#$&*
course Mth 277
3/21 1
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
*********************************************
Question: `q001. Find f_x and f_y when f(x,y) = xy^4*arctan(y).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
f_x = y^4*arctan(y)
f_y = x*((y^4)’*arctan(y) + y^4*(arctan(y))’)
=x*(4y^3arctan(y) + y^4/(1+y^2))
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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).
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I’m not sure why xy^4*arctan(y) is not x(y^4’*artan(y) + y^4arctan(y)’) instead it is subtracted where it is worked out and why is the arctan(y) disappears from equation when mult by y^4’??????
@& I did leave out the arcTan(y) in the simplification.
Corrected:
f_y = x * ( (y^4) ' arcTan(y) - y^4 * (arcTan(y)) ' )
= x * 4 y^3 arcTan(y) - 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 arcTan(y) + 4 x y^2 - y) and denominator (1 + y^2).*@
------------------------------------------------
Self-critique rating:2
*********************************************
Question: `q002. Determine z_x and z_y by differentiating the expression 4x^2 + 2y^2 + 3z^2 = 9 implicitly.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
z_x = 8x + 6z(`dz/`dx) = 0
`dz/`dx = -8x/6z = -4x/3z
z_y = 4y + 6z(`dz/`dx) = 0
`dz/`dy = -4x/6z = -2x/3z
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: `q003. Let f(x,y) = (x^2 + y^2)/(xy), P = (2, -1, -5/2)
• Find the slope of the tangent line of the graph of f parallel to the xz-plane at the point P.
• Find the slope of the tangent line of the graph of f parallel to the yz-plane at the point P.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
tangent line of the graph of f parallel to the xz-plane at point P(x0, y0, z0) has slope f_x(x0, y0)
f_x(2, -1)
f_x = (x^2 + y^2)*(xy)^-1 = (x^2 + y^2)’*(xy)^-1 + (x^2 + y^2)*((xy)^-1)’
(x^2 + y^2)’ = 2x*(xy)^-1
((xy)^-1)’ = f(g(h))’, f(z) = 1/z, g(h) = xy = x’*y + x*y’
f(g)’ = -1/z^2
g(h)’ = y
f(g(h))’ = -1/(xy) + y
I’m confused somewhere I’ve missed a step, I think I didn’t calculate for a composite function somewhere???????????
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):
Even after looking at solution I’m confused.
------------------------------------------------
Self-critique rating:
@& The derivative is easier if you express the function as
f(x, y) = x^2 / (x y) + y^2 / (x y) = x / y + y / x.
The x derivative is 1 / y - y / x^2.
The slope `dz / `dx, at the given point, is therefore -3/4.
It follows that the vector `i - 3/4 `k is tangent to the graph.
Let me know if that doesn't help you clarify the situation.*@
*********************************************
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).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For f(x,y) = cos(yx^2)
f_xy wrt x = f(g(h)), f(z) = cos(z), g(h) = yx^2, f’(z) = -sin(z), g’(h) = 2yx
f(g(h))’ = f(g(h))’*g’(h) = -sin(yx^2)*2yx = -2yxsin(yx^2)
Now derivative of -2yxsin(yx^2) wrt y,
f(g(h)), f(z) = -2yxsin (z), g(h) = yx^2, f’(z) = -2xcos(z), g’(h) = x^2
f(g(h))’ = f(g(h))’*g’(h) = -2xcos(yx^2)* x^2
f_xy = -2x^3cos(yx^2)
f_xy wrt y = f(g(h)), f(z) = cos(z), g(h) = yx^2, f’(z) = -sin(z), g’(h) = x^2
f(g(h))’ = f(g(h))’*g’(h) = -sin(yx^2)*x^2 = -x^2sin(yx^2)
Now derivative of -x^2sin(yx^2) wrt x,
f(g(h)), f(z) = -x^2sin(z), g(h) = yx^2, f’(z) = -2xcos(z), g’(h) = 2yx
f(g(h))’ = f(g(h))’*g’(h) = -2xcos(yx^2)* 2yx = -4yx^2cos(yx^2)
f_yx = -4yx^2cos(yx^2)
NOT equal expressions, -2x^3cos(yx^2) does not match -4yx^2cos(yx^2)??????
????I’m missing something here because I’m not getting equal expressions????
For f(x,y) = (cos^2(x))*(cos y)
f_xy = cos^2(x)’*cos(y) + cos^2(x)*cos(y)’ = -sin^2(x)cos(y) + 0 = -sin^2(x)cos(y)
Now derivative wrt y of -sin^2(x)cos(y)
(-sin^2(x)cos(y))’ = -sin^2(x)’*cos(y) + -sin^2(x)*cos(y)’
= 0 + sin^2(x)sin(y)
f_yx = cos^2(x)’*cos(y) + cos^2(x)*cos(y)’ = 0+ -cos^2(x)sin(y) = -cos^2(x)sin(y)
Now derivative wrt x of -cos^2(x)sin(y)
(-cos^2(x)sin(y) )’ = -cos^2(x)’*sin(y) + -cos^2(x)*sin(y)’
= sin^2(x)sin(y) + 0
Expressions are the same!!!
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
For the first function
f_x = -2 x sin(y x^2) so f_xy = -2 x^3 cos(y x^2)
f_y = -x^2 sin(y x^2) and f_yx = -2 x^3 cos(y x^2)
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I had some trouble with this problem got the second one though, if you see my error please let me know. Thanks
------------------------------------------------
Self-critique rating:
@& The given solution is wrong.
The x derivative of cos(y x^2) is
-2 x y sin(y x^2).
The y derviative of this expression is
-2 x sin(y x^2) - 2 x^3 y cos(y x^2).
The y derivative of cos(y x^2) is
- x^2 sin(y x^2),
and the x derivative of this expression is
-2 x sin(y x^2) - 2 x^3 y cos(y x^2). *@
*********************************************
Question: `q005. In physics the wave equation is given by z_tt = c^2 * z_xx and the heat equation is given by z_t = c^2 * z_xx. In the two following cases, see if z satisfies the wave equation, the heat equation, or neither.
• z = sin(2t)*sin(2cx).
• z = (e^(-t))(sin (x/c) + cos(x/c).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I looked at solution for first equation and now am going to attempt the second
z_x = e^-t[(1/c)cos(x/c) - (1/c)sin(x/c))]
z_xx =e^-t[(-(1/c^2)sin(x/c) - (1/c^2)cos(x/c))] = -(1/c^2)e^-t[sin(x/c) + cos(x/c)]
z_t = -e^-t(sin (x/c) + cos(x/c)
z_tt = e^-t(sin (x/c) + cos(x/c)
This would be the heat equation correct, I’m still a little confused even after working through it.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I also had some trouble understanding Example #8 where f_xx = (f_x)_x and f_xxy = (f_xx)_y and some trouble with just bringing it all together, if you have an idea of which notes cover this section I would love to look at them but I looked through the notes and had trouble seeing them.
------------------------------------------------
Self-critique rating:
@& You have shown that z_xx = -1/c^2 * z_tt, so that
z_tt = - c^2 * z_xx
Except for the - sign this is the same as the wave equation; however because of the - sign it's not so.*@
*********************************************
Question: `q002.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
??????Not sure what question this relates to???????????????????
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@& Me neither; must be left over from some editing.*@
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
@& See my notes.
You had a question about an example in the text; remind me about that when I'm on campus and near the book.*@