Query 113

#$&*

course Mth 277

10/30 5

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

Your solution:

f_x = y^4arctan(y)

f_y, product rule, simplify, left with x(4y^3*arctan(y) + 1/(y^2+1) *y^4)

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):

shouldn't the product rule be (f*g) = f ' * g + f * g' ???

------------------------------------------------

Self-critique rating:3

@& Should be, yes.

Your answer is good.

Simplified you would have

x * (4 y^3 (y^2 + 1) arctan(y) + y^4) / (y^2 + 1) *@

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

OK

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

SIMPLE.

OK

------------------------------------------------

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:

worked on paper. found an error in your derivative, just a typo

OK

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):

derivative should be (-1x^2 +1) / x^2 to get -3/4

?????????????the analagous analysis of the slope doesn't make sense to me yet. I dont see how you can say 2- 1/2 = 3/2 from the point (2,-1,-5/2)

where does the 2-1/2 come from???

------------------------------------------------

Self-critique rating:

@& If x = 2 then the function is (4 + y^2) / (2 y), with derivaive -2 / y^2 + 1/2. For y = 1 this is -2 + 1/2 = -3/2.

Looks like my - sign error persisted.*@

*********************************************

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:

OK

did for both solutions

first I got they both = -2x*sin(yx^2)

then second one i got sin(2x)sin(y)

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):

It would be more clear if you said f_y = this...so f_y_x = that. so you would know that its the partial derivative of x in the paritial derivative of y of f function.

like d/dy f(x,y) = ANSWER. so d/dy * ANSWER = other answer

------------------------------------------------

Self-critique rating:

@& The xy notation is pretty standard; the entire xy is the subscript. *@

*********************************************

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:

z_t = 2cos(2t) * sin(2cx)

z_tt = -4sin(2t) * sin(2cx)

here the ""c"" terms are not the same, but just a multiple. so z would have to be z = z_t * (C)z_tt

C representing constant multiple

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 understand the partial deravation behind this, but i'm not sure where you said ""we can see that z_xx = z_tt *c^2. because i see z_tt / c^2

------------------------------------------------

Self-critique rating:

@&

z_tt = -4 sin(2 t) sin(2 c x)

so

c^2 z_tt = -4 c^2 sin(2 t) sin(2 c x)

This is equal to z_xx.

So

z_xx = c^2 z_tt.*@

*********************************************

Question: `q002.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@& Good, and thanks for the corrections.

Check my notes.*@

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!