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course Mth 277
November 14 around 10:30pm.
Query 11.5*********************************************
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:
1st way:
Plug in x and y into given equation:
=(cos 3t)*(cot 3t)+1
=((cos 3t)^2+(sin 3t))/(sin 3t)
Derivative of that: ?
2nd way:
f_x=y and dx/dt=(-3sin(3t))
f_y=x and dy/dt=(-3/(sin 3t)^2)
=y(-3sin 3t)+ x(-3/(sin 3t)^2)
=(cot 3t)*(-3sin 3t) + (cos 3t)(-3/((sin 3t)^2)
=[-3 cos 3t] + [(-3 cos 3t)/((sin 3t)^2)
=(-3[(sin 3t)^2 +1]*(cos 3t))/((sin 3t)^2)
confidence rating #$&*:
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Given Solution:
z = x y + 1 = cos(3t) * cot(3 t) + 1, so
dz/dt = (cos(3t)) ' cot(3t) + cos(3t) (cot(3t) ) ' = -3 sin(3t) cot(3t) - cos(3 t) sec^2(3 t), which could be simplified.
Using the chain rule
dz/dt = dz/dx dx/dt + dz/dy dy/dt
= y * (-3 sin(3t) ) + x * (-sec^2(3 t))
= -3 sin(3 t) cot(3 t) + cos(3 t) * (-sec^2(3 t) ),
which could also be simplified but is clearly equal to the previous expression.
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Self-critique (if necessary):
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Question: `q002.
Let F(x,y) = x^2 + y^2 where x(u,v) = u cos(v) and y(u,v) = u + v^2. Let z = F(x(u,v),y(u,v)). Find z_u and z_v in the following ways.
Expressing z explicitly in terms of u and v.
Apply the chain rule for two independent parameters.
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Your solution:
1st way:
z=F(u cos v, u+v^2)
=(u cos v)^2 + (u+v^2)
Derivative: 2(u cos v)(u cos v) + (u+v^2)
Good, but you still have to find (u cos v) + (u+v^2), where ' indicates the partial derivative with respect to the appropriate variable.
2nd way:
f_u= 2u*(cos v)^2+1 and du/dt=cos v
f_v= 2v-2 sin (v)* cos (v)* u^2 and dv/dt=2v
=[2u(cos v)^2+1*(cos v)] + [2v-2 sin(v)*cos(v)*u^2*(2v)]
You would then simplify this and it would equal the derivative of the 1st way answer.
confidence rating #$&*:
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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):
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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:
This would be where I apply the chain rule for a function of 3 variables with three parameters.
w_r=(w_x*x_r)+(w_y*y_r)+(w_z*z_r)
=[f_x(e^(x - y + 3z^2))]*[f_r(r+t-s)] + [f_y(e^(x - y + 3z^2))]*[f_r(3r-2t)] + [f_z(e^(x - y + 3z^2))]*[f_r(sin(rst))]
=[1e^(x - y + 3z^2)]*[1] + [-e^(x - y + 3z^2)]*[3] + [6z e^(x - y + 3z^2)]*[st*cos (rst)]
=[e^(x - y + 3z^2)] + [-3e^(x - y + 3z^2)] + [6z e^(x - y + 3z^2)]*[st*cos (rst)]
It probably simplifies more than this, however, after you get it simplified, you will then put it in terms of r, s, t and that will give you w_r.
Good. Compare with the given solution below.
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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):
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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=(1/150)+(1/300)+(1/450)
R1= dec 3ohm/sec
R2= inc 4 ohm/sec
R3= dec 3ohm/sec
I dont even know how to set this problem up.
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):
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Good work overall. Check my notes.
The given solutions were posted for assignments through 11.7 as of Saturday.
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