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Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Resubmitting Practice Test 1
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question form
#$&*
Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Worked Through Practice Exam
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1. Solve the equation y ' - 2 y = -1 for y(1) = 2.
[use integrating factor e^-(2t) to get (y e^(-(2t) ) ' = e^(-2 t) with solution implicit in y e^(-2 t) = 1/2 e^(-2t) + c so that y = 1/2 + c e^(2 t); evaluate c and check that solution works]
Nonhomogeneous
p(t) = 2
y = Ce^(-2t) + e^(-2t) int(e^(2t) -1) dt
y = Ce^(-2t) + e^(-2t) (-.5)(e^(2t))
y = Ce^(-2t) - .5
2 = Ce^(-2) - .5
2.5 = Ce^-2
C = 2.5/e^(-2)
Therefore
y = (2.5/e^(-2))e^(-2t) - .5
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It seems that your solution leads to
y ' = -5 / e^-2 e^(-2 t) - 5 / e^(-2) e^(-2t) + 1 , which does not come out to -1.
I believe an appropriate multiple of e^(2 t) , plus 1, would lead to a solution that works when substituted back into the equation.
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If you go through all the steps of the solution you should get such a function.
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Solutions should always include a check to be sure they work.
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I found a simple calculation error on my part the first time through. This is what I get now:
y’ - 2y = -1
p(t) = -2 int(p(t)) = -2t
y = Ce^(2t) + e^(2t) int(e^(-2t)(-1)) dt
integrating
-(-1/2)e^(2t) * e^(-2t)
= 1/2
y = Ce^(2t) + 1/2
2 = C e^(2*1) + 1/2
1.5 = Ce^2
C = 1.5/(e^2)
y = (1.5/e^2)*e^(2t) + 1/2
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Good.
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2. The temperature of a room approaches the outdoor temperature at a rate proportional to the difference between the two temperatures. The outdoor temperature is -10 Celsius, and the initial room temperature is 20 Celsius. The average rate at which temperature changes during the first 30 minutes is -.2 Celsius / minute.
Write the differential equation for this situation, and use it along with the given conditions to find the temperature of the room as a function of clock time.
temp_out = -10 temp_in = 20
rate = -.2 = K
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k is not equal to the average rate of temperature change.
k is a parameter in the equation and in the solution, that must be evaluated in terms of the given information.
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T’(t) = k[S - T]
T’(t) = k[-10 - T(t)]
integrate
T(t) = Ce^(-kt) -10
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You haven't shown the steps of the integration. This is very risky.
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20 = Ce^(-k(0)) -10
30 = C
So
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Is the average rate of change of this temperature function for the first 30 minutes equal to -.2 C / min?
Would it be so if your -2 was -.2?
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I am still not sure what your comments are getting at to correctly solve both of these problems.
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You got
T(t) = 30e^(-2t) -10
Using this function you can calculate the average rate of change of the temperature with respect to clock time, for the interval from t = 0 to t = 30 minutes.
I do not believe that you'll find the average rate of change to be -.2 Celsius / minute.
Nor do I believe this will be the case if you use -.2 for the value of k (I expect your use of -2 might have been a typo, since you began by assuming k = -.2).
The problem with your solution is that you seem to have assumed k to be the average rate of change of temperature. k the coefficient of t in the exponent of your function, not the average rate of change of temperature.
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Your solution for C is OK, but your assumption regarding k is not.
Your function is
T(t) = 30 e^(-k t) - 10.
You have to use the given conditions to find k.
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Do your best to answer the following question for this room: Any reasonable attempt will be given at least partial credit. A complete solution will get some extra credit. An acceptable full-credit answer would include either of the following:
· A correct differential equation which could be solved to answer the question.
· A good graphical description of the direction field of the equation.
[the temperature T approaches room temperature at a rate proportional to the temperature relative to room temperature, so
dT/dt = k (T - (-10 C) ).
Solve this equation, leaving k as an unknown parameter.
Find the temperature after 30 minutes, and use this along with your solution function to evaluate k.
Plot the resulting curve along with your direction field and verify that the curve and the field are consistent.]
A soft drink is taken from the refrigerator at 3 Celsius, and set in the room. If the room had stayed at 20 Celsius, the temperature of the soft drink would have reached 15 Celsius after 40 minutes. What will be the temperature of the room after 40 minutes, and what will be the temperature of the soft drink at this time?
T’(t) = k[20 - T]
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The 20 C temperature is not constant. To solve the problem you have to replace 20 C by the function that models room temperature.
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integrating you get
T(t) = Ce^(-kt) + 20
3 = Ce^0 + 20
C = -17
So T(t) = -17e^(-.3t) + 20
when t=40 the temp of the room would be
T(40) = -17e^(-.3*40) + 20
T(40) = 19.99 C
The temp of the soft drink would be 15 C ( it states this already in problem)
???Not sure if I solved this problem correctly but I did what I thought it was asking. Please correct me if I was wrong???
3. Find the exact equation corresponding to dF = 0, where F(t, y) = sqrt(t) * y^3. Explain how you would have solved this equation had you not been given the function F(t, y), and apply the condition y(0) = 1 to find a specific solution.
[dF = F_x dx + F_y dy, so the equation dF = 0 would be F_x dx + F_y dy = 0. The partial derivatives of the F function comprise the N and M functions of the exact equation. The equation must pass the test for exactness, since F_x y = F_y x for any differentiable function F(x, y).]
F(t,y) = sqrt(t)y^3
Take the derivative
(sqrt(t)y^3)’ = 0
sqrt(t)’y^3 + sqrt(t) (y^3)’ = 0
(t^(-1/2)/2)y^3 + (sqrt(t)*(2y^2)y’ = 0
1/(2sqrt(t))y^3 dt + (2sqrt(t)y^2) dy = 0
M(t,y)dt + N(t,y)dy = 0
M(t,y) = 1/(2sqrt(t))y^3 taking derivative
M = (y^3*t^(-3/2))/4
N(t,y) = 2sqrt(t)*y^2
N = 4sqrt(t)*y^3
M and N are not equal.
I can’t find my mistake. Maybe I went about solving this problem incorrectly???
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F_yx = F_xy, which ensures that M_y = N_t. This is the test.
Having worked through the derivatives, you don't have F_yx for either of your calculations.
I believe your correct equation will be
1 / (2 sqrt(t)) y^3 dt + 3 sqrt(t) y^2 dy = 0.
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I found my error when solving this problem the first time.
F(t,y) = sqrt(t)y^3
taking derivative
(sqrt(t))’ y^3 + sqrt(t)(y^3)’ = 0
(1/(2sqrt(t))y^3 dt + (sqrt(t)(3y^2))dy = 0
Checking to make sure Mdt + Ndy = 0, take derivative to check for exactness.
M = (1/(2sqrt(t))y^3
treat t as constant
dM/dy = 3y^2/(2sqrt(t))
N = (sqrt(t)(3y^2))
treat y as constant
dN/dt = 3y^2/(2sqrt(t))
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You've verified that the equation
(1/(2sqrt(t))y^3 dt + (sqrt(t)(3y^2))dy = 0
is exact.
Now how would you solve it?
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4. A swimmer pushes off the side of a pool and glides, slowing to a speed of 10 centimeters /second in 5 seconds after her feet lose contact with the wall, during which time she travels 300 centimeters. Assuming that the net force acting on the swimmer is proportional to her velocity, so that her acceleration is proportional to her velocity, what are her initial velocity and the drag constant. This can be solved without using her mass, but if you wish you may assume a reasonable mass.
[The force on the swimmer is of the form F = - c v so the m dv/dt = - c v, or m v dv/dx = - c v. These equations can be solved to get v(t), or v(x), as appropriate to the given information.]
???Not sure if reading through this problem it means that she went 300 cm in 5 sec???
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This is what the statement of the problem says.
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so initial velocity 300cm/5s = 60 cm/s = v_0?? Or would her initial velocity technically be 0, because when she is at the start position up on the wall she is not moving???
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The timing starts just after her feet leave the wall.
60 cm/s is her average velocity. It is not the velocity with which she leaves the wall, and is not particularly important to this solution.
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now trying to solve for k:
mv dv/dx = -kv
int(-m/k) dv = int(dx) the limits of v is 60 to 10 and limit on int for dx is 0 to 300
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Good, except that you don't know the initial value of v.
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-mv/k = x
-m(10-60)/k = (300-0)
50m/k = 300
m/k = 6
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You can treat k/m (or m/k, if you prefer) as a constant. So it would be valid to solve for m/k.
You know that velocity is a linear function of position, and you know the velocity at position x = 300 cm.
You might be better off solving the equation using t as your variable. The application of the known conditions is easier to understand if your variable is time.
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k = m/6???
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if x = 300 cm
mdv/dt = -kv
-m/(kv) dv = dt
integrating
(-m/k)ln|v| = t + C
Now I’m not sure where to go from here to find k, initial gel, or t????
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5. Find the general solution of the equation y ' + t^2 / y - 4 y = 0.
[This can be rearranged into the standard form of a Bernoulli equation and solved accordingly.]
Rearranging order of equation you get:
y’ -y = -t^2y^-1
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You haven't shown the change of variable or the intermediate steps here.
If you remember the formulas well enough this is OK, but it's often easier to understand all the steps than to try to memorize a bunch of formulas.
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v’ - 2v = -2t^2
Now use the linear non homogeneous equation format
v = Ce^(2t) + e^(2t) int(e^(-2t)(-2t^2)
v = Ce^(2t) -2e^(2t) int(t^2 e^(-2t) dt
using integration by parts u = t^2 du = 2t dv=e^(-2t) v= -.5e^(-2t)
v = Ce^(2t) - 2e^2t( -.5e^(-2t)(t^2) + int(te^(-2t)) again integrate by parts u =t du=dt dv= e^(-2t) v = -.5e^(-2t)
v = Ce^(2t) -2e^2t(-.5t^2(e^(-2t) - .5e^(-2t)t - .25e^(-2t))
v = v = Ce^(2t) + t^2 + t + .5
let y = v ^(1/(1-n) = v^(1/3)
y = (Ce^(2t) + t^2 + t + .5)^(1/3)
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I don't believe this works when substituted back into the original equation.
Your procedure is good, but before I can look for errors you need to fill in all the details, without skipping anything by using formulas that bypass the details.
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I actually wrote the problem done wrong the first time trying to solve the problem.
y’ -4y = -t^2y^-1
p(t) = -4 q(t) = -t^2 n=-1
v’ +(1-(-1))(-4)v = (1-(-1)(-t^2)
v’ - 8v = -2t^2
using non homogeneous format
p(t) = -8 int(p(t)) = -8t
v = Ce^(8t) + e^(8t) int(e^(-8t) * (-2t^2)
integrating
-2e^(8t) int(t^2e^(-8t))
use by parts
u = t^2 du=2t dv=e^(-8t) v = -(1/8)e^(-8t)
-2e^(8t)(-(1/8)t^2 * e^(-8t) + (1/4)int(te^(-8t)))
integrate by parts again
v = Ce^(8t) -2e^(8t)(-(1/8)t^2e^(-8t) -(1/32)te^(-8t) - (1/64)e^(-8t))
simplifying
v = Ce^(8t) + (1/4)t^2 + (1/16)t + (1/32)
putting back in terms of y:
y = v^(1/(1-n)) = v^(1/3)
y = (Ce^(8t) + (1/4)t^2 + (1/16)t + (1/32))^(1/3)
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6. Solve the equation y ' - 8 sin(2 t) = 6 y.
[This equation can be rearranged into the form of a linear equation and solved using an integrating factor.]
rearranging equation
y’ - 6y = 8sin(2t)
p(t) = -6
y = ce^(6t) + e^(6t) int(e^(-6t)*(8sin(2t)) dt
y = ce^(6t) + 8e^(6t) int( sin(2t)e^(-6t) )dt
integrating twice by parts
int(sin(2t)e^(-6t) = (-1/6)sin(2t)e^(-6t) - (1/18)cos(2t)e^(-6t) - (1/3)int(sin(2t)e^(-6t)
(4/3)int(sin(2t) e^(-6t) = (-1/6)sin(2t)e^(-6t) - (1/18)cos(2t)e^(-6t)
int(sin(2t) e^(-6t) = ((-1/6)sin(2t)e^(-6t) - (1/18)cos(2t)e^(-6t))/(4/3)
= (-1/8)sin(2t)e^(-6t) - (1/24)cos(2t)e^(-6t)
plugging that into the y = equation you get
y = Ce^(6t) + 8e^(6t) *((-1/8)sin(2t)e^(-6t) - (1/24)cos(2t)e^(-6t))
y = Ce^(6t) - sin(2t) - (1/3)cos(2t)
7. A 1% saline solution flows into a 100-gallon tank initially full of 4% saline solution. A thoroughly mixed solution flows out of the tank at the same rate as the inflow. What is the rate of inflow if after 1 hour the concentration in the tank is 3% saline?
No idea how to set up these types of problems. Can you please set up the problem and work through it???
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Let Q(t) be the quantity of salt in the tank and r the rate of flow.
At what rate is solution flowing in?
At what rate is salt flowing in?
What is the concentration of salt in the tank at time t, in terms of the as-yet unknown function Q(t)?
What is the rate at which mixed solution is flowing out of the tank?
What is the rate at which salt is therefore flowing out of the tank?
Using the answers to these questions, write an equation for dQ/dt, solve it, and evaluate the unknown constants (one of which is r, the other of which is the integration constant).
You might want to consult your text for the general way of setting up such equations.
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Still no idea what is what and where to start even after looking through the book and notes.
I don’t even know how to answer the questions you posed.
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You submitted Assignment 6 the day before yesterday.
This assignment is directly related to the present problem.
You solved the first problem correctly. However, as I noted in my comments on that assignment, you might well have been following the solution to an example problem without completely understanding the motivation for some of the steps (having been a student many years ago, I'm familiar with this step in the learning process).
I'm going to recommend that you try to answer these questions for that problem, so you'll see the reasoning behind the vaious steps in your solution.
Then try to apply what you learn to this problem.
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Note that you submitted that assignment under an incorrect access code, having used 901 for the last three digits of the code. The document is duly posted at the corresponding site, but I'm going to post it along with this one in today's batch. So it will follow this one in your menu.
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8. Find the solution of the equation y ' = y^2 x / 3 for initial condition y(0) = 1. Evaluate your solution at x = 1.
Use an Euler approximation with step 1/4 to estimate the solution at x = 1.
Your estimate will either be high or low. Which is it and why is it so?
1) Solve the equation:
dy/dx = y^2x/3
y^-2 dy = (1/3)x dx
integrate both sides
-1/y = x^2/6 + C
y = - 6/(x^2 + C)
finding C
1 = -6/(0 + C) C = -6
y = -6/(x^2 - 6)
2) find when x=1 with step size .25
k=0
y1 = y0 + .25((y^2_0(x_0))/3)
y_1 = 1
k=1
y_2 = y1 + .25((y^2_1(x_1))/3)
y_2 = 1 + .25(1^2(.25)/3)
y_2 = 1.0208
k=2
y_3 =y_2 + .25((y^2_1(x_1))/3))
y_3 = 1.0643
k=3
y_4 = 1.135
Our estimate will be a low estimate because the actual value at x = 1 is:
y = -6/(1^2 -6) = -6/-5 = 1.2.
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The estimate is low, as you found by evaluating your function.
The reason is is low has to do with the geometry of the situation, more specifically with the concavity of the function.
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Check my notes. You'll have additional questions. You can submit a copy of this document and use #### for insertions.
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