Query115

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course Mth 277

845pm 12/3/2011

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.

11.5

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Question: `q001.

Let z = f(x,y) = xy + 1 where x = cos 3t and y = cot 3t.

Find dz/dt after finding z explicitly in terms of t.

Use the chain rule for one parameter to find dz/dt.

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Your solution:

z = cos(3t)*cot(3t) + 1

'z = cos(3t)*-csc^2(3t) - 3cot(3t)*sin(3t)

= -3csc^2(3t)cos(3t) - 3cot(3t)sin(3t)

confidence rating #$&*:

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Given Solution:

z = x y + 1 = cos(3t) * cot(3 t) + 1, so

dz/dt = (cos(3t)) ' cot(3t) + cos(3t) (cot(3t) ) ' = -3 sin(3t) cot(3t) - cos(3 t) sec^2(3 t), which could be simplified.

Using the chain rule

dz/dt = dz/dx dx/dt + dz/dy dy/dt

= y * (-3 sin(3t) ) + x * (-sec^2(3 t))

= -3 sin(3 t) cot(3 t) + cos(3 t) * (-sec^2(3 t) ),

which could also be simplified but is clearly equal to the previous expression.

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Self-critique (if necessary):

dz/dt = dz/dx dx/dt + dz/dy dy/dt

dz/dx = x

= cos(3t)

@&

dz/dx = y, not x.

*@

dx/dt = -3sin(3t)

dz/dy = y

= cot(3t)

@&

dz/dy = x, not y.

*@

dy/dt = -3csc^2(3t)

dz/dt = -3xsin(3t) + -3ycsc^2(3t)

= -3cos(3t)sin(3t) -3cot(3t)csc^2(3t)

Its not the same IDK??????????

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Self-critique rating:1

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Question: `q002.

Let F(x,y) = x^2 + y^2 where x(u,v) = u cos(v) and y(u,v) = u + v^2. Let z = F(x(u,v),y(u,v)). Find z_u and z_v in the following ways.

Expressing z explicitly in terms of u and v.

Apply the chain rule for two independent parameters.

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Your solution:

f(x,y) = x^2 + y^2

x(u,v) = u cos(v)

y(u,v) = u + v^2.

z = F(x(u,v),y(u,v))

z = [u cos(v)]^2 + (u + v^2)^2

= u^2cos^2(v) + v^4 + u^2 + 2uv^2

z_u = 2ucos^2(v) + 2u + 2v^2

z_v = -2u^2sin(v)cos(v) + 4v^3 + 4uv

dz/du = dz/dx * dx/du + dz/dy * dy/du

= 2x*cos(v) + 2y

= 2ucos(v)cos(v) + 2(u+v^2)

= 2ucos^2(v) + 2u + 2v^2

dz/dv = dz/dx * dx/dv + dz/dy * dy/dv

= -2xuvsin(v) + 2y*2v

= -2ucos(v)*uvsin(v) + 2(u+v^2)*(2v)

= -2u^2vsin(v)cos(v) + (2u + 2v^2)*(2v)

= -2u^2vsin(v)cos(v) + 4uv + 4v^3

CHECK

confidence rating #$&*:

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Given Solution:

z = x^2 + y^2 = (u cos(v))^2 + (u + v^2) ^ 2 = u^2 cos^2(v) + u^2 + 2 u v^2 + v^4.

So

z_u = 2 u cos^2(v) + 2 u + 2 v^2

and

z_v = -2 u^2 cos(v) sin(v) + 4 u v + 4 v^3.

Applying the chain rule:

F_x = 2 x and F_y = 2 y.

dx/du = cos(v), dx/dv = - u sin(v), dy/du = 1 and dy/dv = 2 v. Thus

dz/du = dz/dx * dx/du + dz/dy * dy/du

= 2 x * cos(v) * 2 v + 2 y * 1

= 2 u cos(v) * cos(v) + 2 (u + v^2) * 1.

When simplified this agrees with our previous result z_u = 2 u cos^2(v) + 2 u + 2 v^2 .

dz/dv works out in an analogous manner.

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Self-critique (if necessary):OK

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Self-critique rating:

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Question: `q003. Find w_r where w = e^(x - y + 3z^2) and x = r + t - s, y = 3r - 2t, z = sin(rst).

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Your solution:

dw/dr = dw/dx*dx/dr + dw/dy*dy/dr + dw/dz*dz/dr

= e^(x-y+3z^2)*1 - e^(x-y+3z^2)*3 + 6ze^(x-y+3z^2)*stcos(rst)

= -2e^(x-y+3z^2) + 6stze^(x-y+3z^2)cos(rst)

= 2e^(x-y+3z^2) * [3stzcos(rst) - 1]

= 2e^[r+t-s-3r+2t+3sin^2(rst)] * [3stsin(rst)cos(rst) - 1]

= 2e(-2r+3t-s+3sin^2(rst)) * [3stsin(rst)cos(rst) -1]

confidence rating #$&*:

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Given Solution:

w_r can be written dw/dr, and we have

dw/dr = dw/dx dx/dr + dw/dy dy/dr + dw/dz dz/dr

= e^(x - y + 3 z^2) * 1 - e^(x - y + 3 z^2) * 3 + 6 z e^(x - y + 3 z^2) * (r s) cos(r s t).

Simplifying and substituting for x, y and z we get

(-2 + 6 sin(rst) ) e^(-2 r + 3 t + 3 ( sin^2(rst)) )

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Self-critique (if necessary):

I think in the given solution dw/dz*dz/dr should be equal to 6ze^(x-y+3z^2)*(st) cos(r s t)

also there should be an (-s) in the exponential

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Self-critique rating:3

@&

You are correct:

= e^(x - y + 3 z^2) * 1 - e^(x - y + 3 z^2) * 3 + 6 z e^(x - y + 3 z^2) * (s t) cos(r s t).

x - y + 3 z^2 = -2r + 3 t -s + 3 sin^2(rst)

*@

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Question: `q004. The combined resistance of three resistors R is given by the formula 1/R = 1/(R1) + 1/(R2) + 1/(R3). Suppose that at a certain

instant R1 = 150 ohm, R2 = 300 ohm, and R3 = 450 ohm. R1 and R3 are decreasing at a rate of 3 ohm/sec and R2 is increasing at a rate of 4 ohm/sec.

How fast is R changing at this instant and is it increasing or decreasing?

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Your solution:

1/R = 1/(R1) + 1/(R2) + 1/(R3)

d/dR = R/R1*R1/dR1 + R/R2*R2/dR2 + R/R3*R3/dR3

= 1/150 * -3 + 1/300*-3 + 1/450*4

= -0.02 - 0.01 + 0.009

= -0.021 ohm/s

confidence rating #$&*:

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Given Solution:

Regarding R, R1, R2 and R3 as functions of t, we take the t derivative of both sides and get

-1/R^2 dR/dt = -1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt

so that

dR/dt = (-1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt) * R^2

Substituting the given values for the three resistances and the three rates of change we get

dR/dt = (-1 / (150 ohms)^2 * (-3 ohms/sec) - 1 / (300 ohms) * (+4 ohms/sec) - 1 / (450 ohms) * (-3 ohms/sec) ) * R^2

= 7 / (67500 ohm sec) * (900/11 ohms) ^2

= .69 ohms / sec, approx..

You can verify that for the given values of R1, R2 and R3 we get R = 900/11.

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Self-critique (if necessary):

Had the right idea, just didn't do it with respect to time. 1/R derivative is 1/R^2 I understand the problem.

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Self-critique rating: 3"

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#