Final Flashcards
Unit Vector
V / |V|
Vector of a certain length in the direction of another vector
(V / |V| ) * length
Angle Between Two Vectors
cos(x) = (a.b) / |a||b|
Projection of Vector U onto V (vector)
[V.U / |V|^2 ] * V
Projection of Vector U onto V (scalar)
V.U / |V|^2
Area of a Parallelogram
|U * V|
Volume of a Parallelpipid
(a * b) . c
Requirements for Vectors Being Parallel
1) a = <b>c OR
2) a * b = 0 OR
3) a.b = |a||b|</b>
Vectors are Coplanar
Triple Scalar Product = 0
Parametric
<x,y,z> = <xo, yo, zo> + t<a,b,c>
x = xo + at
y = yo + bt
z = zo + ct
Symmetric
t = (x-xo)/a = (y-yo)/b = (z-zo)/c
If a=0, then x=xo
Length of a Curve
L = (int)_a^b |r’(t)|dt OR
L = (int)_a^B sqrt(f’(t)^2+g’(t)^2+h’(t)^2)dt
Re-parametrize the curve using the length formula
- Set the found length equal to s
- Solve for t
- Plug t into the original r(t) equation
Normal Vector
Of Two Parallel Planes: The slopes
Of Two Vectors: Cross product
Binormal Vector
T(t) = r’(t) / |r’(t)|
Projectile Motion
a(t) = -gj
Equation of a Tangent Plane
z = zo + fx(xo,yo)(x-xo) + fy(xo,yo)(y-yo)
Area of Triangle
A = 1/2 |B * H| (CROSS PRODUCT)
Orthogonal when
dot product = 0
Gradient
F = <fx, fy, fz>
Gradient at point p
find gradient then plug in values
Directional Derivative
Gradient * u
Directional Derivative with given angle
Duf(x, y) = fx(x, y) cos(𝜃) + fy(x, y) sin(𝜃),
Implicit Differentiation
(Df/Dx) * (Dx/Du) + (Df/Dy) * (Dy/Du) (change all values to the given value at the end)
Max Value
D(a,b) > 0 and fxx(a,b) < 0
(a,b) is a maximum
Min Value
D(a,b) > 0 and fxx(a,b) > 0
(a,b) is a mininum
How to find min and max values
Critical Point is when
the gradient of f is equal to 0 OR the partial derivative does not exist
D(a,b) = 0 means
there is not enough information
How to find critical points
Find the partial derivative of each (x,y,z), then set the equation equal to zero.
How to find Min and Max values
D(a, b) = fxx(a, b)fyy(a, b) − (fxy(a, b))2
D(a,b) is a saddle point when
D(a, b) < 0
The procedure to find the maximum and minimum values of f subject to
g(x, y) = k is the following:
- Find all x, y, λ such that ∇f = λ∇g and g(x, y) = k.
- Evaluate f at these points (x, y) and choose the smallest and largest values.
The number λ is called a Lagrange multiplier.
Equation of the tangent plane to the level surface F (x, y, z) = k at P (x0, y0, z0)
is
Fx (x0, y0, z0) (x − x0) + Fy (x0, y0, z0) (y − y0) + Fz (x0, y0, z0) (z − z0) = 0
Equation of the normal line at P (x0, y0, z0) is
(x − x0) / (Fx (x0, y0, z0)) = (y − y0) / (Fy (x0, y0, z0)) = (z − z0) / (Fz (x0, y0, z0)
Line Integral
int f (x(t), y(t))|r′(t)| dt
Line Integral in Space (3D)
int f(x(t), y(t), z(t)) √(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 dt
Work Done
int F(r (t)) * |r′ (t)| dt
Fundamental Theorem for Line Integrals
int ∇f · dr = f (r(b)) − f (r(a)).
F is a conservative vector field if there is a function f such that
F = ∇f
Green’s Theorem
double int (∂Q/∂x) − (∂P/∂y) dA.
Curl
curlF = ∇F × F
Vector Field F is conservative if
curl
