#$&*
course Mth 279
February 8 around 3:30pm. I'm resubmitting with revisions and added &&&& before and after what I added. Thanks for the help!
QAs Feb 7 RESUBMITTING`q001. The direction field shown in the figure below applies to the equation dP/dt = k P, with k = .06.
For P = 2000, then what should be the slope of the direction field, according to the equation dP/dt = .06 P?
****
0.06*2000=120
#$&*
Do the slopes here appear to match the slope you just calculated?
****
All the slopes are different. The higher P goes, the higher the slope gets.
#$&*
The slope you should have obtained was over 100. The slopes on this direction field all appear to be less than 1. We need to either reconcile this, or discard the direction field of the figure.
Let's first check out a solution to the problem. As you can easily verify, dP/dt = k P gives us general solution P(t) = C e^(k t). To get the solution passing through the P axis at 2000, we apply the initial condition P(0) = 2000. Our particular solution for this initial condition is P(t) = 2000 e^(k t). Our direction field is for k = .06; for P(0) = 2000 our function is therefore P(t) = 2000 e^(.06 t). A graph of this function is consistent with the direction field.
Verify this. Following the trend of the direction field, estimate the likely value of P(3). What is your estimate?
****
2500
#$&*
What do you get if you evaluate P(3)? Is this consistent with your estimate?
****
2394.43, close!
#$&*
According to the direction field, does the solution curve through (0, 2000) exit the box to the right or through the top?
****
right
#$&*
Check this out by evaluating P(12). What is your conclusion?
****
=2000*e^(0.06*12)
=4108.86
#$&*
Reasonable answers to these questions will tend to reinforce our confidence in the direction field.
Consider the direction field line through (8, 4000). Based on the scale of your graph, estimate the rise of that segment. It should be obvious that the run of the segment is 2. What therefore is the slope represented by that segment?
****
I estimate approx. 500 for the rise. Slope is 500/2=250
#$&*
The slope appears to be less than 1. Explain why that segment actually represents a slope in the hundreds.
****
Because the scale of the graph is completely different. Y-axis is measured in 1000 units and x-axis is measured in units of 1.
#$&*
Construct the slope field for the equation dP/dt = .06 P, for the interval 10 000 <= P <= 20 000, 0 <= t <= 12. Scale your slope segments appropriately.
Based on your slope field, at what value of t do you estimate the value of P(t) will double, starting with P = 10 000 at t = 0?
****
I don’t understand this question. Double of 10 000 is 20 000, which t is 12.
@& The solution curve will therefore reach the top of the box at the doubling time.
If you start from 10 000 your curve should reach the top of the box before it reaches the right-hand side, allowing you to determine the doubling time.*@
&&&&
OK
&&&&
`q002. A quantity y is directly proportional to a quantity x if there exists a constant k such that y = k x.
The rate, with respect to time, at which the temperature of an object left to cool in a room at temperature T_room is directly proportional to the difference between the temperature of the object, which we will represent by T(t), and the temperature of the room.
Letting T ' (t) stand for the rate at which the temperature T(t) changes with respect to clock time, write down the proportionality.
****
dT/dt = k(T-T_room)
roc of temp = constant*difference in temperature
first order linear non-homogeneous
#$&*
You should have written down a differential equation. That equation is linear. Find its general solution.
****
T’-kT = kT_room (let tau=T-T_room)
Tau’=T’ (since T_room is a constant)
Tau’=k*tau
Tau’-k*tau=0 HOMOGENEOUS!
@& Good.
The general solution to y ' - k y = 0 is y = A e^(k t).
Similarly the general solution to tau ' - k tau = 0 is tau = A e^(k t)*@
&&&&
OK! Got it!
&&&&
#$&*
If the object has an initial temperature of 50 Celsius, and its temperature initially changes at -5 Celsius per minute when placed in a room at temperature 20 Celsius, then what is the specific function that models its temperature as a function of time?
****
T_init = 50 C
dT =-5 C/min when 20 C
T(50) = 20 + Ce^(-5k)?!
@& tau = A e^(k t) so
T - T_room = A e^(k t).
Your conditions are
T(0) = 50
and
T ' (0) = - 5.
The condition on initial temperature leads to
T_room = 50
The condition T ' (0) = -5 allows you to evaluate A.
What is your expression for T ' (t)?
What therefore is your expression for T ' (0)?
What therefore must be the value of k?*@
&&&&
I’m still lost on how to do this problem. I will ask you in class and explain what I’m having trouble at.
&&&&
#$&*
`q003. Consider the equation dy/dt = -.01 y^2.
Explain why we cannot solve this equation using the techniques we have been applying to first-order linear equations.
****
Because of the y^2. It’s not linear and therefore there will be places you can’t go with the solutions.
Also, expressions do not get “undefined” in first order linear equations because it’s always e^( ).
#$&*
We will see why shortly, but for now accept that this equation can be rearranged into the form
dy / y^2 = -.01* dt
and integrate both sides to get a general solution.
****
-1/y=0.01(t^2/2)+c
y=1/(0.005t^2+c)
#$&*
`q004. Consider the equation dy/dt = -.01 t y^2. Algebraically rearrange this equation so that dy is a factor of one side, with all other y factors on that side, and dt, along with all other t factors, is on the other.
Integration both sides of the equation to get a general solution.
1/y = -.005 t^2 + c
y = 1 / (-.005 t^2 + c)
Find the particular solution for which y = 25 when t = 0.
****
Y=1/(-0.005t^2 + c)
Y(0)=25
Y=1/25?!
@& y(0) = 1/(-0.005 * 0^2 + c).
y(0) = 25.
What therefore is the value of c?*@
&&&&
C=(-8) GOT IT!
@& This comes out
y(0) = 1 / (..005 * 0^2 + c)
so that
y(0) = 1 / c.
Since y(0) = 25 we have
25 = 1 / c
so that
c = 1 / 25 = .04.*@
&&&&
#$&*
Show that the function f(t, y) = -.01 t y^2 is defined and continuous for all values of y and t.
****
Well, you could plug in values for t and y and graph the function and see what the graph looks like and if it’s continuous or not. I guess you could also use limits to solve.
#$&*
Show that there exist values of t for which your solution is neither defined nor continuous.
****
If t=0, then:
=-0.01*0*y^2
=0
@& 0 is a perfectly good value.
However your expression
0.01*0*y^2
is for the function f(t, y), not for the solution function 1 / (-.005 t^2 + c).
It is the latter function which, for some values of c, leads to a discontinuity.*@
#$&*
While a linear equation of the form y ' + p(t) y = g(t) always has a solution which is defined and continuous, on any interval for which p(t) and g(t) are defined and continuous, this example shows that the same is not necessarily so for an equation of the form y ' + f(t, y) = 0. Explain this statement.
****
y’ + (3/t)y = 9t
Since p(t)=(3/t), P(t)=ln t^3
Integrating factor: t^3
t^3y’ + 3t^2y = 9t^4 or (t^3y)’ = 9t^4
solve for y, and plugging in C, you can graph the function and see where it exists or use theorem 2.1 that says that the initial value problem has a unique solution defined at a certain point.
@& Your statement is good, and true, but it doesn't address the question of how the current example shows that the same is not necessarily true when the equation is not linear.*@
&&&&
If the equation is non-linear, then there will be places you “can’t go” with the solution and will lead to discontinuity.
&&&&
#$&*
`q005. The rate, with respect to time, at which the velocity of a spherical object moving through water changes is directly proportional to the square of its velocity.
Write this as a differential equation.
****
dV/dt=velocity^2
#$&*
Solve the equation.
@& Lowercase v is better for velocity.
In any case the equation is
dv / dt = v^2.
This equation is easily solved by separating variables and integrating.
See if you can find the general solution.*@
&&&&
I’ll attempt!
Int (dv/v^2) = Int (dt) OR Int (1/v^2, dv) = Int (dt)
(-dv/v) = t+c
(-dv/v) = e^(t+c) OR (-dv/v) = Ae^(t+c)
I know this isn’t the general solution, but I hope it’s close!
&&&&
****
#$&*
If the object is initially moving at 30 cm/s and changing velocity at the rate of 1000 cm/s^2, then what function describes its velocity?
****
@& If you have the general solution then these conditions allow you to evaluate the arbitrary integration constant and obtain the specific velocity function.*@
#$&*
`q006. Recall implicit differentiation. If you need a review of implicit differentiation see the q_a_ document at http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm . If you need a review of the chain rule, see http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm and http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm .
If y is a function of t, then what is the derivative with respect to t of the function H(t, y) = t cos y + 1/y?
****
If H(t,y)=0
[ty^2*sin(y)+1]/[y^2*cos(y)]
@& The derivative is
(t^2) ' cos(y) + t^2 (cos(y)) ' + (1/y) '
= 2 t cos(y) - t^2 sin(y) y ' - (1 / y^2) * y '.
Be sure you understand how those y ' factors come from the chain rule. If necessary, review the indicated documents.*@
&&&&
H’(t, y)= (t)’cos y + t(cos y)’ + (1/y)’
=1 cos y + t(-sin y*y’) + (y’-y’/y^2)
=cos y - t sin y y’ - (y’/y^2)
Y is a function of t.
&&&&
#$&*
`q007. Verify that the derivative of the function H(t, y) = t^2 sqrt(y) is 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y '.
****
H(t, y)=t^2 sqrt(y)
Use product rule: (1st times derivative of the 2nd, plus, 2nd times derivative of the first)
t^2*(1/2*sqrt(y)) + (sqrt (y))*(2t)
=2 t*sqrt(y) + t^2/(2 sqrt(y))*y '
#$&*
Verify that the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0
is of the form
( H(t, y) ) ' = 0,
where H(t, y) = t^2 sqrt(y) and ' indicates the derivative with respect to t.
****
If you re-wrote this it would be:
(t^2 sqrt(y))’
which is a product and can be found by doing what I did on the previous problem.
#$&*
Verify that the equation
( H(t, y) ) ' = 0
is equivalent to the equation
H(t, y) = c,
where c is any constant number.
****
If H(t,y)=c, then (c)’=0
Any constant number, such as 23, for example, has a derivative of 0.
(23)’=0
#$&*
For the given function H(t, y) = t^2 sqrt(y), solve the equation
H(t, y) = c
for y as a function of t.
****
t^2 sqrt(y) = c
t^2*(1/2*sqrt(y)) + (sqrt (y))*(2t) = c?
@& It's only H(t, y) that's equal to the constant, not ( H ( t, y) ) '.
The equation is
t^2 sqrt(y) = c.
Solve for y.*@
&&&&
Y=(c/t^4) where c >= 0
&&&&
Or would I solve this implicitly with c being the independent variable?
#$&*
Show that this function y(t) satisfies the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0.
****
Where does the y’ come in at?!
@& As you said:
Use product rule: (1st times derivative of the 2nd, plus, 2nd times derivative of the first)
t^2*(1/2*sqrt(y)) + (sqrt (y))*(2t)
=2 t*sqrt(y) + t^2/(2 sqrt(y))*y '
*@
&&&&
Now I see.
&&&&
#$&*"""
@& See my one additional note, and let me know if you have more questions.*@