Math 6C - Chapter 12 Quiz

Math B6C – Chapter 12 Quiz – SOLUTIONS

* Fall 2001 *

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1. Find for the function g(x, y) = ln(x² + y²). Then g_x(1, 1) = ?

Thus,

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2. For the function g of problem #1, compute . Then g_yx(1, 2) = ?

Since this function is infinitely differentiable in the vicinity of (1,2), it follows that its mixed partial derivatives are equal. Thus g_yx = g_xy, and so we only need to differentiate g_x (which we found in problem #1) with respect to y:

Thus, .

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3. Find the linearization of f (x, y, z) = xz² – y²cos(z) at (1, 1, 0). L(x,y,z) = ?

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4. For the function h(x, y) = e^2x+y, find the best quadratic approximation at
the point (0, 0). Q(x,y) = ?

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5. f (x, y, z) = x y² – y z + x² z . Find the gradient of f at (1,1,1).

The y-component of this gradient is _____ .

, and so

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6. Find the directional derivative of the f in problem #5,

in the direction of at the point (1,1,1).

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7. Find the equation of the tangent plane to x² – y² + z² = 4 at the point (1,1,2).

Let . 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, 2), which as usual we’ll call L. The tangent plane is then just L(x,y,z) = 4.

Thus our tangent plane is given by , or

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8. Find the linearization of f (x, y, z) = x y z + 1 at (1, 2, 3). L(x, y, z) =

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9. Find the maximum value of f (x, y, z) = x + y z on the sphere x² + y² + z² = 3.

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

and ,

yielding the following equations:

The first of these equations gives us . Substitution the second equation into the third yields , from which we deduce that , and so the only two possible solutions for l are , and hence . So now we know that:

Now we substitute what we know into the equation , since any solution must be on that sphere:

Now we know that:

And we must check to see which of these 8 points (since these “plus or minuses” are all independent) is the maximum:

So the maximum value of f on the given sphere is 2 (and the minimum value is –2).

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10. If a triolic field intensity is described by T(x, y, z) = sin(xyz), then the triolic

field is increasing at (1, 2, p/2) most rapidly in the direction of . . .

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

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11. How fast is the triolic field increasing in the direction you found in #10?

(approximately)

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

which approximately equals .

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12. Find the tangent plane to the surface defined by at (1,0,0).

The equation is given by z = L(x,y), where L is the linearization of G at (1,0).

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13. Let T(x, y, z) = x²yz describe the temperature at (x, y, z). How hot is the hottest spot on the surface implicitly defined by ?

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

and ,

yield the following equations:

There are many ways to proceed. One way is to take ratios of the first two equations and the second two:

which yields

Now substitute (expressing everything in terms of y) into the constraint equation:

Then we have and . All of these “plusses or minuses” are independent, so we must check T out on all these 8 points:

Thus the hottest point on the surface has a temperature of .

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14. Find the maximum of x + y + z subject to the constraint ?

Suppose that x=0 and y=1. The constraint equation is then satisfied regardless of what z equals. Thus z can be made as large as we like, and so can the sum x + y + z. This means there is no maximum of x + y + z subject to the given constraint.

Given any (large) real number R, there are points on the surface , the sum of whose coordinates surpass R.

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15. Find the equation of the tangent plane to the surface at (1,1,1).

Let . We seek the tangent plane to the level surface defined by . One way to accomplish this is to find the linearization of F (which as usual we’ll call L) at (1,1,1). The tangent plane will then be L(x,y,z) = 4.

Thus our tangent plane is given by , or

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16. 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 = 0, y = –1. Thus a possible extremum or saddle point is (0, –1).
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 –40 – 16 = –56 < 0, hence our point (0, –1) is a saddle point.

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17. Find any local extrema and saddle points of      over the open
       rectangle defined by    0 < x < 3,    0 < y < 3.   Give the value of the function at this
       point, and classify it as a saddle point, local maximum or local minimum:

Extrema and saddle points occur at any points where .

This yields the following nonlinear system:

The first equation has solutions x =1 with y=anything, or y=0 with x=anything. The second equation has solutions x=0 with y=anything, or x=2 with y=anything, or with x=anything. The conjunction of the first set of solutions with the second yields the following possible local extrema or saddle points:

To see whether these points are maxima, minima, or saddle points, we invoke the multivariable second derivative test. The second derivative matrix, the Hessian, is

The determinant of this matrix is

Evaluating this Hessian determinant function at each of our candidate points, we get:

Thus the first point is a possible max or min, while the second two points are saddle points. To see whether the first point is a max or a min, we evaluate g_xx at this point:

Since this second derivative is positive, the point is a relative minimum of g.

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18. Find the absolute maximum of over the rectangle defined
by . The highest elevation on the graph of z = g(x,y) above

this closed rectangle in the xy-plane is _____ .

In problem #17, we learned that there are no local maximum points within the given rectangle. Thus we only need to check the edges and corners for any maximum or minimum points. The four edges can be parametrized as follows:

Taking the composition of g over each of these paths, we get:

Notice that g is identically 0 along the x and y axes.

Taking the derivative (and setting it equal to zero) of the second of these compositions, we get

which has a solution for every integer value of n. The only odd multiple of that lies within the interval 0 < t < 3 is . Thus could be a maximum point for g (i.e., the input that yields g’s maximum over the given square in the xy-plane).

Taking the derivative (and setting it equal to zero) of the third of these compositions, we get

which has solution t = 1. Thus could be a maximum point for g (i.e., the input that yields g’s maximum over the given square in the xy-plane).

Thus we have the four corners and the above two points upon which to evaluate g to locate the absolute maximum of g over the given square.

The maximum of g over the given square is thus 3, the highest elevation of g over the square.

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19. Find the absolute minimum of over the rectangle defined by . The lowest elevation on the graph of z = g(x,y) above this
closed rectangle in the xy-plane is _____ .

All we need do is to compare the minimum found in problem #17 with the values just tabulated in problem #18:

Comparing this with the previous values, we see that the relative minimum of g is also the absolute minimum over the given square.

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20. How many local extrema and saddle points has ?

Such points occur when , i.e. when

This yields the following nonlinear system:

which simplifies to

The first equation has solutions x = 0 with y=anything, or x = –2 with y=anything. The second equation has solutions y = 0 with x=anything, or y=2 with x=anything. The conjunction of the first set of solutions with the second yields the following possible local extrema or saddle points:

To see whether these points are maxima, minima, or saddle points, we invoke the multivariable second derivative test. The second derivative matrix, the Hessian, is

The determinant of this matrix is

Evaluating this Hessian determinant function at each of our candidate points, we get:

Thus (0, 0) and (–2, 2) are both saddle points (since the determinant is negative for these points). To see whether the remaining two points are max or min, we evaluate h_xx at these points:

and so (0, 2) yields a relative minimum and (–2, 0) yields a relative maximum of h.

So h has two saddle points, one local minimum and one local maximum point; in total h has 4 local extrema and saddle points.