Math 119 - Weeks 1-5 Combined

Directional derivative =

The dir. derivative at point a→ = (a, b) in the direction of UNIT VECTOR u→

D_u→ f(a, b) = ∇f(a→) · u→

Tangent plane using gradient vector =

Eqn of tangent plane to the level surface at the point a→ = (a, b, c) can be expressed as

∇f(a→) · (x→ − a→) = 0

How to find rate of change + special cases using dir. derivatives =

At point a→, rate of change = ‖∇f(a→)‖ × cos θ

Steepest ascent/increase = ||∇f||

cos θ = 1

Steepest descent/decrease = -||∇f||

cos θ = -1

No change = 0, when ⟂∇f(a→)

cos θ = 0

ex.) ∇f(a→) = [6, 6], then ⟂∇f(a→) = [6, -6] or [-6, 6]

Define critical point =

∇f(a, b)→ = (0, 0)

What does it mean if D(a, b) > 0 and f_xx(a, b) > 0?

(a, b) is a local minimum of f.

What does it mean if D(a, b) > 0 and f_xx(a, b) < 0?

(a, b) is a local maximum of f.

What does it mean if D(a, b) < 0?

(a, b) is a saddle point.

What does it mean if D(a, b) = 0?

The second derivative test is inconclusive, and (a, b) is a degenerate critical point.

Determinant of Hessian Matrix =

D(x, y) = f_xx(x, y) · f_yy(x, y) − [f_xy(x, y)]².

How to find abs max/min =

Find critical points f_x = 0 and sub into f_y = 0 to get critical pts → evaluate f at crit points

For boundary points: substitute constraint → differentiate the 1D function, set = 0, check endpoints → evaluate f at boundary points

Abs max/min will be the largest/smallest values

Gradient vector =

A vector with all single order partial derivatives, acts as the derivative for multi-variable functions

Given function f :

∇f = [∂f/∂x, ∂f/∂y]

Multi variate chain rule equation =

∇f(r(t))*r'(t)

or

dz/dt = ∂z/∂x(dx/dt) + ∂z/∂y(dy/dt)

What does parametrizing curves do =

Allows curves to be split up into multiple single-variable functions, expressed as a path travelled over time instead of a static curve

Tangent plane approximation =

To approximate (x, y, z)

df = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy

Where (x_0, y_0, z_0) is the centre of approx (easy value to calculate)

dx = (x - x0) and dy = (y - y0)

z = z_0 + df

How to calculate greatest percentage error =

Given V = πr^2h and r has max 2% error and h has max 0.5% error

Calculate dV/V and sub in dr/r and dh/h error values into the inequality for maximum

dV/V = 2dr/r + dh/h <= 0.045 or 4.5% error max

differential formula =

Let f(x,y) be the function and with the point (a, b)

df = f_x(a, b)dx + f_y(a, b)dy

Change-of-variable formula (double integral) =

∬_{D_xy} f(x, y) dxdy = ∬_{D_uv} f(g(u, v), h(u, v)) det(J) dudv

Where J = [∂(x, y)] / [∂(u, v)],

det(J) = (∂x/∂u)(∂y/∂v) - (∂y/∂u)(∂x/∂v)

ONLY IF J = [∂(x, y)] / [∂(u, v)] ≠ 0

Inverse property of Jacobian =

[∂(x, y)] / [∂(u, v)] = 1 / [(∂(u, v)) / (∂(x, y))]

What does dxdy become after change of variables?

dxdy = det([∂(x, y)] / [∂(u, v)]) dudv

Jacobian of transformation for polar coords =

dxdy = rdrdθ

Change-of-variable for polar coords =

∬_{D_xy} f(x, y) dxdy = ∬_{D_rθ} f(rcosθ, rsinθ) rdrdθ

Polar coordinates =

(r, θ)

Where r = radius from origin, θ = angle with positive x-axis

Radius in polar coordinates ALWAYS =

Positive

Contour plot =

Take multiple level curves to construct the full function

Always 1 dimension below actual function dimension

Level curve =

A "slice" of a 3D multi-variable function at a specific constant/height to get a 2D image

Clairaut's Theorem =

Let f_x and f_y be partial derivatives

if f_x and f_y and f_xy exists near (a, b) AND f_xy is continuous near (a, b) then f_yx must also exist

also: f_xy(a, b) = f_yx(a, b)

Tangent plane equation =

z = z_0 + [f_x(x_0, y_0)](x - x_0) + [f_y(x_0, y_0)])(y - y_0)

Where (x_0, y_0, z_0) is the point of tangency

Define mini-max/saddle point =

point that can be local max or min depending on the angle

partial derivative =

∂

Derivative of one variable in a multi-variable function, rest of the variables are constant

Lagrange Multiplier formula and purpose =

λ = lagrange multiplier. Optimizes a function constrained by a function g(x, y) = k ONLY.

∇f→ = λ∇g→ ONLY IF ∇g→ ≠ 0→

How to find abs max/min with lagrange multipliers =

IF INEQUALITY: Find critical points ∇f→ = 0→ and sub into f(x, y). (Must satisfy constraint)

Solve ∇f→ = λ∇g→ and g(x, y) = k to get points, sub in f(x, y)

Solve ∇g→ = 0→ and g(x, y) = k, sub in f(x, y) [IF THERE EXISTS SOLUTIONS]

If the constraint g(x, y) = k has endpoints, evaluate f(x, y) at these points

max = largest computed f

min = smallest computed f

Average value of a 2D function over a region D =

[∬_D f(x, y) dA] / [∬_D dA]

If D is difficult to integrate whole (∬_D f dA), what to do =

Split up into two non overlapping regions.

∬_D f dA = ∬_D₁ f dA + ∬_D₂ f dA

ONLY IF D₁ ∩ D₂ = ∅ and D = D₁ ∪ D₂

Plane intercept formula =

[x / a] + [y / b] + [z / c] = 1

Line intercept formula =

[x / a] + [y / b] = 1

How to calculate volume under a surface z =

Given z = h(x, y) and region D in the xy-plane

∬_D h(x, y) dx dy

ONLY IF h(x, y) ≥ 0

If m ≤ f(x, y) ≤ M on region D =

m ∬_D dA ≤ ∬_D f(x, y) dA ≤ M ∬_D dA

also: mA ≤ ∬_D f(x, y) dA ≤ MA