Lecture 011

Partial Derivative

There can be many derivative on a surface. (because we have a tangent plane)

Partial Derivatives: the partial derivatives of z = f(x, y) is

  1. \frac{\partial z}{\partial x} = \lim_{h \rightarrow 0} \frac{f(x+h, y) - f(x, y)}{h} (with respect to x, treating all other variables constant) and
  2. \frac{\partial z}{\partial y} = \lim_{h \rightarrow 0} \frac{f(x, y+h) - f(x, y)}{h} (with respect to y, treating all other variables constant) provided limit exists. (with respect to x, sometimes denoted \frac{\delta z}{\delta x}, \frac{\delta f}{\delta x}, f_x, z_x)

Example: we set everything other than x direction constant

Higher Order Partial Derivatives

f_{xx} = \frac{\partial^2 z}{(\partial x)^2} f_{xy} = \frac{\partial^2 z}{(\partial x)(\partial y)} = \frac{\partial^2 z}{(\partial y)(\partial x)} = f_{yx} f_{yy} = \frac{\partial^2 z}{(\partial y)^2}

Clairaut's Theorem: if f(x, y) is defined on an open disk and f_{xy} and f_{yx} are continuous, then f_{xy} = f_{yx}

Tangent Plane: z(x, y) = f(x_0, y_0) + (f_x(x_0, y_0))(x - x_0) + (f_y(x_0, y_0))(y-y_0) is the equation of the tangent plane to S at P_0 = (x_0, y_0)

// TODO: what does it mean? Differentiability: all tangent lines at (x_0, y_0) are at the same tangent plane. If f(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + E(x, y) where \lim_{(x, y) \rightarrow (x_0, y_0)}\frac{E(x, y)}{\sqrt{(x-x_0)^2 + (y-y_0)^2}} = 0

2D Linear Approximation: y(x) \simeq f(x_0) + f'(x_0) (x-x_0) for x close to a 3D Linear Approximation: z(x, y) \simeq z(x_0, y_0) + z_x(x_0, y_0)(x - x_0) + z_y(x_0, y_0)(y - y_0)

(Total) Differentials: dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy is the differential of z = f(x, y) at (x_0, y_0) (where \Delta z = dz, \Delta x = dx, \Delta y = dy)

Higher Order Chain Rule

Tree Diagram for x=g(t), y=h(t)

Tree Diagram for x=g(t), y=h(t)

Chain Rule: f(g(t), h(t)) where f, g, h differentiable then \frac{d(f(g(t), h(t)))}{dt} = \frac{\partial f(g(t), h(t))}{\partial (g(t))} \cdot \frac{d(g(t))}{dt} + \frac{\partial f(g(t), h(t))}{\partial (h(t))} \cdot \frac{d(h(t))}{dt}

Example: A = \pi \cdot a \cdot b, find the percent change in area when a increase by 2\% and b increase by 1.5\%

\begin{align*} dA &= \frac{\partial A}{\partial a} \cdot da + \frac{\partial A}{\partial b} \cdot db\\ &= \pi b \cdot 0.02a + \pi a \cdot 0.015b\\ &= 0.035 \pi ab\\ &= 0.035 A\\ \frac{dA}{A} &= 3.5\%\\ \end{align*}

Implicit Differentiation

2D Implicit Differentiation: when equation is not a function, apply \frac{d}{dx} on both side to separate \frac{dy}{dx} = f(x, y) given point (x, y) on line. 3D Implicit Differentiation: to get \frac{\partial ?}{\partial z} differentiate both side with \frac{\partial}{\partial x}, \frac{\partial }{\partial y} and treat y, x as constant while write out \frac{\partial ?}{\partial z} instead of assuming it is 1.

Example 1:

Find tangent plane to x^2 + \frac{y^2}{9} + 2z^2 = 1 at point (\frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}}).

Let f = x^2 + \frac{y^2}{9} + 2z^2, g = 1.

  1. we find \frac{\partial z}{\partial x} by assuming only z depends on x and treat other variables (other than x, constant).
  2. we find \frac{\partial f}{\partial x} and \frac{\partial g}{\partial x} (usually g' = 0) in terms of x, y, z, \frac{\partial z}{\partial x}
\begin{align*} \frac{\partial}{\partial x} [x^2 + y^2/2 + 2z^2] &= \frac{\partial}{\partial x}[1]\\ 2x + 0 + 4z\frac{\partial z}{\partial x} &= 0\\ \frac{\partial z}{\partial x} &= -\frac{2x}{4z}\\ \frac{\partial z}{\partial x}|_{(\frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}})} &= \frac{1}{2}\\ \end{align*}

Similarly: we find \frac{\partial z}{\partial y} = -\frac{y}{18z} and get tangent plane by z = z(x_0, y_0) + z_x(x, y)(x - x_0) + z_y(x, y)(y - y_0)

Example 2:

Find tangent plane to x^2 + \frac{y^2}{9} + 2z^2 = 1 at point (\frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{3}{\sqrt{29}}).

Let f = x^2 + \frac{y^2}{9} + 2z^2 by assuming x, y, z independent. Find \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}. Then let f(x, y, z) = 1, this will reduce a dimension, WLOG let z depends on x, y, then

\begin{align*} \frac{\partial f(x, y, z(x, y))}{\partial x} &= \frac{\partial f}{\partial z} \frac{\partial z}{\partial x} + \frac{\partial f}{\partial x}\\ 0 &= \frac{\partial f(x, y, z)}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial f(x, y, z)}{\partial x}\\ \frac{\partial z}{\partial x} &= -\frac{\partial f(x, y, z) / \partial x}{\partial f(x, y, z) / \partial z}\\ \end{align*}

(partial derivative is composed with 2 routes)

// QUESTION: why do we assume x, y independent? significance? // QUESTION: do we have interdependence in calculus? // QUESTION: find partial derivative of leaf over leaf requires rearranging tree?

Example 3:

Find \frac{\partial z}{\partial x} for x^2 - 8xyz + y^2 +6z^2 = 0.

  1. Since we want \frac{\partial z}{\partial x}, we have to assume z depends on x. So we have z(x).
  2. Let f(x, y, z(x)) = x^2 - 8xyz + y^2 +6z^2, notice that f also depends on x.
  3. By chain rule


\begin{align*} x^2 - 8xyz + y^2 +6z^2 &= 0\\ \frac{\partial}{\partial x}(x^2 - 8xyz + y^2 +6z^2) &= 0\\ \frac{\partial f(x, y, z)}{\partial x} &= 0\\ \frac{\partial z(x)}{\partial x} &= 0\\ \frac{\partial z(x)}{\partial x} &= \frac{-\frac{\partial f(x, y, z(x))}{\partial x}}{\frac{\partial f(x, y, z(x))}{\partial z(x)}}\\ \frac{\partial z(x)}{\partial x} &= -\frac{f_x}{f_z}\\ \end{align*}

// TODO: 10.8 Notes

Directional Derivative

Directional derivative:

D_{\overrightarrow{u}} f(a, b) = \lim_{h \rightarrow 0} \frac{f(a + h \cos \theta, b + h \sin \theta) - f(a, b)}{h}

provided the limit exist and \overrightarrow{u} = \langle{\cos \theta, \sin \theta}\rangle (guaranteed to be unit vector because the property of \cos, \sin).

Partial Derivative:

\begin{cases} \frac{\partial f}{\partial x} = D_{\langle{1, 0}\rangle} f_{\theta = 0}\\ \frac{\partial f}{\partial x} = D_{\langle{1, 0}\rangle} f_{\theta = \frac{\pi}{2}}\\ \end{cases}

If \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} both defined:

\begin{align*} D_{\langle{\cos \theta, \sin \theta}\rangle}f &= \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta\\ &= \triangledown f \cdot \langle{\cos \theta, \sin \theta}\rangle\\ &= \|\triangledown f\| \|\langle{\cos \theta, \sin \theta}\rangle\| \cos \varphi\\ &= \|\triangledown f\| \cos \varphi \tag{where $\varphi$ is angle between gradient and directional derivative}\\ \end{align*}

As the result, the maximum direction derivative is the magnitude of gradient. Gradient is a 2D vector pointing in the direction of maximum increase and whose magnitude is its directional derivative


\triangledown f(x, y) = \langle{f_x(x, y), f_y(x, y)}\rangle


  1. \triangledown f(x_0, y_0) = \overrightarrow{0} \implies (\forall \overrightarrow{a})(D_{\overrightarrow{a}} f(x_0, y_0) = 0) and (x_0, y_0) is a critical point
  2. \triangledown f \neq \overrightarrow{0} \implies D_{\overrightarrow{u}} f(a, b) \text{ is maximized when } \overrightarrow{u} = \frac{\triangledown f}{\|\triangledown f\|}
  3. \triangledown f(x_0, y_0) is normal to the level curve f(x, y) = c of z = f(x, y) at point (x_0, y_0). (by \triangledown f(x_0, y_0) \cdot \langle{\frac{dx}{dt}, \frac{dy}{dt}}\rangle = 0)

// TODO: how to find the acceleration unit vector of a ball on a surface

Gradient of 3 variable: \triangledown f = \langle{f_x, f_y, f_z}\rangle

Directional: D_{\overrightarrow{u}}(f) = \triangledown f \cdot \overrightarrow{u} for \langle{\cos \alpha, \cos \beta, \cos \gamma}\rangle \text{ where } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1

Maximum, Minimum

Fermat's Theorem: If y = f(x) has a local maximum or local minimum at x_0, then x_0 is a critical point.

Classifying Critical Points

// TODO: intuition of Hessian Matrix More on Hessian Matrix

Hessian of f: H(f) = \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\ \end{pmatrix} when f(x, y) differentiable, and second order partial derivatives continuous

Second Derivative Test

Absolute Maxima and Minima (Extreme Value Theorem)

2D Extreme Value Thorem: Continuous function y = f(x) on closed, bounded interval [a, b] has an absolute maximum and an absolute minimum.

3D Extreme Value Thorem: Continuous function z = f(x, y) on a closed, bounded domain D has an absolute maximum and an absolute minimum on D.

Find Extrema:

Lagrouge Multiplier with 2 Constraints

Substitution: will be complicated.

Example: Maximize f(x, y) = \frac{1}{20}(28x + 96y - x^2 - 2xy - 9y^2) with constraint g(x, y) = 20x+4y = 216.

\begin{align*} \begin{cases} \triangledown f = \lambda \triangledown g\\ g(x, y)\\ \end{cases} \implies &\begin{cases} f_x(x, y) &= \lambda g_x(x, y)\\ f_y(x, y) &= \lambda g_y(x, y)\\ g(x, y) &= 216\\ \end{cases} \implies &\begin{cases} \frac{1}{20}(48 - 2x - 2y) &= \lambda 20\\ \frac{1}{20}(96 - 2x - 18y) &= \lambda 4\\ 20x + 4y &= 216\\ \end{cases} \end{align*}

Solution will give you critical candidate to be local maximum or minimum.

// QUESTION: is it possible instead of getting critical points, I get a critical line Yes. Possible.

Lagrouge Multiplier with 3 Constraints

Visual Proof of Lagrouge Multiplier

Optimize w = f(x, y, z) to satisfy \begin{cases} g(x, y, z) = c_1\\ h(x, y, z) = c_2\\ \end{cases}

Then we find critical point by:

\begin{cases} \triangledown f = \lambda_1 \triangledown g + \lambda_2 \triangledown h\\ g(x, y, z) = c_1\\ h(x, y, z) = c_2\\ \end{cases}

// WARNING: Lagrouge Multiplier does not guarantee there always be a maximum and minimum with 2 critical points. It was guaranteed by Extreme Value Thorem. There can be two local minimum. (For example, there exists no maximum distance between a point and a curve towards infinity)

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