hat u} f(x,y,z,\dots) = \nabla f(x,y,z,\dots) \cdot \hat u$$
 We can rewrite this using the definition of the dot product as $$D_{\hat u} f(x,y,z,\dots) = \Vert \nabla f \Vert \Vert \hat u \Vert \cos\theta =\Vert \nabla f \Vert \cos\theta $$
Maximizing the Directional Derivative #
Suppose .$f$ is a differentiable function of two or three variables. The maximum value of the directional derivatives .$D_\hat{u} f(\vec x)$ is .$\Vert \nabla f(\vec x ) \Vert$ and it occurs when .$\hat u$ has the same direction as the gradient vector .$\nabla f(\vec x)$
 We can see from the definition above that since the max of .$\cos\theta$ is .$1$ when .$\theta = 0$, therefore the max of the directional derivative occurs at the same angle (when .$\hat u$ has the same direction of .$\nabla f$ )
 TL;DR: The maximal value of .$D_\hat{u} f(\vec x)$ is .$\Vert \nabla f \Vert$
Tangent Planes to Level Surfaces #
 If .$f(x,y) = k$ is a curve, then .$F(x,y,z) = k$ is a surface
 Let .$\vec r(t)$ be a space curve on the surface .$F(x,y,z) = k$
 Let .$\vec r(t_0) = \langle x_0, y_0, z_0 \rangle$ for some .$t_0$ (at some time, the space curve passes through some point which is on the surface)
 We know .$F(\vec r(t)) = k$, thus, using the chain rule: $$0 = \frac{\delta F}{\delta x} \frac{\delta x}{\delta t} + \frac{\delta F}{\delta y} \frac{\delta y}{\delta t} + \frac{\delta F}{\delta z} \frac{\delta z}{\delta t} = \nabla F(x_0, y_0, z_0) \cdot (\vec r(t_0))'$$
 The gradient vector at .$P$, .$F(x_0, y_0, z_0)$, is perpendicular to the tangent vector .$\vec r'(t)$ to any curve .$C$ on .$S$ that passes through .$P$
 In English: The direction of the gradient is always perpendicular to the level surface at every point
 We can then define the tangent plane to the level surface .$F(x,y,z) = k$ at .$P(x_0, y_0, z_0)$ as the plane that passes through .$P$ and has normal vector .$\nabla F(x_0, y_0, z_0)$: $$\nabla F\big\vert_{(x_0, y_0, z_0)} \cdot \langle xx_0, yy_0, zz_0 \rangle = 0$$
 We can also write this in the symmetric equation form: $$\frac{xx_0}{F_x(x_0, y_0, z_0)} = \frac{yy_0}{F_y(x_0, y_0, z_0)} = \frac{zz_0}{F_z(x_0, y_0, z_0)}$$
14.7 Maximum and Minimum Values #
Critical Points #
A function of two variables has a local maximum at .$(a,b)$ if .$f(x,y) \leq f(a,b)$ when .$(x,y)$ is near .$(a,b)$ [This means that for .$f(x,y) \leq f(a,b)$ for all points .$(x,y)$ in some disk with center .$(a,b)$.] The number .$f(a,b)$ is called a local maximum value. If .$f(x,y) \geq f(a,b)$ when .$(x,y)$ is near .$(a,b)$, then .$f$ has a local minimum at .$(a,b)$ and .$(x,y)$ is a local minimum value.
 If the inequalities above hold for all points .$(x,y)$ in the domain of .$f$, then .$f$ has an absolute maximum (or absolute minimum) at .$(a,b)$
 If .$f$ has a local maximum or minimum at .$(a,b)$ and the firstorder partial derivatives of .$f$ exist there, then .$f_x(a,b) = 0$ and .$f_y(a,b) = 0$.
 If the graph of .$f$ has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal
 .$(a,b)$ is a Critical Point of .$f$ if .$f_x(a,b) = 0$ and .$f_y(a,b) = 0$, or if one of these partial derivatives does not exist
 Thus, if .$f$ has a local maximum or minimum at .$(a,b)$, then .$(a,b)$ is a critical point of .$f$
 In other words, at a critical point, a function could have a local maximum or a local minimum or neither (saddle point).
Second Derivatives Test: Suppose the second partial derivatives of .$f$ are continuous on a disk with center .$(a, b)$, and suppose that .$f_x(a, b) = 0$ and .$f_y(a, b) = 0$ (that is, .$(a, b)$ is a critical point of .$f$ ). Let $$D = D(a, b) = \begin{vmatrix}f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\end{vmatrix} = f_{xx}(a, b) f_{yy}(a, b)  (f_{xy}(a, b))^2$$
 If .$D > 0$ and .$f_{xx}(a, b) > 0$ (or .$f_{yy}(a, b) > 0$), then .$f(a, b)$ is a local minimum.
 If .$D > 0$ and .$f_{xx}(a, b) < 0$ (or .$f_{yy}(a, b) < 0$), then .$f(a, b)$ is a local maximum.
 If .$D < 0$, then .$f(a, b)$ is a saddle point
 If .$D = 0$, the test gives no information: .$f$ could be any of the above
 Note that in the first two tests, it’s implied that .$f_{xx}$ and .$f_{yy}$ have the same sign
 The determinant is a Hessian matrix
Extreme Points #
 Just as a closed interval contains its endpoints, a closed set in .$\mathbb{R}^2$ is one that contains all its boundary points.
 For instance, the disk .$D = \{(x,y) \vert x^2 + y^2 \leq 1 \}$ is a closed set because it contains all of its boundary points (which are the points on the circle .$r=1$)
 But if even one point on the boundary curve were omitted, the set would not be closed
 A bounded set in .$\mathbb{R}^2$ is one that is contained within some disk – it is finite in extent
Extreme Value Theorem for Functions of Two Variables: If .$f$ is continuous on a closed, bounded set .$D$ in .$\mathbb{R}^2$, then .$f$ attains an absolute maximum value .$f(x_1, y_1)$ and an absolute minimum value .$f(x_2, y_2)$ at some points .$(x_1, y_1)$ and .$(x_2, y_2)$ in .$D$.
 If .$f$ has an extreme value at .$(x_1, y_1)$, then .$(x_1, y_1)$ is either a critical point of .$f$ or a boundary point of .$d$.
 To find the absolute maximum and minimum values of a continuous function .$f$ on a closed, bounded set .$D$:
 Find the values of .$f$ at the critical points of .$f$ in .$D$.
 Find the extreme values of .$f$ on the boundary of .$D$.
 The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
14.8 Lagrange Multipliers #
 We use Lagrange Multipliers to find critical points of a surface .$f$ given some constraining surface .$g$
Method of Lagrange Multipliers To find the maximum and minimum values of .$f(x,y,z)$ subject to the constraint .$g(x,y,z) = k$ [assuming that these extreme values exist and .$\nabla \neq 0$ on the surface .$g(x,y,z) = k$]:
 Find all values of .$x, y, z$, and .$\lambda$ such that $$\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$$ $$g(x,y,z) = k$$
 Evaluate .$f$ at all the points .$(x, y, z)$ that result from the first step. The largest of these values is the maximum value of .$f$; the smallest is the minimum value of .$f$.
 We can decompose the first equation and use the second equation to get $$f_x = \lambda g_x;\ \ f_y = \lambda g_y;\ \ f_z = \lambda g_z;\ \ g(x,y,z) = k$$
 Notice we don’t care what .$\lambda$ is, only that it exists
Two Constraints #
 We can use Lagrange multipliers for two constraints .$f$ and .$g$ too: $$\nabla f(x_0, y_0, z_0) = \lambda \nabla g(x_0, y_0, z_0) + \mu \nabla h(x_0, y_0, z_0)$$
 Likewise, we can decompose the equation above to get the following five equations:
 $$f_x = \lambda g_x + \mu h_x$$
 $$f_y = \lambda g_y + \mu h_y$$
 $$f_z = \lambda g_z + \mu h_z$$
 $$g(x,y,z) = k$$
 $$h(x,y,z) = c$$