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f(x,y) = 1 + √(4 - y2), Whats the Domain
-2 ≤ y ≤ 2
f(x,y) = 1 + √(4 - y2), Whats the Range
[1,3]
f=ln(x), whats the domain of f?
x > 0
rewrite in form for graphing x2 + 9y2 = 3
x2 / √(3)2 + y2 / √(⅓)2 = 1
lim(x,y) - > (3,1) F(x,y) = 6. what can you say about the point (3,6)
nothing
lim(x,y) - > (3,1) F(x,y) = 6. what can you say about the point (3,6) if F is cont.
it is equal to 6
When finding if a limit exists, and fn is undefined at point what do you do?
find line(s) that go through the point you trying to find the limit of and substitute.
another partial notation for Fxy(X,Y)
∂2F/∂y∂x
14.3 #10
know it
What conditions must be true for Fxyz = Fxzy = Fzxy …
Fn must be continuous
do one: 14.3 15, 17, 20, 21, 25, 28, 31, 34, 36, 42, 43, 50, 64, 66, 68
yeahhhh
eq of tangent plane
Z - Zo = Fx(Xo,Yo) (X - Xo) + Fy(Xo,Yo) (Y - Yo)
explain why f(x,y)=x/(x+y) is differentiable at (2,1)
because Fx and Fy are continuous
How to find linearization of tangent plane
find eq of tangent plane, solver for Z and rewrite Z as L(x,y)
if we have plane Z - Zo = Fx(Xo,Yo) (X - Xo) + Fy(Xo,Yo) (Y - Yo), how do we find the normal vector to the plane?
rewrite as 0 = -(Z - Zo) + Fx(Xo,Yo) (X - Xo) + Fy(Xo,Yo) (Y - Yo) so normal vector is < Fx(Xo,Yo), Fy(Xo,Yo), -1 >
what is L(x,y) and how do we use it?
L(x,y) is the linearization of the plane, simply the tangent plane solved for z. with this we can estimate points close to the a known point.
write eq for F(2.1,1.9) if F(2,2)=10, Fx(2,2) = 5 Fy(2,2) = 3
L(X,Y) = 10 + 5(.1) + 3(-.1) = 10.2
L(x+∆x,y+∆y,z+∆z)
= F(x,y,z) + Fx(x,y,z)∆x + Fy(x,y,z)∆y + Fz(x,y,z)∆z
Differential of f(x,y) (what's dz)
dz= Fx(x,y)dx + Fy(x,y)dy
∆y vs dy
∆y = actual change in y when x changes by ∆x, dy = change in y from tangent line when x changes by dx
Find the differential of m=p5 q3
dm = (5q3p4)(dp) + 3p5q2(dq)
if z = 5x2 + y2 and (x,y) changes from (1,2) to (1.05,2.1), compare the values of dz and ∆z
dz = .9 ∆z = .9225
when do you use d vs ∂ ?
When your function is of more than one variable
z=f(x,y) where x and y are functions of s and t. write chain rule notation.
∂z/∂s = ∂f/∂x ∂x/∂s + ∂f/∂y ∂y/∂s and ∂z/∂t = ∂f/∂x ∂x/∂t + ∂f/∂y ∂y/∂t
z=f(x,y) where x and y are functions of s. write chain rule notation.
dz/ds = ∂f/∂x dx/ds + ∂f/∂y dy/ds
r = f(x,y,z,t) where x, y z and t are fns of u v and w. write out the chain rule wrt w
∂r/∂w = ∂r/∂x ∂x/∂w + ∂r/∂y ∂y/∂w + ∂r/∂z ∂z/∂w + ∂r/∂t ∂t/∂w
14.5 14 & 15
do one
14.5 36, 37, 38, 45
do one
14.5 21, 22, 23
do one
f(x,y) = y3 + x√(y) Find the direction of max increase at (8,1), and the value of that rate of increase
direction is DEL(F) (8,1) = < 1,-1 > and value is √(2) (14.6 notes)
f(x,y) = y3 + x√(y) Find the direction of min increase at (8,1), and the value of that rate of decrease? note the direction of max ROC is < -1,1 >
direction is -DEL(F) (8,1) = < 1,-1 > and value is -√(2) (14.6 notes)
f(x,y) = y3 + x√(y) Find the direction of the level curve at (8,1)? note the direction of max ROC is < -1,1 >
direction is perp to DEL(F) (8,1) … so < 1,1 >
how do you find the vector perpendicular to < 2,-3 >
M=rise/run = -3/2 Negative reciprocal = ⅔ so perp vector is < 3,2 >
f(x,y) = y3 + x√(y) Find for V = < -2,5 > at (8,1)
-7/ (29)^1/2
Du f(Xo,Yo) (w/out del) =
Fx(Xo,Yo)a + Fy(Xo,Yo)b
DEL (F) =
< ∂f/∂x, ∂f/∂y >
DuF(Xo,Yo) (w/ del) =
delF(Xo,Yo) DOT U
DuF (w/ cos) =
|DEL(F)| |u| cos(ø)
the max increasing directional derivative is when
ø = 0
the max decreasing directional derivative is when
ø = π
DuF = 0 when _________ and this means ________
ø = π/2 or 3π/2, your on a level curve.
why are DEL(F) vectors always perp to level curves
because their pointing in the steepest direction.
The value of max increasing of a directional derivative is? why?
| DEL(F) | , b/c the direction of max increase is when DEL(F) and U are parallel.
do a 14.6 problem
if you want
The find the arc length you take the integral of ______ which is ___ or ___
Speed, |r’(u)| or √([x’(t)]2 + [y’(t)]2 + [z’(t)]2 )
formula for arc length for 0 ≤ t ≤ 2
∫(0 to 2) √([x’(t)]2 + [y’(t)]2 + [z’(t)]2 ) dt
arc length to any t value
∫(0 to T) |r’(u)|du
s(t) = ∫(0 to T) |r’(u)|du, how to reparametrize wrt arc length
solve for S(t), then rewrite in terms of t, then you have T as a fn of S (arc length)
13.3 # 15
handleee ittt
meow
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