# calc #3 exam2

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1. 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 y-axis?
• lim (3xy)/(x^2+2y^2)
• x-->0

• =(3(0)y)/(0^2)
• =(0)/(2y^2)
• =0 Zero
2. B) Evaluate thelimit as (x,y) approaches the (0,0) along the line y=ax?
• Limit (3xy)/(x2+2y2)
• y-->ax

• =(3x(ax))/(x2+2(a2)(x2))
• =(3ax2)/(x2)+(2(a2)(x2))

• = X _(3a)__
• X (1+2a2)

• = __3a__
• (1+2a2)
3. C)
Does the limit (3xy)/
(x^2+2y^2)Exist?
(x,y)-->(0,0)
• the limit (3xy)/(x2+2y2)
• (x,y)-->(0,0)

• THE LIMIT DOES NOT EXIST
• Because 0 IS NOT EQUAL TO (3a)/(1+2a^2)

4. Question # 2
Let F(x,y) = x^2 - 2xy

• f(x,y) = x2 - 2xy
• fx = 2x - 2y
• fy= -2x

Gadient of F(x,y) = (2x - 2y)i - (2x)j
5. What is the formulaof the directional derivative of f a a point (x0, y0) ??
• Formula
• where u = the direction over the (direction)2 with square root!
6. B) Find the direction derivative of f at the point (1,1) in the direction of -3i + 4j?

• gradient of f =(2x - 2y)i - (2x)j
• (x0, y0)--> (1,1)
• =(2(1) - 2(1))i - (2(1))j
• =(2-2)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
• = ((0)i - 2j ) (-1/3i + 4/5j)
• =(0)(-3/5) + (-2)(4/5)
• = 0 - 8/5
• = -8/5
7. 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
8. How to find the stationary points and determine the local extreme values of the function f(x,y)?
• Suppose that f has continuous second-order partial derivatives in a neighborhood of (x0, y0) and that ∇f (x0, y0) = 0. Set
• A = fxx (x0,y0),
• B = fyx (x0,y0),
• C = fyy (x0,y0)

• and form the discriminant D = AC − B2

• 2. IfD>0,thenf takes on
• a local minimum at (x0, y0) if A > 0,
• a local maximum at (x0, y0) if A < 0
9. D) Find the stationary point and determine the local extreme value of the function f(x, y)?
• f(x,y)=x2 - 2xy
• fx=2x-2y
• fxx=2-->A
• fy=-2x
• fyy=0-->C
• fxy=-2-->B

• AC-B2=D
• 2(0)-(-2)=-4
• D=-4
• D<0
• It is a sattle point!

10. wha is the formula for the equation of the tangent plane of a surface at a point?
df/dx(x-x0) + df/dy(y-y0) + df/dz(z-z0) = 0
11. 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)--> (1-1)--> 0
• df/dy = cosy - cos(x+y)-->P=(0,0,0)
• cos(0)-cos(0)-->(1-1)--> 0
• df/dz = -1-->P=(0,0,0)--> -1
• Aply formula
• df/dx(x-x0) + df/dy(y-y0) + df/dz(z-z0) = 0
• 0(x-0) + 0(y-0) + (-1)(z-0)=0
• -z = 0
• Z = 0
12. What would be an equation for a scalar parmetric equation for the normal line of a surface at a point?
• X = x0 + df/dx(x0,y0,z0)t
• Y = y0 + df/dy(x0,y0,z0)t
• Z = z0 + df/dz(x0,y0,z0)t
13. 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
14. 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")

15. 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)