Calc 3 - everything. :(

angle between vectors

cos(theta) = dot product divided by product of the magnitudes

area of a triangle with three points

make two vectors, and multiply .5 by the magnitude of their cross products

equation for tangent LINE, given a curve r(t)

for x y and z individually r(point given) + r’(point given)*t

arc length of curve r(t)

integral of magnitude of r’(t) dt

equation of tangent plane

take partial derivatives and plug in points. z - z0 = Fx(x-x0) + Fy(y-y0).

curvature r(t)

magnitude of r’(t) cross product with r''(t) divided by the magnitude of r’(t) cubed.

given DuF(points) of function f(x,y,z). Find the directional derivative given in direction v.

find gradient vector at points given. Find u with formula (u/||v||). Then do GV * u at x, y, and z. Add these together.

direction leading to maximal rate

gradient vector with points plugged in.

given a function f(x,y,z) and a constraint equation using variables x, y, and z.

Find the gradient vectors of both, f and g. Set the two equal to each other but multiply g values by lambda. Divide to isolate the x, y, and z. Plug the x, y, and z values with lambda into the original function to find the value of lambda, and then the value of x, y, and z. Then plug those values into the original f function. If the sum is less than the constraint, min, more, max.

double integral bounded by the curves (x functions).

Set up two integrals. the outside is the values that equal the curves given. The inside integral is the variable functions given as bounds. If they are in terms of x, they are the dy integral and vice versa.

polar coordinates, given bounds

Use integral given and bounds given. dA=rdrdtheta

x²+y², x, and y in polar coordinates

r², rcos(theta), rsin(theta)

maximal rate (directional derivatives)

magnitude of gradient vector with points plugged in

Surface area, double integrals

square root of (1 + (Fx)² + (Fy)²). dA is rdrdtheta

Volume, double integrals

Usual dxdy double integral

basic triple integral

bottom bound is values given. For upper bounds set each value equal to which integral you’re doing. (for x+y+z=1), dz bounds would be 1-x-y, for dy it would be 1-x, and dx would be 1.

cylindrical substitution dV

r*dz*dr*dtheta

cylindrical coordinates, x y and z

z=rcos0, y=rsin0, z=z, z²+y²=r²

spherical coordinates dV

(rh)²*sin(ph)*drh*dph*d0

cylindrical coordinates, x,y, and z

x=rhsin(ph)cos(0), y=rhsin(ph)sin(0), z=rhcos(ph), x²+y²+z²=rh²

work done by force field F(x,y,z) along the path r(t)

integral of F * dr

wok done by the force field F(x,y,z) along the straight path from point 1 to point 2

set up parametric equation and plug those values into F(x,y,z). Take the integral.

find conservativeness

cross product of the field and the partial derivatives. If it equals zero, it is conservative!

find a function f(x,y,z) such that F=gradient vector

find partial derivatives of F. take integral of first equation, add C(y,z). Then take the partial derivative with respect to y (C w/ respect to y at the end. Compare with second equation to see what y equals. Then lowkey just work backwards….

Greens theorem

double integral with bounds given of Qx - Py (Q is second equation, P is first equation)

Type 2 surface integral

double integral of [-Fx(df/dx) - Fy(df/dy) + Fz] dA

divergence

gradient vector * F