Math B6C – Exam 3 – SOLUTIONS

 

*  Fall 2002 *

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1.   Find  for the function  g(x, y) = ln(x³ + xy²).      Then   g_x(1, 1) = ?

 

 

Hence,  

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2.   Find the linearization of  f (x, y, z) = xy³ + z²cos(xy)    at  (1, 1, 0).      L(x,y,z) = ?

 

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3.   For the function  h(x, y) = e^3x–2y,  find the best quadratic approximation at the point   (0, 0).   

     Q(x,y) = ?

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4.    Find the gradient of   f (x, y, z) =  5xy² + 2z³     at the point  (1,1,3).   The y-component of this 
       gradient is  _____ .

 

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5.   Find the directional derivative of the  f  in problem #4, in the direction of  at the point  (1,1,3).
    

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6.  Find the equation of the tangent plane to   x³ – xyz + z³ = 1    at the point  (1,1,1).

 

Let   F(x,y,z) = z³x³ – xyz + z³.   The easiest way to get the tangent plane to a level surface of F uses the gradient:  

 

The point (1,1,1) is on the surface F(x,y,z) = 1;   suppose that  (x,y,z) is an arbitrary point on the tangent plane to the surface at (1,1,1).   Then the gradient, being perpendicular to the surface and the tangent plane, must be perpendicular to the vector

 

(x,y,z) – (1,1,1) =  (x–1, y–1, z–1).

Thus,

.

 

Another solution, using linearization

 

We seek the tangent plane to the level surface defined by.     One way to do this is to find the linearization of F at the point (1, 1, 1), which as usual we’ll call L.   The tangent plane is then just L(x,y,z) = 1.

 

 

Thus our tangent plane is given by , or

 

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7.   Find the maximum value of    on the part of the unit circle, x ² + y ² = 1,  where
      x>0 and y>0.  

 

We use the method of Lagrange Multipliers for this problem.   Let  .
Then extrema of f are to be found at points where, for some l,   .   Then

 

,

yield the following equations:

The 2^nd of these equations gives us  .    Substituting this into the first equation, we get:

 

One solution to this is x=0.  In this case, the first equation implies that y=0 too, but the point (0,0) isn’t on the unit circle.  So we may assume that x doesn’t equal 0, and we can divide out the x and solve for l.

When  l = –1/2,  from equation * we obtain:

Substituting this result into the equation for the unit circle, we get:

 

This has yielded the four points,  only one of which is in
 Quadrant I.   

 

When  l = 3/2,  from equation * we obtain:

Substituting this result into the equation for the unit circle, we get:

 

This has yielded the four points,      only one of which is in Quadrant I.

 

Finally we check to see which of these 2 points (in Quadrant I) is the maximum:       

 

 

So the maximum value of  f on the part of the unit circle that lies in Quadrant I is 3/2.

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8.  If a tachyon field intensity is described by  T(x, y, z) = 3e^xy–2e^yz,  then the tachyon field is increasing at (2, 0, 1) most rapidly in the direction of . . .

 

A scalar field like T increases most rapidly in the direction of its gradient:

 

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9.  How fast is the tachyon field increasing in the direction of maximum increase that you found in #8?   

 

The length of the gradient vector we just found in #10 is precisely what we seek:

 

.

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10.  Find any local extrema and saddle points of    .

 

Extrema and saddle points occur at any points where.

 

 

This yields the following linear system:

 

Which has solution  x = –1,  y = –2.   Thus a possible extremum or saddle point is (–1, –2).

To see which it is, we invoke the multivariable second derivative test.   We need the second derivative matrix, the Hessian:

 

 

The determinant of this matrix is  –12 – 4 = –16 < 0,  hence our point (–1, –2) is a saddle point.

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