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
Example: we set everything other than x direction constant
\frac{\partial z}{\partial x}|_{(x=a, y=b)} = z_x(x, y) |_{(x=a, y=b)}
so the tangent vector is \langle{1, 0, z_x(a, b)}\rangle
notice we replace everything other constant component with 0 (since we assume they don't change)
notice we replace the component we are taking derivative against with 1 (since we want our base fraction to be 1)
notice we relace the z component (output component) to be the derivative (since base is 1)
tangent line: \langle{x, y, z(x, y)}\rangle + t \langle{0, 1, z_x(a, b)}\rangle
tangent plane: plane formed by two tangent lines \overrightarrow{N}(\langle{x, y, z}\rangle - \overrightarrow{P}) = 0
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
Continuous but not differentiable: the derivative from two direction is not the same (a cusp).
Not Differentiable: if it is not differentiable in one direction, then it is not differentiable.
Differentiable: if f(x, y), f_x(x, y), f_y(x, y) all exists and continuous near (x_0, y_0), then z = f(x, y) is differentiable at (x_0, y_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)
a derivative line is total if the parent only depends on one variable
a derivative line is partial if the parent depends more than one variables
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\%
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.
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.
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)
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
(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?
Find \frac{\partial z}{\partial x} for x^2 - 8xyz + y^2 +6z^2 = 0.
// TODO: 10.8 Notes
Directional derivative:
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:
If \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} both defined:
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
\triangledown f = \overrightarrow{0} \implies (\forall \overrightarrow{a})(D_{\overrightarrow{a}}f = 0)
\triangledown f \neq \overrightarrow{0} \implies (D_{\overrightarrow{u}}f(a, b) \text{ is maximized over } \overrightarrow{u} \text{ when } \overrightarrow{u} = \frac{\triangledown f}{\| \triangledown f \|})
Theorem:
// 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
Fermat's Theorem: If y = f(x) has a local maximum or local minimum at x_0, then x_0 is a critical point.
// 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
D > 0 \land f_{xx} < 0 \implies (x_0, y_0) \text{ is local max}
D > 0 \land f_{xx} > 0 \implies (x_0, y_0) \text{ is local min}
D < 0 \implies (x_0, y_0) \text{ is a saddle point}
D = 0 \implies \text{inconclusive}
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.
Bounded: not infinite
Closed: boundary value included
Find Extrema:
Find critical points on the interror of D
Find critical points along the boundary of D (both endpoint of the line and interror of the line)
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.
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.
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:
// 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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