Calculus problem sets

[MikeSwim07]

Hi,

I am new to this forum so I hope I am doing this right.

The rate at which a rumor spreads through a high school of 1000 can be modeled by the differential equation dH/dt=kH(1000-H), where k is a positive constant and H is the number of students that have heard the rumor t hours after 8AM.

1) Explain the meaning of the expression (1000-H) in the differential equation.
2) How many students have heard the rumor when the rumor is spreading the fastest?
3) Suppose that 2 students know the rumor at 8am and that 100 have heard it by 10am. Write the formula for H as a function of t.
4. At what time of day has half of the student body heard the rumor?

A particle moves along a line in such a way that at time t, 1<=t<=8, its position is given by s(t) = integral from (1,t) of (1-x*(cosx)-(lnx)(sinx))dx.

1) Write a formula for the velocity of the particle at time t.
2) At what instant does the particle reach its maximum speed?
3) When is the particle moving to the left?
4) Find the total distance traveled by the particle from t=1 to t=8.

Let functions f and g be defined by f(x)=x and g(x)=x+(k/x), where k is a positive constant and let R be the region between the graphs of f and g on the interval [1,3]

1) Find the average value of g on the interval [1,3] if k=2. I got this one.
2) Find in terms of k, the volume of the solid generated when R is rotated about the x-axis.
3. Set up but do not evaluate an integral expression in terms of k for the volume of the solid generated when R is rotated about the horizontal line y=-2.

Thanks

 

[Deleted member 4993]

Hi,

I am new to this forum so I hope I am doing this right.

The rate at which a rumor spreads through a high school of 1000 can be modeled by the differential equation dH/dt=kH(1000-H), where k is a positive constant and H is the number of students that have heard the rumor t hours after 8AM.

1) Explain the meaning of the expression (1000-H) in the differential equation.
2) How many students have heard the rumor when the rumor is spreading the fastest?
3) Suppose that 2 students know the rumor at 8am and that 100 have heard it by 10am. Write the formula for H as a function of t.
4. At what time of day has half of the student body heard the rumor?

[quote:3j9rsmu5]A particle moves along a line in such a way that at time t, 1<=t<=8, its position is given by s(t) = integral from (1,t) of (1-x*(cosx)-(lnx)(sinx))dx.

1) Write a formula for the velocity of the particle at time t.
2) At what instant does the particle reach its maximum speed?
3) When is the particle moving to the left?
4) Find the total distance traveled by the particle from t=1 to t=8.

Let functions f and g be defined by f(x)=x and g(x)=x+(k/x), where k is a positive constant and let R be the region between the graphs of f and g on the interval [1,3]

1) Find the average value of g on the interval [1,3] if k=2. I got this one.
2) Find in terms of k, the volume of the solid generated when R is rotated about the x-axis.
3. Set up but do not evaluate an integral expression in terms of k for the volume of the solid generated when R is rotated about the horizontal line y=-2.

Thanks[/quote:3j9rsmu5]

In this forum we help you to do your homework - not do the homework.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[MikeSwim07]

Oh I'm sorry. I forgot to tell you what I did already.

First problem.
1) I think it represents the number of people who haven't heard the rumor yet.
2) fastest at 500 students because this is where the inflection point it, greatest slope. Half way between min and max
3) I think I need to use the logistic equation but I am not sure what to do.
4) need answer to 3 I think

Second Problem
1) By the second fundamental theorem of calc, t can simply be plugged into the equation, so the equation is the same for v(t) except with ts instead of xs.
2) Not sure. I tried graphing this on my calculator and I couldn't find the greatest slope.
3. ?
4) Integral from 1 to 8 of the equation gives 25.229

Third Problem:
1) Use the average value formula and you get 3.0986
2) I think it is like pi* integral of (X+(K/x)^2)
3) ?

 

[galactus]

The rate at which a rumor spreads through a high school of 1000 can be modeled by the differential equation dH/dt=kH(1000-H), where k is a positive constant and H is the number of students that have heard the rumor t hours after 8AM.

1) Explain the meaning of the expression (1000-H) in the differential equation.

Yes, you are correct. It is the number who have NOT heard the rumor.

2) How many students have heard the rumor when the rumor is spreading the fastest?

Yes, right again....500

3) Suppose that 2 students know the rumor at 8am and that 100 have heard it by 10am. Write the formula for H as a function of t.

Upon solving the DE, we get\(\displaystyle H=\frac{1000e^{1000kt}}{e^{1000kt}+1000C}\)

Using the initial conditions: \(\displaystyle H(0)=2, \;\ H(2)=100\), we can solve for the values of k and C.

Sub in t=0 and set H=2and solve for C. Then, sub in the newly found C as well as t=2 and H=100 and solve for k.

4. At what time of day has half of the student body heard the rumor?

After k and C are placed into the DE, set it equal to 500 and solve for t. Yes, you're correct. It is around 3, but a little more than that if you want it down to the minute.

 

[MikeSwim07]

Hi,

I am not sure what you are doing for question one part 3. Don't you have to use the logistic equation? like H = L/(1+be^-kt)?

So for at time 0, 2 know it it would be

1000/1+be^-k*0 so b = 499. then you can use the other info to get k.

EDIT: nevermind, I figured this problem out.

Can you please help me with the second problem? the one with the particle?

 

[galactus]

A particle moves along a line in such a way that at time t, 1<=t<=8, its position is given by \(\displaystyle s(t) = \int_{1}^{t}(1-x*(cosx)-(lnx)(sinx))dx.\)

1) Write a formula for the velocity of the particle at time t.

Velocity is the derivative of position, so

Use the second fundamental theorem of calculus. \(\displaystyle v(t)=\frac{d}{dt}\int_{1}^{t}[1-xcos(x)-ln(x)sin(x)]dx=1-tcos(t)-ln(t)sin(t)\)

2) At what instant does the particle reach its maximum speed?

Graph it and you can see the extrema. There are 3 of them along with the endpoints at t=1 and t=8.
Then, sub them into v(t) and see which is the highest value for the max speed. Remember, Speed is the absolute value of velocity.

3) When is the particle moving to the left?

It moves left when the graph heads into negative territory. Or set v(t)=0 and solve for t. It moves right on one side of the root and negative on the other side of the root. On the graph, look at where it crosses the x-axis. If it moves below the x-axis, it is heading left. When the graph is above the x-axis, it is heading right. Look close and you can see it begins to head left at around t=5.2 seconds, then heads right again at t=8.

4) Find the total distance traveled by the particle from t=1 to t=8.

We can integrate from 1 to the value found in part c, then from part c to 8. Add them up.

If we integrated from 1 to 8, we would get the difference of the negative and positive distances. Since there is a root between the

domain of 1 and 8, we have to break it up.

\(\displaystyle \int_{1}^{5.2}[1-tcos(t)-ln(t)sin(t)]dt+\int_{5.2}^{8}[1-tcos(t)-ln(t)sin(t)]dt\).

Use some tech to do it. the ln(t)sin(t) is rough to integrate on a elementary basis. The second integral will have a negative value, so take the absolute value and add it to the other one for the total distance.

Let me know what you come up with.

 

[MikeSwim07]

Thank you so much. I got around 21.5 for the last part.

Now so for the last one, the average value is 3.0986?

And I am confused about parts 2 and 3.

 

[galactus]

For part 2, look at the graph and follow my instructions. Notice the highs and lows on the graph.

There are extrema at t=1.234, t=3.73, t=6.7, and the endpoints at t=1 and t=8.

Enter these values into the function and see which gives the largest value. That is the max speed.

For part 3, the particle moves left when the graph goes into negative range. It crosses the x-axis at t=5.2 and is below the x-axis from 5.2 to 8.

At 8 it crosses back into positive and moves right again.

 

[MikeSwim07]

I'm sorry, I wasn't clear enough. I understand that problem now.

This is the problem I was referring to

Let functions f and g be defined by f(x)=x and g(x)=x+(k/x), where k is a positive constant and let R be the region between the graphs of f and g on the interval [1,3]

1) Find the average value of g on the interval [1,3] if k=2. I got this one.
2) Find in terms of k, the volume of the solid generated when R is rotated about the x-axis.
3. Set up but do not evaluate an integral expression in terms of k for the volume of the solid generated when R is rotated about the horizontal line y=-2.

 

[Deleted member 4993]

Are these take home tests?

I am seeing same set of problems in other web-sites also!!!

 

[MikeSwim07]

No, it's not a test. Just a homework assignment

 

[galactus]

Let functions f and g be defined by \(\displaystyle f(x)=x\) and \(\displaystyle g(x)=x+\frac{k}{x}\), where k is a positive constant and let R be the region between the graphs of f and g on the interval [1,3]

2) Find in terms of k, the volume of the solid generated when R is rotated about the x-axis.

Use washers/disks. Nothing fancy.

3. Set up but do not evaluate an integral expression in terms of k for the volume of the solid generated when R is rotated about the horizontal line y=-2.

You can still use washers. In this case, g(x) has f(x) as its oblique asymptote.

Try setting it up and let me know what you get. Remember it is being revolved around y=-2...a horizontal line.

The set up is not that much different than part 2, but the offset of -2 must be considered in the integral.

 

[MikeSwim07]

well for 2 I got Pi*Integral from 1 to 3 of (x+(k/x)^2)-x^2

And for 3 i am getting:

Pi*integral for 1 to 3 of (-2-(x+(k/x)^2))-(-2-x^2))

 

[MikeSwim07]

Nevermind, I figured it out.

Thanks for your help