query 115

#$&*

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.

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

z explicitly in terms of t

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

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

Thus,

dz/dt = cos(3t)’*cot(3t) + cos(3t)*cot(3t)’

d/dt(cos(3t)) = -3sin(3t)

d/dt(cot(3t)) = d/dt(1/tan(3t)), composite fn f(g(h)) f(z)=1/tan(z), g(h) = 3t

f’(z) = 1’*tan(z) - 1*(tan(z))’/(tan^2(z))

= 0 - (1*sec^2(z))/ tan^2(z) = -sec^2(z)/tan^2(z)

= -csc^2(z)

g’(h) = 3

f(g(h))’ = f’(z)*g’(h) = -csc^2(3t)*3 = -3csc^2(3t)

dz/dt = cos(3t)’*cot(3t) + cos(3t)*cot(3t)’ = -3sin(3t)cot(3t) + -3cos(3t)csc^2(3t)

where the +1 has a derivative of 0.

By using the Chain Rule

z = xy + 1 and x = cos(3t) and y = cot(3t)

dz/dx =y, dz/dy = x, dx/dt = -3sin(3t), dy/dt = -3csc^2(3t)

Using the Chain Rule for one independent parameter:

dz/dt =(dz/dx)*(dx/dt) + (dz/dy)(dy/dt)

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

=cot(3t)*-3sin(3t) + cos(3t)*-3csc^2(3t), same result we received before!!!

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):

????Not sure if I’m correct but wouldn’t derivative of cot(3t) = -3csc^2(3t) not sec^2(3 t)???

I think tan(t) = -sec^2(t) but cot(3t) = -3csc^2(3t)??????

@& You're right. Good work.*@

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

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.

• Expressing z explicitly in terms of u and v.

• Apply the chain rule for two independent parameters.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First find the partial derivatives:

dz/dx = d/dx(x^2 + y^2) = 2x

dz/dy = d/dy(x^2 + y^2) = 2y

dx/du = d/du(u cos(v)) = cos(v)

dx/dv = d/dv(u cos(v)) = -usin(v)

dy/du = d/du(u + v^2) = 1

dy/dv = d/dv(u + v^2) = 2v

Therefore, the chain rule for 2 independent parameters gives

dz/du = (dz/dx)(dx/du) + (dz/dy)(dy/du)

2x* cos(v) + 2y*1 = 2ucos(v)*cos(v) + 2(u + v^2)*1

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

dz/dv = (dz/dx)(dx/dv) + (dz/dy)(dy/dv)

2x*-usin(v) + 2y*2v = 2ucos(v)*-usin(v) + 2(u + v^2)* 2v

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

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.

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Self-critique (if necessary):

I messed up somewhere in my approach or logic I’m not sure which one????

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

w_r = (dw/dx)(dx/dr) + (dw/dy)(dy/dr) + (dw/dz)(dz/dr)

(dw/dx) = d/dx(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = 1f(g(h))’ = 1* e^(x - y + 3z^2) = e^(x - y + 3z^2)

(dw/dx) = d/dx(e^(x - y + 3z^2)) = e^(x - y + 3z^2)

(dx/dr) = d/dr(r + t - s) = 1

(dw/dy) = d/dy(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = -1f(g(h))’ = -1* e^(x - y + 3z^2) = -e^(x - y + 3z^2)

(dw/dy) = d/dx(e^(x - y + 3z^2)) = -e^(x - y + 3z^2)

(dy/dr) = d/dr(3r - 2t) = 3

(dw/dz) = d/dz(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = 6zf(g(h))’ = 6z* e^(x - y + 3z^2) = 6ze^(x - y + 3z^2)

(dz/dr) = d/dr(sin(rst)) = f(g(h))f(z) = sin(z), g(h) = rst

f(z)’ = cos(z), g(h)’ = st

f(g(h))’ = f(z)’*g(h)’ = st*cos(rst)

(dz/dr) = d/dr(sin(rst)) = st*cos(rst)

Now to put it al together, w_r = (dw/dx)(dx/dr) + (dw/dy)(dy/dr) + (dw/dz)(dz/dr)

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

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

= 6ze^(x - y + 3z^2)*(st*cos(rst)), where z = sin(rst), x = r + t - s, and y = 3r - 2t

6(sin(rst))e^((r + t - s) - (3r - 2t) + 3 sin(rst)^2)*(st*cos(rst))

= 6st*sin(rst)cos(rst)e^(-2r + 3t - s + 3 sin(rst)^2)

So our final answer is

w_r = 6st*sin(rst)cos(rst)e^(-2r + 3t - s + 3 sin(rst)^2)

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):

?????I’m confused about how dw/dy dy/dr = - e^(x - y + 3 z^2) * 3, and isn’t just e^(x - y + 3 z^2)*-1 ?????

Apart from that my solution is almost the same as yours I believe.

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

Self-critique rating:

@& I can't spot your error right off. I'll have to take another look.

However I do believe that

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

The v^4 definitely follows from the y^2 with y = u + v^2.

*@

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I have a really hard time being given info and have to put it together to form an equation, is this normal or should I be able to put this info into an equation fairly quickly by now???

Have to look at 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.

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Self-critique (if necessary):

Ok, so first step take derivative of all values wrt t, getting

dR/dt(1/R) = -1/R^2

dR1/dt( 1/(R1)) = -1/R1^2

dR2/dt(1/(R2)) = -1/R2^2

dR3/dt( 1/(R3)) = -1/R3^2

So we know that the change in R wrt t will be the combined changes in R1,R2 and R3 wrt t. Giving us the equation…

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

Now to find roc in R wrt t we mult. thru by -R^2 and get

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

Now plugging in values for R1,R2 and R3 as well as the roc in R1,R2 and R3 which is given we get..

dR/dt = [-1/(150ohm)^2*(-3ohm/s)-1/(300ohm)^2*(4ohm/s) -1/(450)^2*(-3ohm/s]R^2

=approx. [.00013ohm/s - .00004ohm/s + .000015ohm/s]*-R^2

= approx. .000105ohm/s*-R^2

?????I understand everything till this point but don’t understand how you obtained a value for R^2????

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Self-critique rating:

@& 1/R = 1 / R1 + 1 / R2 + 1 / R3

Muliply both sides by R * R1 * R2 * R3, to get everything in the numerator.

Factor R out of the right-hand side, etc..

Few students will get this problem on the first try, but it should help prepare you for similar problems.*@

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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

query 115

#$&*

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.

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

z explicitly in terms of t

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

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

Thus,

dz/dt = cos(3t)’*cot(3t) + cos(3t)*cot(3t)’

d/dt(cos(3t)) = -3sin(3t)

d/dt(cot(3t)) = d/dt(1/tan(3t)), composite fn f(g(h)) f(z)=1/tan(z), g(h) = 3t

f’(z) = 1’*tan(z) - 1*(tan(z))’/(tan^2(z))

= 0 - (1*sec^2(z))/ tan^2(z) = -sec^2(z)/tan^2(z)

= -csc^2(z)

g’(h) = 3

f(g(h))’ = f’(z)*g’(h) = -csc^2(3t)*3 = -3csc^2(3t)

dz/dt = cos(3t)’*cot(3t) + cos(3t)*cot(3t)’ = -3sin(3t)cot(3t) + -3cos(3t)csc^2(3t)

where the +1 has a derivative of 0.

By using the Chain Rule

z = xy + 1 and x = cos(3t) and y = cot(3t)

dz/dx =y, dz/dy = x, dx/dt = -3sin(3t), dy/dt = -3csc^2(3t)

Using the Chain Rule for one independent parameter:

dz/dt =(dz/dx)*(dx/dt) + (dz/dy)(dy/dt)

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

=cot(3t)*-3sin(3t) + cos(3t)*-3csc^2(3t), same result we received before!!!

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):

????Not sure if I’m correct but wouldn’t derivative of cot(3t) = -3csc^2(3t) not sec^2(3 t)???

I think tan(t) = -sec^2(t) but cot(3t) = -3csc^2(3t)??????

@& You're right. Good work.*@

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

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.

• Expressing z explicitly in terms of u and v.

• Apply the chain rule for two independent parameters.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First find the partial derivatives:

dz/dx = d/dx(x^2 + y^2) = 2x

dz/dy = d/dy(x^2 + y^2) = 2y

dx/du = d/du(u cos(v)) = cos(v)

dx/dv = d/dv(u cos(v)) = -usin(v)

dy/du = d/du(u + v^2) = 1

dy/dv = d/dv(u + v^2) = 2v

Therefore, the chain rule for 2 independent parameters gives

dz/du = (dz/dx)(dx/du) + (dz/dy)(dy/du)

2x* cos(v) + 2y*1 = 2ucos(v)*cos(v) + 2(u + v^2)*1

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

dz/dv = (dz/dx)(dx/dv) + (dz/dy)(dy/dv)

2x*-usin(v) + 2y*2v = 2ucos(v)*-usin(v) + 2(u + v^2)* 2v

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

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):

I messed up somewhere in my approach or logic I’m not sure which one????

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

w_r = (dw/dx)(dx/dr) + (dw/dy)(dy/dr) + (dw/dz)(dz/dr)

(dw/dx) = d/dx(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = 1f(g(h))’ = 1* e^(x - y + 3z^2) = e^(x - y + 3z^2)

(dw/dx) = d/dx(e^(x - y + 3z^2)) = e^(x - y + 3z^2)

(dx/dr) = d/dr(r + t - s) = 1

(dw/dy) = d/dy(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = -1f(g(h))’ = -1* e^(x - y + 3z^2) = -e^(x - y + 3z^2)

(dw/dy) = d/dx(e^(x - y + 3z^2)) = -e^(x - y + 3z^2)

(dy/dr) = d/dr(3r - 2t) = 3

(dw/dz) = d/dz(e^(x - y + 3z^2)) = f(g(h))’ f(z) = e^(z), g(h) = x - y + 3z^2

f(z)’ = e^(z), g(h)’ = 6zf(g(h))’ = 6z* e^(x - y + 3z^2) = 6ze^(x - y + 3z^2)

(dz/dr) = d/dr(sin(rst)) = f(g(h))f(z) = sin(z), g(h) = rst

f(z)’ = cos(z), g(h)’ = st

f(g(h))’ = f(z)’*g(h)’ = st*cos(rst)

(dz/dr) = d/dr(sin(rst)) = st*cos(rst)

Now to put it al together, w_r = (dw/dx)(dx/dr) + (dw/dy)(dy/dr) + (dw/dz)(dz/dr)

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

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

= 6ze^(x - y + 3z^2)*(st*cos(rst)), where z = sin(rst), x = r + t - s, and y = 3r - 2t

6(sin(rst))e^((r + t - s) - (3r - 2t) + 3 sin(rst)^2)*(st*cos(rst))

= 6st*sin(rst)cos(rst)e^(-2r + 3t - s + 3 sin(rst)^2)

So our final answer is

w_r = 6st*sin(rst)cos(rst)e^(-2r + 3t - s + 3 sin(rst)^2)

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):

?????I’m confused about how dw/dy dy/dr = - e^(x - y + 3 z^2) * 3, and isn’t just e^(x - y + 3 z^2)*-1 ?????

Apart from that my solution is almost the same as yours I believe.

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

Self-critique rating:

@& I can't spot your error right off. I'll have to take another look.

However I do believe that

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

The v^4 definitely follows from the y^2 with y = u + v^2.

*@

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I have a really hard time being given info and have to put it together to form an equation, is this normal or should I be able to put this info into an equation fairly quickly by now???

Have to look at 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.

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Self-critique (if necessary):

Ok, so first step take derivative of all values wrt t, getting

dR/dt(1/R) = -1/R^2

dR1/dt( 1/(R1)) = -1/R1^2

dR2/dt(1/(R2)) = -1/R2^2

dR3/dt( 1/(R3)) = -1/R3^2

So we know that the change in R wrt t will be the combined changes in R1,R2 and R3 wrt t. Giving us the equation…

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

Now to find roc in R wrt t we mult. thru by -R^2 and get

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

Now plugging in values for R1,R2 and R3 as well as the roc in R1,R2 and R3 which is given we get..

dR/dt = [-1/(150ohm)^2*(-3ohm/s)-1/(300ohm)^2*(4ohm/s) -1/(450)^2*(-3ohm/s]R^2

=approx. [.00013ohm/s - .00004ohm/s + .000015ohm/s]*-R^2

= approx. .000105ohm/s*-R^2

?????I understand everything till this point but don’t understand how you obtained a value for R^2????

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Self-critique rating:

@& 1/R = 1 / R1 + 1 / R2 + 1 / R3

Muliply both sides by R * R1 * R2 * R3, to get everything in the numerator.

Factor R out of the right-hand side, etc..

Few students will get this problem on the first try, but it should help prepare you for similar problems.*@

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Self-critique (if necessary):

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#*&!

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