If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

The upside-down Delta symbol is called 'del', and will by typed accordingly.  This symbol stands for the 'del' operator, which takes the partial derivative with respect to each variable, and multiplies it by the unit vector in the direction of that variable.

Thus for example if we have f(x, y), a function of the 2 variables x and y:

• its x derivative can be denoted f_x, and the unit vector in the x direction is i, giving us f_x i
• its y derivative can be denoted f_y, and the unit vector in the y directionis j, giving us f_y j
• del f is therefore equal to f_x i + f_y j.
• note that since the 'del' operator gives us a vector, it is denoted in boldface type.  We write

del f = f_x i + f_y j.

For a function of three or more variables, we must apply a similar process for each variable.  For example if we have the function g(x, y, z) we would get

del g = f_x i + f_y j + f_z k.

The expression del f is called the 'gradient' of f.  The gradient has a number of very useful properties:

• The gradient of a function, evaluated at a given point, gives us a vector in the direction of maximum increase of the function at that point.  Its magnitude is the maximum rate at which the function changes.
• If we take the dot product of the gradient, evaluated at a point, with any unit vector, we get the rate at which the function changes as we move in the direction of that unit vector.
• The equation f(x, y) = c describes a curve in the xy plane.  If (x, y) is a point on that curve, then the gradient, evaluated at that point, is normal to the curve ((i.e., perpendicular to the tangent line at that point).
• The equation f(x, y, z) = c describes a surface in xyz space.  If (x, y, z) is a point on that surface, then the gradient, evaluated at that point, is normal to the surface ((i.e., perpendicular to the tangent plane at that point).

Question:  `q001.  Let f(x, y) = x^2 y^3.  Find del f.

Your solution: 

Confidence rating:
 

Given Solution:  del f = f_x i + f_y j = 2 x y^3 i + 3 x^2 y^2 j

Self-critique (if necessary):

Self-critique rating:

Question: 

`q002.  Let f(x, y, z) = x^2 + x cos y + e^(yz).  Find del f.

Your solution: 

Confidence rating:
 

Given Solution:  del f = f_x i + f_y j + f_z k = ( 2 x + 3 cos(y) ) i + (-x sin(y) + z e^(yz) ) j + y e^(yz) k.

Self-critique (if necessary):

Self-critique rating:

Question: 

`q003.  Find the directional derivative of f(x, y) = x^2 y^3, at the point (2, 1) in the direction of the unit vector u = .6 i + .8 j. 

• If you move .01 unit in the direction of that vector by how much does your directional derivative indicate that the value of f(x, y) will change? 
• What will be your x and y coordinates after the move?
• What is the value of f(x, y) at the point (2, 1)?
• What is the value of f(x, y) at the new point?
• Did the function change by the amount you predicted?

The gradient is

del f = 2 x y^3 i + 3 x^2 y^2 j

At (2, 1) the gradient takes value 4 i + 12 j.

The directional derivative in the direction of the unit vector u is

(del f) dot (u) = 1.2 x y^3 + 2.4 x^2 y^2.

At (2, 1) the directional derivative takes value 12.

Had we take the dot product of the gradient 4 i + 12 j at the point (2, 1), with the unit vector u, we would have obtained the same result, 2.4 + 9.6 = 12.

Thus when we move in the direction of u, the rate of change of f(x, y) with displacement is 12.  That is, in the near vicinity of (2, 1), the value of f(x, y) increases at or close to the rate of 12 units for every unit of distance in the direction of u.

Thus if we move distance .01 in the direction of u we expect f(x, y) to change by 12 * .01 = .12.

At (1, 2) the value of f(x, y) is easily found to be 4. 

If we move .01 unit in the direction of u, we easily see that our position changes by .01 u = .006 i + .008 j, so our x coordinate changes by .06 and our y coordinate by .08.  Starting from point (1, 2) we therefore end up at point (2.006, 1.008).

Evaluating f(x, y) at (2.006, 1.008) we obtain value (2.006)^2 * (1.008)^3 = 4.1214.

We conclude that as a result of moving .01 unit in the direction of u, the function changes its value from 4 to 4.1214, a change of .1214.

The change predicted by the differential is .12, which differs from the actual change by about a 1% of that change.

Your solution: 

Confidence rating:
 

Given Solution: 

Self-critique (if necessary):

Self-critique rating:

Question: 

`q004.  The ellipse x^2 / 4 + y^2 / 9 = 1 can be expressed as the curve f(x, y) = 1, with f(x, y) = x^2 / 4 + y^2 / 9.  The point (1, 3 sqrt(3) / 2) lies on the curve.  Find a vector in the direction normal to the curve at that point.

The gradient of f(x, y), evaluated at a point, gives the direction in which f(x, y) is increasing most rapidly in the vicinity of that point. 

The level curves of f(x, y), i.e., the curves f(x, y) = c for constant c, have the property that for any point on a given level curve, a vector tangent to the curve lies in the direction in which the value of the function remains unchanging.

The direction perpendicular to the direction of maximum increase is the direction in which the value of the function remains constant.

So the tangent vector at a point is perpendicular to the gradient at that point.

It follows that the gradient is normal to the level curve.

The gradient of f(x, y) is

del f = f_x i + f_y j = x/2 i + 2 y / 9 j.

At the point (1, 3 sqrt(3) / 2) the gradient is

1/2 i + sqrt(3) / 3 j , approximately .5 i + .57 j,

This vector is therefore normal to the level curve f(x, y) = x^2 / 4 + y^2 / 9 = 1, i.e., is normal to the ellipse, at the given point.

The figure below depicts this ellipse in the first quadrant, with the lines x = 1 and y = sqrt(3) / 2 intersecting at the given point.  The line passing through this point with slope (y - 3 sqrt(3)/2) / (x - 1) = 2 sqrt(3)/3.  This slope matches that of the vector 1/2 i + sqrt(3) / 3 j , so this line is perpendicular to the curve at the given point.

Your solution: 

Confidence rating: