Calculus II
Asst # 19
Question:
**** Query problem 11.5.12 4th edition 11.5.8 5th edition 11.5.12 was 10.5.8
$1000 at rate r
what differential equation is satisfied by the amount of money in the account at time t since the original investment?
Your solution:
Confidence Assessment:
Given Solution:
dM/dt = 0.05M
** The equation is dM/dt = r * M.
The question is not posed for the specific value r = .05. **
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11:39:48
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**** What is the solution to the equation?
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11:40:08
M = 1000e^(0.05*t)
t is in years
The equation is dM/dt = r * M.
We separate variables to obtain
dM / M = r * dt so that
ln | M | = r * t + c and
M = e^(r * t + c) = e^c * e^(r t), where c is an arbitrary real number.
For any real c we have e^c > 0, and for any real number > 0 we can find c such
that e^c is equal to that real number (c is just the natural log of the
desired positive number). So we can replace e^c with A, where it is
understood that A > 0.
We obtain general solution
M = A e^(r t) with A > 0.
Specifically we have M ( 0 ) = 1000 so that
1000 = A e^(r * 0), which tells us that 1000 = A. So our function is
M(t) = 1000 e^(r t).
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11:40:09
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**** Describe your sketches of the solution for interest rates of 5% and 10%.
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11:40:34
They're just exponential graphs
The graph of 1000 e^(.05 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.05), or approximately (1, 1051.27).
The graph of 1000 e^(.10 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.10), or approximately (1, 1105.17).
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11:40:34
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**** Does the doubled interest rate imply twice the increase in principle?
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11:40:42
No
We see that at t = 1 the doubled interest rate r = .10 results in an increase of $105.17 in principle, which is more than twice the $51.27 increase we get for r = .05.
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Question:
Query problem 11.5.25 was 11.5.22 (3d edition 11.5.20) At 1 pm power goes out with house at 68 F. At 10 pm outside temperature is 10 F and inside it's 57 F.
Give the differential equation you would solve to obtain temperature as a function of time.
Solve the equation to find the temperature at 7 am.
**** What is your prediction of the temperature at 7 am and how did you get it? Would you revise your estimate up or down in considering the
assumption you made in writing the differential equation?
Your solution:
Confidence Assessment:
Given Solution:
Assuming that dT / dt = k * (T – 10) we find that T(t) = 10 + A e^(k t).
Counting clock time t from 1 pm we have
T(0) = 68 and
T(9) = 57
giving us equations
68 = 10 + A e^(k * 0) and
57 = 10 + A e^(k * 9).
The first equation tells us that A = 58. The second equation becomes
57 = 10 + 58 e^(9 k) so that
e^(9 k) = 47 / 58 and
9 k = ln(47 / 58) so that
k = 1/9 * ln(47/58) = -.0234, approx..
Our equation is therefore
T(t) = 10 + 58 * e^(-.0234 t).
At 7 am the clock time will be t = 18 so our temperature will be
T(18) = 10 + 58 * e^(-.0234 * 18) = 48, approx..
At 7 a.m. the temperature will be about 48 deg.
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11:41:36
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**** What is your differential equation?
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11:41:47
dT/dt = -k(T - 10)
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11:41:48
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**** How did you solve your differential equation?
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11:42:03
I just worked it out by a general solution given in the book
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Question:
**** 11.6.15 was 11.6.11 (3d edition 11.6.6). 20 cal/day maintains weight; rate of wt change is prop to difference with prop const 1/3500
Your solution:
Confidence Assessment:
Given Solution:
**** What is the differential equation you would solve to get W(t) for intake I cal/day?
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11:42:27
dW/dt = w/3500
** The rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight.
The number of daily calories required to maintain weight is weight * 20, or using the notation W(t) for weight, the number of calories required to maintain weight is W(t) * 20.
If the daily intake is I then the difference between the number of calories consumed and the number required to maintain weight is I - W(t) * 20.
Thus to say that rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight is to say that
dW/dt = k ( I - 20 * W).
We are told that k = 1/3500 so that
dW/dt = 1/3500 ( I - 20 W).
The equation is solved by separation of variables:
dW / (I - 20 W) = 1/3500 * dt. Integrating both sides we get
-1/20 ln | I - 20 W | = t / 3500 + c. We solve for W. First multiplying both sides by -20 we get
ln | I - 20 W | = -20 t /3500 + c (-20 * c is still just an arbitrary
constant so we still call the result c).
| I - 20 W | = e^(-1/175 * t + c) or
| I - 20 W | = A e^(-1/175 * t), with A > 0.
If I > 20 W then we have
I - 20 W = A e^(-1/175 * t), with A > 0 so that
W = I / 20 - A e^(-1/175 * t), with A > 0.
Thus if the person consumes more than 20 W calories per day weight will approach the limit I / 20 from below.
If I < 20 W then we have
-(I - 20 W) = -A e^(-1/175 * t), with A > 0 so that
W = I / 20 + A e^(-1/175 * t), with A > 0.
Thus if the person consumes fewer than 20 W calories per day weight will approach the limit I / 20 from above.
If W = 160 when I = 3000 then I < 20 W and we have
W = I / 20 + A e^(-1/175 * t), with A > 0. At t = 0 we have
160 = 3000 / 20 + A e^(-1/175 * 0), or so that
160 = 150 + A so we have
A = 10.
Now the weight function is
W(t) = 150 + 10 e^(-1/175 * t).
This graph starts at W = 150 when t = 0. The 'half-life' for e^(-1/175 t) occurs when -1/175 t = ln(1/2), at t = -175 * ln(1/2) = 121.3 approx.. At this point 10 e^(-1/175 * t) = 10 e^(-1/175 * 121.3) = 5 and W(t) = 150 + 5 = 155.
The graph will therefore exponentially approach W = 160, passing thru points (0,150) and (121.3,155). **
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Question:
**** (no longer present as of 4th edition but should be worked and self-critiqued) Query problem 11.6.16 (was 10.6.10)
A chemical C is formed when a single molecule of substance A combines with a single molecule of substance B to form a single molecule of substance C. The rate at which molecules of C are formed is proportional to product of the amount of chemical A present, and the amount of chemical B present. The initial quantities are respectively a and b, and x is the amount of C present at time t.
What is the differential equation for x = quantity of C at time t?
If a and b are equal, then in terms of the unspecified proportionality constant k, what is the solution of this equation for x(0) = 0?
Your solution:
Confidence Assessment:
Given Solution:
The problem states that C is formed by the combination of a molecule of A and a molecule of B, and that the rate of combination is proportional to the product of the numbers of molecules of A and B present.
If x molecules of C have been formed, then x molecules of A and of B will have been used up.
If the initial numbers of A and B are respectively a and b, then if x molecules of C have been formed there will be a – x molecules of A and b – x molecules of B.
This product of the numbers of molecules of A and of B present will be (a – x) * (b – x), so the equation will be
dx/dt = k (a - x)*(b - x).
If a and b are the same then a = b and we can write
dx / dt = k ( a - x) ( b - x) = k ( a - x) ( a - x).
The equation is therefore
dx / dt = k ( a-x)^2. Separating variables we have
dx / (a - x)^2 = k dt. Integrating we have
1/(a-x) = k t + c so that
a - x = 1 / (k t + c) and
x = a - 1 / (kt + c).
If x(0) = 0 then we have
0 = a - 1 / (k * 0 + c) so that
c = 1 / a.
Thus
x = a - 1 / (k t + 1/a) = a - a / (a k t + 1) = a ( 1 - 1/(akt + 1) ). **
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Question:
Query problem 11.6.29 was 11.6.25 (3d edition 11.6.24) F = m g R^2 / (R + h)^2.
Find the differential equation for dv/dt and show that the Chain Rule dv/dt = dv/dh * dh/dt gives you v dv/dh = -gR^2/(R+h)^2.
Solve the differential equation, and use your solution to find escape velocity.
Give your solution.
Your solution:
Confidence Assessment:
Given Solution:
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RESPONSE -->
Good student
solution:
F = m*a and a = dv/dt, so
F = m*(dv/dt)
F = mgR^2/(R+h)^2
Substituting for F:
m*(dv/dt) = -mgR^2/(R+h)^2 ( - because gravity is a force of attraction, so that the acceleration will be in the direction opposite the direction of increasing h)
Dividing both sides by m:
dv/dt = -gR^2/(R+h)^2
Using dv/dt = dv/dh * dh/dt, where dh/dt is just the velocity v:
dv/dt = dv/dh * v
Substituting for dv/dt in differential equation from above:
v*dv/dh = -gR^2/(R+h)^2 so
v dv = -gR^2/(R+h)^2 dh.
Integral of v dv = Integral of -gR^2/(R+h)^2 dh
Integral of v dv = -gR^2 * Integral of dh/(R+h)^2
(v^2)/2 = -gR^2*[-1/(R+h)] + C
(v^2)/2 = gR^2/(R+h) + C.
Since v = vzero at H = 0
(vzero^2)/2 = gR^2/(R+0) + C
(vzero^2)/2 = gR + C
C = (vzero^2)/2 – gR.
Thus,
(v^2)/2 = gR^2/(R+h) + (vzero^2)/2 - gR
v^2 = 2*gR^2/(R+h) + vzero^2 - 2*gR
As h -> infinity, 2*gR^2/(R+h) -> 0
So
v^2 = vzero^2 - 2*gR.
v must be >= 0 for escape, therefore vzero^2 must be >= 2gR (since a negative vzero is not possible).
Minimum escape velocity occurs when vzero^2 = 2gR,
Thus minimum escape velocity = sqrt (2gR).
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Question:
Query problem 11.6.20 THIS IS THE FORMER PROBLEM, VERY UNFORTUNATELY OMITTED IN THE NEW EDITION. rate of expansion of universe: (R')^2 = 2 G M0 / R + C; case C = 0
Your solution:
Confidence Assessment:
Given Solution:
what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0?
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RESPONSE -->
I am assuming R' = dR/dx so I have a variable of integration.
dR/dx = `sqrt( 2 G M0 / R )
dR/dx = 'sqrt( 2 G M0) / 'sqrt R
dR / 'sqrt R = 'sqrt( 2 G M0) dx
2 'sqrt R = 'sqrt( 2 G M0) x
4 R = ( 2 G M0) x^x
R = ( 2 G M0 x^2 ) / 4
This then satisfies for x=0, R=0.
R ' = `sqrt( 2 G M0 / R ) gives you
dR/dt = 'sqrt( 2 G M0) / 'sqrt R so that
dR * sqrt(R) = sqrt(2 G M0) dt and
2/3 R^(3/2) = sqrt(2 G M0) t + c.
Since R(0) = 0, c = 0 and we have
R = ( 3/2 sqrt( 2 G M0) t)^(2/3).
This can be simplified, but the key is that R is proportional to t^(2/3), which increases without bound as t -> infinity.
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Question:
Problem (omitted from 4th and 5th editions but should be worked and self-critiqued) 11.6.13 from 3d edition
Light intensity is defined to be light energy per unit of area. You don't need this definition to solve the following:
If light intensity
is absorbed by water in proportion to the amount of intensity
present, with the rate measured in
Your solution:
Confidence Assessment:
Given Solution:
The rate of absorption, in units of intensity per unit of distance, is proportional to the intensity of the light.
If I is the intensity then the equation is
dI/dx = k I.
This is the same form as the equation governing exponential population growth and the solution is easily found by separating variables. We get
I = C e^(-k x).
If the initial intensity is I0 then the equation becomes
I = I0 e^(-k x).
If 50% is absorbed in the first 10 ft then the intensity at that position will be .50 I0 and we have
.5 I0 = I0 e^(-k x) so that
e^(-k * 10 ft) = .5 and
k * 10 ft = ln(2) so that
k = ln(2) / (10 ft).
Then at x = 20 ft we have
I = I0 * e^(- ln(2) / 10 ft * 20 ft) = I0 * e^-(2 ln(2) ) = .25 I0, i.e., 25% will be left
At x = 25 ft we have
I = I0 * e^(- ln(2) / 10 ft * 25 ft) = I0 * e^-(2.5 ln(2) ) = .18 I0 (approx), i.e., about 18% will be left.
Note that the expressions e^(-2 ln(2)) and e^(-2.5 ln(2)) reduce to .5^2 and .5^(2.5), since e^(-ln(2)) = 1/2.
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16:50:31
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
Apparently the last problem given is missing some information.
Very interesting stuff. It's amazing what can be done with differential equations (and I'm sure this stuff is just the tip of the iceberg).
I'm afraid that the tip of that iceberg is absorbing way too much energy and is melting way too fast.
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