Math 2401

Quiz #2   Feb 16 Tom Morley 8:00 am

Open Book and Notes.  Carefully explain your proceedures and answers. Calculators allowed, but answers mush be exact.

Problem 1 (9 points)

George Mallory needs to know the following : Consider the surface : xy^2 + x^2 y^3 + z = 85.

Find a normal to the surface at x = 3, y = 2, z = 1.

Find an equation for the tangent plane to this surface at this same point .

Ans

f[x_, y_, z_] := x y^2 + x^2 y^3 + z

f[3, 2, 1]

85

gradf = {D[f[x, y, z], x] , D[f[x, y, z], y] , D[f[x, y, z], z]}

{y^2 + 2 x y^3, 2 x y + 3 x^2 y^2, 1}

gradf /. {x → 3, y →2, z → 1}

{52, 120, 1}

(x - 3) 52 + (y - 2) 120 + (z - 1) == 0 //ExpandAll

-397 + 52 x + 120 y + z == 0


Quiz #2   Feb 16 Tom Morley 8:00 am

Problem 2 (9 points)

Find the angle bwtween the two surfaces:
x^2 - y^2 + z^2 = 1 and    x + 2 y^2 -z^2 = 0  at the point: (1/2, -1/2,1)
(Either the obtuse or acute angle is correct)

Ans

In[90]:=

Clear[f, g, x, y, z]

In[91]:=

f := x^2 - y^2 + z^2

In[92]:=

g := x + 2 y^2 - z^2

In[93]:=

gradf = {D[f, x], D[f, y], D[f, z]}

Out[93]=

{2 x, -2 y, 2 z}

In[94]:=

a = %/. {x→ 1/2, y→ - 1/2, z → 1}

Out[94]=

{1, 1, 2}

In[96]:=

gradg = {D[g, x], D[g, y], D[g, z]}

Out[96]=

{1, 4 y, -2 z}

In[97]:=

b = %/. {x→ 1/2, y→ -1/2, z → 1}

Out[97]=

{1, -2, -2}

In[98]:=

a . b /(Sqrt[a . a] Sqrt[b . b])

Out[98]=

-5/(3 6^(1/2))

In[99]:=

N[ArcCos[%] 180/Pi]

Out[99]=

132.876


Quiz #2   Feb 16 Tom Morley 8:00 am

Problem 3 (9 points)

Parabolic coordinates are u,v and θ. In terms of these coordinates, x, y and z are:

x = u v cos(θ)     
y == u v sin(θ)     
z = (u^2 - v^2)/2
If f is a function of  u, v, and θ , express∂f/∂u and  ∂f/∂θ,  in terms of derivatives with respect to x,y,z, that is ∂f/∂x, ∂f/∂y and  ∂f/∂z. The answer(s)  will involve u,v, and θ also.

Ans

In[103]:=

Clear[f, u, v, θ]

In[104]:=

D[f[u v Cos[θ], u v Sin[θ], (u^2 - v^2)/2], θ]//TraditionalForm

Out[104]//TraditionalForm=

u v cos(θ) f^(0, 1, 0)(u v cos(θ), u v sin(θ), 1/2 (u^2 - v^2)) - u v sin(θ) f^(1, 0, 0)(u v cos(θ), u v sin(θ), 1/2 (u^2 - v^2))

Or in (sort of) English :

  ∂f/∂θ   = u v sin(θ)   ∂f/∂x + u v cos(θ) ∂f/∂y

In[106]:=

D[f[u v Cos[θ], u v Sin[θ], (u^2 - v^2)/2], u]//TraditionalForm

Out[106]//TraditionalForm=

  ∂f/∂u   = v cos(θ)   ∂f/∂x + v sin(θ) ∂f/∂y + u∂f/∂z

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