Math 223 Homework 3 Solutions

  1. As you move in the x direction from (-6,14), the values on the contours decrease, so tex2html_wrap_inline79 . Also, in this direction, the contours get closer together while decreasing, so the surface must be concave down, which means tex2html_wrap_inline81 . In the y direction, you stay on the contour z=159 (not shown), so tex2html_wrap_inline87 . Likewise, tex2html_wrap_inline89 . To figure out tex2html_wrap_inline91 , look at the slope in the x direction as you move to increasing y. The contours do not change at all, so tex2html_wrap_inline97 .
  2. To estimate tex2html_wrap_inline99 use the points (-6.2,14,158) and (-6,14,159). Thus tex2html_wrap_inline105 , so

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    and for tex2html_wrap_inline107 use the points (-6,18.2,158) and (-6,14,159), and

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  3. tex2html_wrap_inline113

    tex2html_wrap_inline115 , and

    tex2html_wrap_inline117

    tex2html_wrap_inline119 .

  4. The tangent plane is tex2html_wrap_inline121 which is equivalent to

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    Using the tangent plane to estimate f(-7,15) yields an approximate value of z=163.04 which compared to the actual value of f(-7,15)=160.45 gives a relative error of tex2html_wrap_inline129 . A relative error of 1.6% is a good approximation.

  5. The value of tex2html_wrap_inline131 and tex2html_wrap_inline133 give the change of f over x and y. Since f is measured in concentration of oil (moles per liter) and x and y are measured in distance (km), tex2html_wrap_inline131 gives the rate of change of concentration per kilometer traveled in the x direction. tex2html_wrap_inline133 gives the rate of change of concentration per kilometer traveled in the y direction. The gradient gives you the direction in which the rate of change of concentration is increasing the fastest, and its magnitude gives the maximum rate of change of concentration.
  6. tex2html_wrap_inline155 .
  7. To move toward the highest concentration, we should move in the direction of maximum increase of f, which is in the direction of the gradient. Thus, we should move along tex2html_wrap_inline159 for one unit. The unit vector in that direction is approximately tex2html_wrap_inline161 .
  8. If we move one unit along tex2html_wrap_inline163 , we are at the new point (-6-0.99,14+0.01)=(-6.99,14.01). The gradient at the new point is

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  9. The magnitude of the gradient at each point is

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    so,

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    Since this is greater than 0.05, we must move one unit in the direction of tex2html_wrap_inline167 from the point (-6.99,14.01). The unit vector in the direction of tex2html_wrap_inline167 is tex2html_wrap_inline173 , so we should move one unit in the positive y direction to the point (-6.99,15.01). The gradient at this point is

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    The difference in the magnitudes of the gradients from (-6.99,14.01) and (-6.99,15.01) is

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    which is less than 0.05, so we can stop.

  10. We are now at the point (-6.99,15.01) where the gradient is approximately the zero vector. Thus there is no direction of maximum rate of increase. There is also no maximum rate of decrease since this is the direction of tex2html_wrap_inline185 . Looking at the contours we see that we are at a point that is above all the other contours, thus, we must be at the highest point of the function (local max.) Since this point is where the concentration of oil is the highest, we should drop the algae here.