Multivariable Calc Final

center of mass (x bar, y bar, z bar)

(Myz/m, Mxz/m, Mxy/m) M stands for moment m stands for mass

general equation for a sphere in 3D with a center at P(a, b, c) and radius r

(x - a)² + (y - b)² + (z - c)² = r²

ellipse properties

(x-h)²/a² + (y-k)²/b² = 1 or (x-h)²/b² + (y-k)²/a² = 1 because a is always the length of the major/longer axis, and b is the length of the shorter one c² = a² - b² focus is (h ± c/a, k) vertex are along the major axis (h ± a, k) or (h, k ± a) depending on which axis, x or y, is the major axis

notes for 12.5

... to find the equation of a plane, you need a point on the plane, P(x₁, y₁, z₁) the normal vector, or two vectors on the plane so that you can find the cross product, or the direction vector of another plane that is parallel to it (which is its normal vector) the normal vector for the plane is {a, b, c} the general equation for a plane is a(x - x₁) + b(y - y₁) + c(z - z₁) = 0 ... find distance of a point to a plane have a point not on the plane, P(x₁, y₁, z₁) equation of the plane in the form ax + by + cz + d =0 use the distance equation D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²) OR D = |A x B| / |A|, where vector a is a vector on the line and vector be is the vector from the line to the point (a and b both have the same initial point) ... find the distance between two parallel planes have a point not on the "base" plane, P(x₁, y₁, z₁) equation of the "base" plane in the form ax + by + cz + d = 0 use the distance equation D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²) use the distance equation for two parallel lines D = |d - d₁| / √(a² + b² + c²) ... find the equation of a parametric function need a point on the line, P(x₁, y₁, z₁) need the direction numbers of the line in the form of a vector {a, b, c} x = at + x₁, y = bt + y₁, z = ct + z₁ ... how to tell if parametric functions are parallel, skew, or intersecting parallel: their direction numbers are scalar multiples of each other intersecting: if there is a value of t that makes both of the functions have equal x, y, and z values (set both of the x, y, and z functions equal to each other to solve for t, check if the t's are all of the same values) skew: if not parallel or intersecting

12.5 notes continued

...two planes point of intersection find both planes normal vectors, N₁ and N₂ find the cross product of the two to get the direction numbers for the line (this works because it will give the direction numbers for a line and not a plane, so we can take the result of the cross product as it is because we need the direction numbers as the parallel direction numbers, not normal vector direction numbers like for planes) find a point where the lines intersect, P(x₁, y₁, z₁) use the general equation for a parametric function ...need a line perpendicular to a given parametric equation given the equation, L, of the line and a point, P, on the perpendicular line arbitrarily name the point on the perpendicular line P(x₁, y₁, z₁) the vector, v, from the point P to the curve is taking the component of P - the respective component of the equation the direction vector of the equation and the v need to have a dot product = 0 (property of the dp where two perpendicular vectors' dp = 0), so solve for t to find the point on the line that makes a perpendicular vector with P and the point on L (which was once arbitrary but now we know what t equals and can plug it in to find the exact point) use the vector from P to the point on L at t to find the direction vector, c, for the perpendicular line use c and P to create the parametric equation of the perpendicular line

finding the domain and range of a function with 2 inputs

1. f(x,y) = √(9-x²-y²) D = {(x,y) | (9-x²-y²) ≥ 0} the range is all the possible values of the output (which in this situation would be in the z axis) so the range is all the possible values of z given that z = √(9-x²-y²) R = {z | z = √(9-x²-y²), (x,y)∈D} *when the D and R are graphed, it is a circle with r = 3 shaded in and with a solid perimeter (since the domain is inclusive) shaded in the center. when we consider that a square root of the range can never be negative, the graph that is extended into the z-axis can only occupy the positive direction, so the full graph will be half a sphere

example of finding the domain of a function with 2 inputs

1. f(x,y) = √(x+y+1)/(x-1) x≠1 because that makes the denominator 0 (x+y+1) ≥ 0 because whatever is under the square root has to be 0 or positive D = {(x,y) | (x+y+1) ≥ 0, x≠1} 2. f(x,y) = xln(y²-x) only limitation is that y²-x > 0, because 0 can never be the result of raising e to a power D = {(x,y), y²-x > 0}

example of finding the domain of a function with 3+ inputs

1. f(x,y,x) = ln(z-y)+ xysinx only constraint is that z-y > 0, so... D = {(x,y,z)∈R³ | z-y > 0}

steps for finding the double integral of a surface that lies over a domain that isn't a box

1. sketch the domain in the xy plane 2. write the "cap/main" equation in terms of x and y in order to solve for z (so that you can integrate with respect to x and then y) 3. if the bounds have y written in terms of x, integrate with respect to y first so that you can replace all y's with x's and make your life easier when you go to the second integral **this is where picking the domain in order to integrate easiest first matters. if you get stuck switch the order of integrals (and the ways in which the bounds are written)

the average value of a function over an integral is

1/area(integral), because you're finding the average height of the solid given the base already (dimensions of the rectangle)

octants

8 "quadrants" of the 3D graph, first quadrant is where x, y, and z are positive

differential of a function (14.4) *********** from here on, what are each useful for???

= derivatives of each input variable added together df = (df/dx)dx + (df/dy)dy ...

make a parametric equation of the line tangent to a graph at a given point

a helix, r(t), has the equations x = 2cos(t) y = sin(t) z = t and intercepts the point (0, 1, π/2), so t = π/2 r'(t) = {-2sin(t), cos(t), 1} r'(t) = {-2sin(π/2), cos(π/2), 1} r'(t) = {-2, 0, 1} so the tangent line at (0, 1, π/2) in parametric equation form is x = -2t, y = 1, z = t + π/2

to find a parametric equation, we need

a point on the line we are trying to find, and a parallel vector

scalar equation of a plane through the point P₀ (x₀, y₀, z₀) that has a normal vector {a, b, c}

a(x - x₀) + b(y - y₀) + c(z - z₀) = 0 this is a rewritten example of the dot product... a(x component of vector) + b(y component of vector) + c(z component of vector) = 0, and replace x, y, and z with other known points form the plane

finding the level curves of a function with 3+ inputs

again, set the equation = k but now your output space will be in 3D because there are 3 different input

determinant of a matrix (3x3)

alternating signs for the terms (-1)^(row # + column #), then take the determinant of the smaller matrices made

r(t) is just taking its components (which is a parametric equation)

and making it a vector starting from the origin and extending to the point given by the component functions, so the resulting vector will just be the vector of the point

cross product

answer is a VECTOR!! A x B = |A||B|sinθ (0<θ<π) or A x B, you set up a matrix with i, j, k on the top row and the vectors A and B below, and then find the determinant (if its A x B, vector A goes above B in the matrix and vice versa) when switching from A x B to B x A, it will essentially be the "same" vector but with the opposite direction find the direction of the cross product by using the right hand rule (if A x B, the direction of A is the "x-axis", B is the "y-axis," and the direction of the cross product is wherever your thumb points, see "right hand rule" quizlet card if you need more help) **if cross product is 0, then the vectors are parallel

for the derivative of the dot product or the cross product of two vectors,

apply the "regular" multiplication rule and then find the derivative of the resulting equation ie d/dx(r(x) dot q(x)) = d/dx(r'(x) dot q(x) + r(x) dot q(x))

line integrals

area under a particular curve on the surface ∫ f(x(t),y(t))√((dx/dt)²+(dy/dt)²) dt from a to b

a plane can be in the form

ax + by + c = z

distance of a point to a plane

b = vector from any arbitrary point P (x, y, z) on the plane to a point outside of the plane P₀ (x₀, y₀, z₀) n = {x₀ - x, y₀ - y, z₀ - z} |b * n| / |n| or |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²) ex. (2, 0, 0), 4x - 6y + 2z - 3 = 0 |4(2) - 6(0) + 2(0) - 3| / √(4² + (-6)² + 2²)

properties of cross product

both cross products, A x B and B x A are orthogonal to the "original plane" created by vectors A and B A x B = -B x A the area of the parallelogram that's made up from two vectors is = |A x B| (magnitdue of a cross b)

limit of a multi variable function

check that as you approach the point for that limit, the value of the limit is the same regarles of the way that you get there ex: check along the x/y/z-axis by holding the other variables constant, y=x line by replacing either all of the x with y in the equation or y with x, y=x=z, etc *check along multiple "paths" or lines

scalar projection

component of B along N signed MAGNITUDE of the projection (not a vector like vector projection) compₙB = (N dot B)/|N| (dot product of N and B) *if you have the angle between the two vectors, just take the magnitude of the "hypotenus" and multiply it by the cosine of your angle to give you the adjacent length deriving the equation for the scalar projection... suppose v is projected onto u like in the picture, and θ is the angle between u and v (ignore the fact that it says the red part is the projection of v onto u, I just wanted a picture) to find the length of v on u, use the equation for cosθ = adjacent/hypotenus (we are trying to solve for adjacent, which is the scalar projection) in this case, the adjacent is (call it v onto u,) and the hypothenus is the magnitude of v (|v|) so |v|cosθ = v onto a this is still not helpful though, because we have two unknowns, θ and v onto a, so we need to do more work from the identity of the dot product, we know that the cosine of the angle between two vectors is cosθ = (a dot b)/|a||b| therefore, the cosing of our angle (θ) between our two vectors is cosθ = (v dot u)/|v||u| putting this all back into the trig equation for cosθ, we get |v|cosθ = |v|(v dot u)/|v||u|, which can be simplified to (v dot u)/|u| ***"dot product over the magnitude of N"

linear equation in x, y, and z

d = -(ax₀ + by₀ + cz₀) plane through the point P₀ (x₀, y₀, z₀) that has a normal vector {a, b, c}

example of finding the derivative of the dot product of two vectors

d/dt[u(t) dot v(t)] u(t) = {sin(4t), cos(5t), t} v(t) = {t, cos(5t), sin(4t)} first, apply the usual multiplication rule, but replace the multiplication symbol with the dot u'(v) dot v(t) + u(t) dot v'(t) = {4cos(4t), -5sin(5t), 1} dot {t, cos(5t), sin(4t)} + {sin(4t), cos(5t), t} dot {1, -5sin(5t), 4cos(4t)} 4tcos(4t) + -5sin(5t)cos(5t) + sin(4t) + sin(4t) + -5sin(5t)cos(5t) + 4tcos(4t) = 8tcos(4t) -10sin(5t)cos(5t) + 2sin(4t)

partial derivatives

derivative of a multivariable function with respect to one variable (keep other variables "fixed" and acting like constants)

find the magnitude of a vector by using

distance formulas

we use all of our regular differentiation rules on vector functions as we do with regular functions, other than the

divisibility rule

vector valued function

domain is the set of real numbers but the range is a set of vectors written as r(t) = {f(t), g(t), h(t)} = f(t)i + g(t)j + h(t)k component functions of vector valued functions are f(t), g(t), and h(t) components are parametric functions

to figure out the equation of a plane, you need the vector that is perpendicular (normal) to it and a point on the plane

dot product between a vector on the plane and the normal vector = 0 n * (terminal point of vector - initial point of vector) = 0 n * (terminal point of vector) - n * (initial point of vector) = 0 n * (terminal point of vector) = n * (initial point of vector) if given two vectors on the plane, you can use the cross product of the two to find the normal vector **you cannot figure out the equation of a plane when you are given a parallel vector because if imagining it like a pencil (vector) and paper (plane,) the paper can be shifted and still be parallel to the pencil

countour map

drawing of the level surfaces of a function

linear approximation of multi variable functions (14.4)

dx(x-x₀) + dy(y-y₀) ... d_(_-_₀)

hyperbola

ex. (x-h)²/a² - (y-k)²/b² = 1 is shown in the picture one variable is positive and one is negative opens on the axis whose variable is positive (shown in the picture opening on the x axis)

example of finding a partial derivative

f(x,y) = x³+x²y³-2y² the derivative with respect to x is 3x²+2xy³ the derivative with respect to y is 3y²x²-4y

example of finding the level curves of a function with 3+ inputs

f(x,y,z) = x²+y²+z² k=x²+y²+z² this is the equation of a sphere with radius √k therefore, the level curves will be a family of concentric spheres with radius √k on a 3D graph

when two space curves intersect

figure out where the x, y, and z components overlap ex. find the vector function that represents the curve of intersection of the cylinder x² + y² = 1, and the plane y + z = 2 the cylinder is a circle with a radius of 1 extended in the z direction, so it can be described with the parametric equations x = cost, y = sint the plane cuts through at specific z values, so we get the equation for z to be z = 2 - y, which y can be replaced with the parametric equation sint to get z = 2 - sint so the vector valued function for the intersection of the two things intersect is r(t) = {cost, sint, 2-sint}

maximum and minimum values of multi variable functions

find fx, fy, fxx, fxy, fyy. the critical points of f(x,y) is when fx = fy = 0 **all possible x values with all possible y values D = fxx*fyy - (fxy)² if D < 0, the point is a saddle point, so it's not a minimum or a maximum if D > 0, it can be a maximum or a minimum depending on its fxx value ... if fxx > 0, point is a minimum ... if xx < 0, point is a maximum

finding the domain of a function with 2 inputs (and sketch it)

find the domain of all the independent variables sketch on a graph of the independent variables as the axis (ie if the input of a function is both x and y, you graph the domain on an xy-plane, versus when x is the only input, the domain is on a "number line")

find the direction numbers from the intersection of two planes

find the normal vector of both planes (coefficients of the variables) take the normal vectors' cross product

triple scalar product

finding the volume of a 3D parallelogram (parallelepiped) or finding the determinant of putting all three vectors into a 3x3 matrix will give the volume I don't like this way, but also: the magnitude of the dot product of one vector with the cross product of the other two vectors to find the volume of the parallelepiped (essentially is going to tell you to multiply each component of the cross product by the compenent of the dot product, which is just finding the determinant) **if tsp = 0, then the volume is 0 and the points are COPLANAR

to solve a double integral

first solve the equation with respect to the inner variable, then find the integral of the result of the inner integral with respect to the second variable

multiple partial derivatives

found the same way, but done multiple times fxy means first differentiate with respect to x and then y

find parametric equations for the tangent line to the curve with the given parametric equations at the specified point

given the point, solve for t, so you know what value of t was plugged into the original equation to get the point that was given solve for the derivative of the component functions, and plug in t to find all three of the direction numbers for the tangent vector of that point. use the direction numbers of the tangent and the originally given point to create the parametric equation

example of finding the level curves of a function with 2 inputs

h(x,y) = 4x² + y² + 1 the equation for the level curves is k = 4x² + y² + 1 k = 4x² + y² + 1 means that its level curves are ellipses because the level curves are equal to those of k = 4x² + y² shifted up 1 on the z-axis

zero vector

has a magnitude of 0 and no specific direction

what is a partial derivative graphically

have the surface S in the 3D coordinate plane with the point P on it hold either x or y constant at b, which means you will have a slice of the graph at either x = b or y = b ex. hold y constant at y = b, meaning you will have a slice perpendicular to the y axis at (x, b, 0), and you can find the derivative of the graph at the point with respect to x

link for trig identities

https://www.google.com/search?q=trig+identities&tbm=isch&ved=2ahUKEwjArIm9t_bvAhUHI1kKHeN6CP0Q2-cCegQIABAA&oq=trig+identities&gs_lcp=CgNpbWcQAzIFCAAQsQMyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADICCAA6BAgAEENQq1lYv2JggGNoAHAAeACAAVSIAc8DkgEBNpgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=2gtzYMDXLIfG5ALj9aHoDw&bih=425&biw=1024&rlz=1CAQRFK_enUS909#imgrc=fSwYD3P8ApAVTM

*another way to think about it* when two space curves intersect

if one equation is a parametric equation, then you have it in terms of x, y, and z and and you substitute those values into your other equation to find the values of t that satisfy that equation replug in t to find the points in space OR GENERALLY solve for one variable (even if it's in terms of another variable) and plug it into the equations to try to figure out the other variables PARAMETRIC EQUATIONS! ex. the cone z = √(x² + y²) and the plane z = 3 + y intersect. find the vector function that represents the curve of intersection √(x² + y²) = 3 + y x² + y² = 9 + 6y + y² x² - 9 = 6y y = (x² - 9)/6 NOW SOLVE FOR THE PARAMETRIC EQUATIONS x = t y = (t² - 9)/6 z = 3 + (t² - 9)/6 = (t² + 9)/6

properties of the triple scalar product

if the determinant is 0, then all of the vectors are coplanar if the determinant isn't 0, the determinant is the area of a parallelepiped

if we want to add the vectors u and v (parallelogram law)

if we start at u, we first move the tip of v to the tip of u, and then create a parallelogram using the parallels of vectors u and v. the resulting vector begins at the intersection of the tips and ends at the opposite corner

if we want to add the vectors u and v (triangle law)

if we start at u, we first move u so that its tail coincides with the tip v of and define the sum of u + v as the tip of u to the tail of v

integrals of functions with multiple variables

integrate regularly, but integral will still be in vector form

joint density function *****

is equal to 1 everywhere that it's defined

reimann sum

is the same as for "regular" functions (midpoint, left, right)

evaluating an integral by reversing the order of integration

is useful when you run into a problem integrating the inner function (ex. eˣ^² cannot be easily integrated with respect to x first because u-substitution and other ways to integrate are not helpful here) therefore, you rewrite the bounds to be able to integrate with respect to y (in terms of x) first in hopes that when you then integrate with respect to x there will be "more" terms of x to work with to integrate the same rules of double integration apply (sketching the domain, etc.) after you find out if you have to reverse the order of integration or not

the derivative of a vector function

lim h → 0 (r(t + h) - r(t)) / h take the derivative of each component function ex. r(t) = (1 + t³)i + (te⁻ᵗ)j + (sin(2t))k d/dt (1 + t³) = 3t² d/dt (te⁻ᵗ) = e⁻ᵗ - te⁻ᵗ d/dt (sin(2t)) = 2cos(2t) r'(t) = (3t²)i + (e⁻ᵗ - te⁻ᵗ)j + (2cos(2t))k r'(0) = 0i + 1j + 2k

r(t) (a vector valued function) is continuous at a if

lim t → a = r(a) *same concept as with "regular" functions **the limit of a vector valued function is equal to a vector

12.6 notes

most important things: traces, standard form traces are when you hold one of the variables of x, y, or z at 0 so that you can see what it looks like when it intersects the other two variable's plane (ex. holding z = 0 shows me what the graph looks like at the xy-plane) standard form is when you match the way the equation is written to the way the equation is in the "TABLE" (more below) **tab starred as "TABLE" for the other common shapes of 3D shapes of the second degree

parametric equation and vector equation of a line that passes through the point (5, 1, 3) and is parallel to the vector i + 4j - 2k

parametric equation x = t + 5 y = 4t + 1 z = -2t + 3 vector equation r = (t + 5)i + (4t + 1)j + (-2t + 3)k

determinant of a matrix (2x2)

photo

to prove that two parametric equations are skew

prove that they aren't parallel (check that their direction vectors aren't scalars of each other) set the x, y, and z equations of the first line and the second line equal to each other, and see if there are any common values of t that work for all three statements

right hand rule

put your arm in the direction of the x-axis, curl your hand to the y-axis, and wherever your thumb points is the z-axis (use for cross product too to determine where the normal vector will be pointing)

vector

quantity that has both a magnitude and direction, and is denoted either by boldface, → over it, "triangle" brackets, or in terms of i, j, and k

if |r(t)| = c (a constant) for all values of t, then

r'(t) dot r(t) = 0, so that means that the position vector and the tangent vector are perpendicular (which makes sense because it's a sphere)

unit tangent vector T(t)

r'(t)/|r'(t)|

Lagrange multipliers

set the gradient of f (what you are looking for) = the gradient of g times some constant (λ) (what the constraint is) *if there are multiple equations that the function is constraint to (h), you will also + the gradient of h times some constant (µ) to the right side so fx = gxλ, fy = gyλ (derivatives are equal to each other at some point) and so on (for however many variables there are) solve the system of equations with the constraints at the gradients plug in numbers back into the original equation to find the point which satisfies the question

to find the rectangular equation of a parametric equations

set the parametric equations equation to t and then set the two equations equal to each other ex. x=5t, y=t x/5=t,y=t y=x/5

level curves

setting a multi-variable equation equal to a constant (k) drawing the curve at that specific k (like 0,1,2, etc.) projection of the output onto the plane of the inputs try to get it in a form you know so you know the shapes of your level curves (match to standard equation from the starred tab "TABLE!!") **addition or subtraction of a constant is probably the shifting of a standard curve in the output axis

vector projection (vector projected onto another vector)

shadow of another vector on a vector, so result is a VECTOR component vector of a vector (B) onto another vector (N) in the direction of N projₙB => projection of B onto N projₙB = (N * B)/|N| * (N / |N|) the vector projection = scalar projection times the unit vector of n, which makes the scalar projection a vector now because the unit vector gives a direction while not adding anything to its magnitude (because it has a magnitude of 1)

sketching a space curve example

sketch the curve whose vector equation is r(t) = (cos(t))i + (sin(t))j + (t)k (called a helix) no matter what value you put in t (think value of an angle,) it will always lie on the unit circle. therefore, the x and y values will always be in a circle with radius 1 centered at the origin, but spiral up higher and higher as t increases in the direction of the z axis

Find the volume of the solid that lies under the paraboloid z = 3x² + 3y², above the xy-plane, and inside the cylinder x² + y² = 2x.

sketch the domain because the domain is a circle, think about writing the bounds and equation in polar coordinates the equation for the domain is (x-1)² + y² = 1 after completing the square and taking the shift into account to find the equation in terms of r, replace x=rcosθ and y=rsinθ the domain in terms of polar coordinates is now (rcosθ-1)² + (rsinθ)² = 1 expanding this, we have... r²cos²θ - 2rcosθ +1 + r²sin²θ = 1 r²(cos²θ + sin²θ) -2rcosθ = 0 r²(1) = 2rcosθ r = 2cosθ therefore, 0≤r≤2cosθ since the leftmost point of the circle is on the origin, -π/2≤θ≤π/2 because the angle needed to capture the whole circle ranges from ±π/2 now, we can start doing the integral with the equation in terms of z, which becomes z = 3(rcosθ)² + 3(rsinθ)² = 3r²(cos²θ + sin²θ) = 3r²(1) = 3r² since r is in terms of theta, first integral with respect to r. ∫∫(3r²)r* drdθ from 0≤r≤2cosθ and -π/2≤θ≤π/2 = 9π/2 (integration uses double angle identity)

symmetric equations

solve for t for x, y, and z and then set all three equations equal to each other we can then plug in, for example, a z = 0 into the new equation to find where the equation intersects the xy-plane

how to find a line integral

step 1. parametrize f(x,y) to get it in terms of x(t) and y(t) (or have the vector equations) step 2. find

unit tangent vector

t(t) = r'(t) / |r'(t)| ex. r(t) = (1 + t³)i + (te⁻ᵗ)j + (sin(2t))k d/dt (1 + t³) = 3t² d/dt (te⁻ᵗ) = e⁻ᵗ - te⁻ᵗ d/dt (sin(2t)) = 2cos(2t) r'(t) = (3t²)i + (e⁻ᵗ - te⁻ᵗ)j + (2cos(2t))k r'(0) = 0i + 1j + 2k |r'(0)| = √5 t(t) = (1/√5)j + (2/√5)k

direction cosine

the cosine of the angle a vector makes with the x, y, and z axis cosθ = (component of a/|a|) will give the direction cosine for that specific axis example: a = i + 2j + 3k |a| = √(14) direction cosine for x is cosθ = (1/√(14)) (solving for the angle gives you the direction angles) all cosines squared add up to 1

finding the domain of a function with 3+ inputs

the domain of functions with 3+ inputs is (x,y,z)∈R³ then you need to find the domain of all the independent variables

example of find parametric equations for the tangent line to the curve with the given parametric equations at the specified point

the equation of a curve is: x = √(t²+63), y = ln(t² + 63), z = t. the point (8, ln(64), 1) lies on the curve. find the equation of the tangent line at this point. t = 1 (because z = t and t = 1) dx/dt = t/√(t²+63) @ t = 1... dx/dt = 1/8 dy/dt = 2t/(t²+63) @ t = 1... dy/dt = 1/32 dz/dt = 1 so the direction numbers for the tangent vector when t = 1 at (8, ln(64), 1) are {1/8, 1/32, 1} now make the parametric equation for r'(0) = {t/8+8, t/32 +ln(64), t+1}

triple integral for the density function gives

the mass

if the determinant of the matrix isn't 0, then

the matrix has an inverse (if it is, then it doesn't have an inverse)

dot product

the number/scalar you get from multiplying corresponding components of two vectors 3i + 2j -2k * 4i - 3j + 5k = 3*4 + 2*-3 + -2*5 = 12 -6 -10 = -4 |a||b| cosθ, when θ is the angle between a and b (a and b are vectors) cosθ = a * b / (|a||b|) (dot product / magnitudes) rewritten definition of dot product, is helpful when given two vectors and asked to find the angle between them dot product is communitive angle between two vectors is ____ if... 1. acute if a dot b > 0 2. perpendicular is a * b = 0 3. obtuse if a dot b < 0 **two vectors are parallel if the vectors are negative, scalar multiples of each other, so check for that first

space cruve

the parameters of space curves are the parametric functions of vector-valued functions space curve C is traced out by moving the tip of the position vector, which is given by the vector-valued function

moments (need to find the center of mass)

the triple integral of the density function multiplied by one of the variables (x,y,z) as the integrand *put x if you're trying to find the moment for the x coordinate that will then be divided by the mass to get the center of mass this will then give the moment when integrated with the same bounds as were used for the triple integral for finding the mass

vectors are equivalent if

they have the same magnitude and direction (don't need to start in the same place)

finding the double integral in terms of polar coordinates

this is helpful when the domain is bounded in the shape of a circle (domain is in 2D!!!!, meaning that the integrand must be in terms of x and y only!!) double angle identities and other trig identities are useful to solve these add a *r to the integrand when doing the integral integrate with respect to r first

vector vocabulary

tip/initial point (where the vector begins) tail/terminal point (where the vector ends, where the arrow head is)

gradient is the partial derivatives of a function in vector form

true

integration for vector-valued functions use the same rules as regular functions

true (integral is a vector!!)

chain rule with multi variable functions

use tree diagrams z /\ zy, and z and y are both functions of t dz/dt = dz/dx*dz/dt + dz/dy*dy/dt and so on

notes for chapter 12

vector starts at initial point and ends at terminal point compₙB = (N * B) / |N| projₙB = (N * B) / |N| * (N / |N|) (compₙB * unit vector of N, the vector that it's projected onto) finding the angle between two vectors is a rewritten form of the dot product, A * B = |A||B|cosθ... θ = arccos(A * B / (|A||B|)) the angle between planes is really similar to the angle between two vectors, except the vectors used are the normal vectors of the planes... θ = arccos((N₁ * N₂) / |N₁||N₂|)) if you need to find the equation if a sphere TOUCHES **key word!!** the plane, remember that you know the radius because the radius is = to how far away the center is from that plane cross product is used to find the area of a parallelogram and triangle (parallelogram / 2) finding the equation of a line equidistance from two points, you use the distance formulas set equal to each other

double integral gives

volume ∫∫f(x,y)dxdy = volume because dxdy = area of one of the little rectangles that the domain plane is broken up into, multiplied by the "height" (f(x,y)) of the curve so ∫∫f(x,y)dAd **integrals can be negative!!

triple integral for the function 1 gives the

volume over those bounds

find the angle of intersection of two planes

we need two vectors that lie on the curves because then we can use the property of the dot product (a dot b = |a||b|cosθ) to find the angle between the curves at that specific point in this case, we need the direction numbers of the tangent vectors for both curves at the point of intersection then, find the magnitudes of both of those tangent vectors and solve for the angle in between them

orthogonal

when two vectors are perpendicular (dot product is = 0)

scalar multiplication

when you multiply a vector by a number (vector v by c) and cv has c times the original magnitude but has the same direction. if c is negative, the vector will point in the complete opposite way, but will still be scaled by |c|

projection

where a point's "shadow" is on a plane ex. the projection of P(1, 2, 3) on the xy-plane is (1, 2, 0) because the z-component disappears to find the point's projection

the domain of the vector valued functions is

where the domain of the component functions overlap ex. r(t) = {f(t), g(t), h(t)}, f(t) = t³, g(t) = ln(3 - t), h(t) = √t D of f(t) is (-∞, ∞) D of g(t) is (-∞, 3) D of h(t) is [0, ∞) so the D of r(t) is [0, 3)

when the surface lies over a domain that isn't a box,

write one of the bounds in terms of the other variable and try to solve

parametric equation of a line passing through (x₁, y₁, z₁) and parallel to the directions vector {a, b, c}

x = x₁ + at y = y₁ + bt z = z₁ + ct *a, b, and c are direction numbers

when the surface lies over a domain in the shape of a box

you can plug in the limits directly (15.1)

when you multiply the dot product by a unit vector,

you get the directional derivative

ex. of finding the vector function of an intersection

you have the equation x²+y²=16 and the plane 2y+z=9. find the vector-valued function where they intersect vector functions are basically parametric equations, so you have to rewrite everything to be in terms of t looking at the equation for the cylinder, the ellipse that is the intersection of the equations would be x=4cost and y=4sint, because x and y can never be off the projection of x²+y²=16 on the xy-plane. all that's left to find is z. z = -2y+9 from the second equation, so plugging y=4sint into this equation we get z = -8sint+9 so the answer is {4cost,4sint,-8sint+9}

the limit of r(t) = {f(t), g(t), h(t)} as t approaches a is

{lim t→a f(t), lim t→a g(t), lim t→a h(t)} it means that the length and direction of the vector r(t) approaches the length and direction of the limits of each of the component functions, provided that the limit of each of the component functions exists ex. limt t → 0 r(t), r(t) = (1 + t³)i + (te⁻ᵗ)j + (sin(t)/t)k need the limit of each component function limt t → 0 (1 + t³) = 1 limt t → 0 (te⁻ᵗ) = 0 limt t → 0 (sin(t)/t) = 0/0 => cos(t)/1 = 1 so the limit of r(t) as t goes to 0 is the vector {1i, 0j, 1k}

r(t) dot r(t)

|r(t)|²

distance between two points in 3D

√((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)

example of reversing the order of integration

∫∫ₙ11eˣ^² where n is x| 3y≤x≤6 and y| 0≤y≤2 since x is written in terms of y, you figure that you integrate with respect to x first however, there is a problem, because 11eˣ^² cannot be easily integrated with respect to x first we know we must switch the order of integration the looking at my sketched domain of n, I can rewrite x| 0≤x≤6 and y| 0≤y≤(x/3) now I do a double integral with respect to y first ∫∫ₙ11eˣ^² where n is x| 0≤x≤6 and y| 0≤y≤(x/3) = 11(e³⁶-1)/6

example of double integral of a surface

∫∫ₙ6xydA over the triangular region n with vertices (0,0), (1,2), and (0,3) 1. sketch the domain. when sketching the domain, there are two lines (y=3-x and y=2x) that make up the side of the triangle. therefore, the limits of the integral are x| 0≤x≤1 and y| 2x≤y≤3-x 2. the main equation (6xy) is already in terms of x and y 3. since the curve is bounded by y and x, but y is written in terms of x, integrate with respect to y first, then x ∫∫6xy dydx over the region y| 2x≤y≤3-x and then x| 0≤x≤1