calc #3 exam2

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edulalo999
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calc #3 exam2
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2011-12-09 15:03:48
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It is a review of calculus 3.
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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?
    • Answer:
    • 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
    A)Find the gradient of F?




    • 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
    • (Gradient of f)(u)
    • 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?
    Formula= (gradient of f)(u)

    • 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
    • (gradient of f)(u)
    • = ((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
    • 1. IfD<0,then(x0,y0)isasaddlepoint.

    • 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)

    • gradient of f(x,y)
    • =- ((cosx-cos(x+y))i+(cosy-cos(x+y))j
    • =-(cosx-cos(x+y))i - (cosy-cos(x+y))j

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