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
Question: `q001.
Let z = f(x,y) = xy + 1 where x = cos 3t and y = cot 3t.
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
Self-critique rating:
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.
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `q003. Find w_r where w = e^(x - y + 3z^2) and x = r + t - s, y = 3r - 2t, z = sin(rst).
Your solution:
Confidence rating:
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)) )
Self-critique (if necessary):
Self-critique rating:
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?
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
Self-critique rating: