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. 

Question:  `q001.  Find f_x and f_y when f(x,y) = xy^4*arctan(y).

Your solution: 

Confidence rating:
 

Given Solution: 

f_x = y^4 arctan(y)

f_y = x * ( (y^4) ' arcTan(y) - y^4 * (arcTan(y)) ' ) = x * 4 y^3 - y^4 * 1 / (1 + y^2) = y^3 ( 4 x - (y / (1 + y^2) ).  Simplification should be continued until the result is a single fraction with numerator y^3 ( 4 x + 4 x y^2 - y) and denominator (1 + y^2).

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Question:  `q002.  Determine z_x and z_y by differentiating the expression 4x^2 + 2y^2 + 3z^2 = 9 implicitly.

Your solution: 

Confidence rating:
 

Given Solution: 

Self-critique (if necessary):

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Question:  `q003.  Let f(x,y) = (x^2 + y^2)/(xy), P = (2, -1, -5/2)

• Find the slope of the tangent line of the graph of f parallel to the xz-plane at the point P.
• Find the slope of the tangent line of the graph of f parallel to the yz-plane at the point P.

Your solution: 

Confidence rating:
 

Given Solution: 

In a plane parallel to the xz plane, y is constant and z is a function of x only.  If y = -1, then the function becomes -(x^2 + 1) / x, and its derivative is -1 + 1 / x^2.  At x = 2 the slope is therefore -3/4.

Alternatively, f_x = 1/y - y / x^2.  At P = (2, -1, -5/2) we get f_x = -1 + 1/2^2 = -3/4.

Analogous analysis of the slope in a plane parallel to the yz plane indicates a slope at (2, -1, -5/2) of 2 - 1/2 = 3/2.

Thus the vectors i - 3/4 k and j + 3/2 k are tangent to the plane at (2, -1, -5/2).  We can calculate a cross product to get a normal vector, and knowing the coordinates of P we can then easily find the equation of the tangent plane.

Self-critique (if necessary):

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Question:  `q004.  For the two following functions, show that f_xy = f_yx.

• f(x,y) = cos(yx^2).
• f(x,y) = (cos^2(x))*(cos y).

Your solution: 

Confidence rating:
 

Given Solution: 

For the first function

f_x = -2 x sin(y x^2) so f_xy = -2 x^3 cos(y x^2)

f_y = -x^2 sin(y x^2) and f_yx = -2 x^3 cos(y x^2)

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Question:  `q005.  In physics the wave equation is given by z_tt = c^2 * z_xx and the heat equation is given by z_t = c^2 * z_xx. In the two following cases, see if z satisfies the wave equation, the heat equation, or neither.

• z = sin(2t)*sin(2cx).
• z = (e^(-t))(sin (x/c) + cos(x/c).

Your solution: 

Confidence rating:
 

Given Solution: 

For the first function

z_t = (sin(2t)) ' sin( 2 c x) = 2 cos (2 t) sin( 2 c x), and z_tt = -4 sin(2 t) sin(2 c x)

z_x = 2 c sin(2 t) * cos(2 c x) and z_xx = -4 c^2 sin(2 t) sin(2 c x).

The two second partials are identical except for the c^2 in z_xx. 

So we see that z_xx = z_tt * c^2.

This is close, but not quite, of the same form as the wave equation.  However the wave equation has the c^2 on the z_xx term, not the z_tt term.

Had the function been z = sin(2 x) * sin( 2 c t), it would have satisfied the wave equation.

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Question:  `q002. 

Your solution: 

Confidence rating:
 

Given Solution: 

Self-critique (if necessary):

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