The flashcards below were created by user
edulalo999
on FreezingBlue Flashcards.

Question #1
A) Consider lim (3xy)/(x^2+2y^2)
(x,y)>(0,0)
Evaluate the limit as (x,y) approaches the (0,0) along the yaxis?
 Answer:
 lim (3xy)/(x^2+2y^2)
 x>0
 =(3(0)y)/(0^2)
 =(0)/(2y^2)
 =0 Zero

B) Evaluate thelimit as (x,y) approaches the (0,0) along the line y=ax?
 Limit (3xy)/(x^{2}+2y^{2})
 y>ax
 =(3x(ax))/(x^{2}+2(a^{2})(x^{2}))
 =(3ax^{2})/(x^{2})+(2(a^{2})(x^{2}))

C)
Does the limit (3xy)/(x^2+2y^2)Exist?
(x,y)>(0,0)
 the limit (3xy)/(x^{2}+2y^{2})
 (x,y)>(0,0)
 THE LIMIT DOES NOT EXIST
 Because 0 IS NOT EQUAL TO (3a)/(1+2a^2)

Question # 2
Let F(x,y) = x^2  2xy
A)Find the gradient of F?
 f(x,y) = x^{2 } 2xy
 f_{x }= 2x  2y
 f_{y}= 2x
Gadient of F(x,y) = (2x  2y)i  (2x)j

What is the formulaof the directional derivative of f a a point (x_{0}, y_{0}) ??
_{ }
 Formula
 (Gradient of f)(u)
 where u = the direction over the (direction)^{2} with square root!

B) Find the direction derivative of f at the point (1,1) in the direction of 3i + 4j?
Formula= (gradient of f)(u)
 gradient of f =(2x  2y)i  (2x)j
 (x_{0}, y_{0})> (1,1)
 =(2(1)  2(1))i  (2(1))j
 =(22)i 2j
 =(0)i 2j
 find U:
 u= (3i + 4j)/(square rt of ((3)^{2}+(4)^{2})
 = (3i + 4j)/square rt of (25)
 =(3i + 4j)/5
 = (3/5)i + (4/5)j
 Aply formula
 (gradient of f)(u)
 = ((0)i  2j ) (1/3i + 4/5j)
 =(0)(3/5) + (2)(4/5)
 = 0  8/5
 = 8/5

C) If x=scost and y=tsins find df/ds and df/dt??
 we set the function
 f=(scost)i + (tsins)j
df/ds= (cost)i + (cos(s))j
df/dt= (s(sint))i + (sin(s))j

How to find the stationary points and determine the local extreme values of the function f(x,y)?
 Suppose that f has continuous secondorder partial derivatives in a neighborhood of (x_{0}, y_{0}) and that ∇f (x_{0}, y_{0}) = 0. Set
 A = f_{xx} (x_{0},y_{0}),
 B = fyx (x_{0},y_{0}),
 C = f_{yy }(x_{0},y_{0})
 and form the discriminant D = AC − B^{2}
 1. IfD<0,then(x0,y0)isasaddlepoint.
 2. IfD>0,thenf takes on
 a local minimum at (x_{0}, y_{0}) if A > 0,
 a local maximum at (x_{0}, y_{0}) if A < 0

D) Find the stationary point and determine the local extreme value of the function f(x, y)?
 f(x,y)=x^{2 }^{ }2xy
 f_{x}=2x2y
 f_{xx}=2>A
 f_{y}=2x
 f_{yy}=0>C
 f_{xy}=2>B
 ACB^{2}=D
 2(0)(2)=4
 D=4
 D<0
 It is a sattle point!

wha is the formula for the equation of the tangent plane of a surface at a point?
df/dx(xx_{0}) + df/dy(yy_{0}) + df/dz(zz_{0}) = 0

Question # 3
Let z = sinx +siny  sin(x + y) be an equation defining a surface in the space R3 and P = (0,0,0) a point in this surface.
A) Find the equation of of the plane of this surface at the point P?
f(x,y)=sinx + siny  sin(x+y) z
 formula:
 df/dx = cosx  cos(x+y)>P=(0,0,0)
 cos(0)cos(0)> (11)> 0
 df/dy = cosy  cos(x+y)>P=(0,0,0)
 cos(0)cos(0)>(11)> 0
 df/dz = 1>P=(0,0,0)> 1
 Aply formula
 df/dx(xx_{0}) + df/dy(yy_{0}) + df/dz(zz_{0}) = 0
 0(x0) + 0(y0) + (1)(z0)=0
 z = 0
 Z = 0

What would be an equation for a scalar parmetric equation for the normal line of a surface at a point?
 X = x_{0} + df/dx(x_{0},y_{0},z_{0})t
 Y = y_{0 }+ df/dy(x_{0},y_{0},z_{0})t
 Z = z_{0 } + df/dz(x_{0},y_{0},z_{0})t

B) Find the scalar parametric equation for the normal line of this surface at P=(0,0,0)?
 X = x0 + df/dx(x0,y0,z0)t>0 + (0)t = 0
 Y = y0 + df/dy(x0,y0,z0)t>0 + (0)t = 0
 Z = z0 + df/dz(x0,y0,z0)t>0 + (1)t> Z = t

how to find in which direction a function of "f" decreases most rapidly at P?
 Just Find the negative Gradient of the function of "f".
 Direction W. D. M. R. at P = (gradient of "f")
 decreas most rapidly = (gradient of "f")
 increase most rapidly = +(gradient of "f")

In what direction does the function f(x,y) = sinx + siny sin(x+y) decrease most rapidly at P?
f(x,y) = sinx + siny  sin(x+y)
 df/dx = cosx  cos(x+y)
 df/dy = cosy  cos(x+y)
 gradient of f(x,y)
 = ((cosxcos(x+y))i+(cosycos(x+y))j
 =(cosxcos(x+y))i  (cosycos(x+y))j

