Question

Can anyone help me solve on my Calculus problems?

Answer 1

Post the problems =)

Answer 2

In a certain medical treatment, a tracer dye is injected into a human organ to measure its function rate and the rate of change of the amount of dye is proportional to the amount present at any time. If a physician injects 0.5 g of dye and 30 minutes later 0.1 g remains, how much dye will be present in 1.5 hours?

Answer 3

The other one:
Assume the room temperature is 70*F. If it takes 10 minutes for a can of soda to warm up from 30*F to 35*F , find how long it takes to warm up a can of soda from 30*F to 40*F

Answer 4

Okay so every 30 minutes 1/5 of the previous amount remains. So, you would do 0.5(1/5)^(t/30) where t is in minutes so the answer should be 0.5(1/5)^(90/30)= 0.004

Answer 5

I think that's right.

Answer 6

Thank you
But I still have few more problems><
Could you please solve them also??

Answer 7

Yes, post away just might be a few seconds before I answer.

Answer 8

Suppose that a population grows according to a logistic model.
1. Write the differential equation for this with k=0.01 and carrying capacity of 60 thousand. 
2. Solve the differential equation with the initial condition P(0)=1000 where time is measured in hours.
3. Find the population after 10 hours.
4. After how many hours does the population reach 2 thousand?
5. As the time t increases without bound, what happens to the population?

Answer 9

Okay this is probably over my head but I am going to make an attempt anyways I just need to do a little reading hopefully someone else comes along.

Answer 10

Thank you
Did you see the second problem that I've posted under the fist problem?

Answer 11

No problem and oh no I did not one second I will try that one first.

Answer 12

I think for the soda problem you would start with this 40 = 30*(1+(1/6))^(t/10) because 35/30 = 1+1/6 so ever 10 minutes you know? This should work if the temperature is increasing exponentially I don't know if it is though. This is just my best guess at a solution let me know if it turns out right if you can haha =) . Anyway then you would solve for t by taking the log of both sides to get log(40/30) = log(1+(1/6))^(t/10) then use log property to get log(40/30) = (t/10)log(1+(1/6)) then do 10*((log(40/30))/(log(1+(1/6))))= t to get t = 18.66 or 19. I think that's what you do.

Answer 13

That should be the right value for t minutes if my method is correct.

Answer 14

Okay so the differential equation you are supposed to be working with for the logistic growth model question is dP/dt = kP(1-(P/M)) right?

Answer 15

I think if I know what I am starting with I can work my way through the problem hopefully.

Answer 16

Oh...
I'm not sure

Answer 17

I'm not sure either I have never done a problem like this and I am only trying this because I watched a video on differential equations earlier today. So the chances that I solve this correctly are slim, but I like a challenge and it would be all the more rewarding if I actually got it right =).

Answer 18

Watch a video? You didnt learned this? You are good

Answer 19

Lol I just like a challenge and if I make a mistake then no I am not good haha, but thank you anyways.

Answer 20

I'm going to close "this open question " because I want to make another post on Openstudy. That doesn't matter that we discuss here.

Answer 21

Alright sounds good.

Answer 22

I will post on your next post or I will message you or something when/if I get the answer.

Answer 23

You can still post here. The "close" doesn"t mean close.

Answer 24

ohic haha thanks =)

Answer 25

I think I got the third problem

Answer 26

Did I have the right initial equation?

Answer 27

I mean was my answer to part 1 correct? I want to keep trying to solve it just so I can learn haha.

Answer 28

I think the part one is
P(t)=60/(1+Ae^-0.01t)
Part 2:
A=59
3:
1.1 thousand
4:
71.02 hours
5:
60 thousand

Answer 29

I am confused the part 1 you wrote is not a differential equation is it? P(t) = 60/(1+Ae^(-0.01t)) does not look like a differential equation to me but I am very new to this.

Answer 30

Can you do this?

Consider the following predator-prey system where x and y are in millions of creatures and t represents time in years:
dy/dt=2x-xy
dy/dt=-4y+xy
1. Show that (4,2) is the nonzero equilibrium solution .
2. Find an expression for dy/dx

Answer 31

I don't know I will try oh btw have all these problem been DE problems cuz the first few I did not solve using DE's I just made some assumptions and tried to solve them sort of like half life problems

Answer 32

Yes they are all DE

Answer 33

oh dude you better not use my answers for those first 2 then because I didn't know they were DE.

Answer 34

I think that will be just fine

Answer 35

Are you sure dude? If they required using DE's I didn't use them at all I just tried to look at them like half life problems. Why are you doing these problems anyway?

Answer 36

For bonus point Haha

Answer 37

oh good =) when is this stuff due?

Answer 38

Tomorrow 0.0

Answer 39

Damn I really want to get the rights answers for these but I just don't think I can learn this much that fast.

Answer 40

Okay I found a problem that is similar to your problem and I am basically trying to model my answer after that answer. I think it should work the same way. When you see the rate of change of "x" is proportional to "x" I think the general idea is you will have something like this dx/dt = Kx where K is a proportionality constant (I am stealing this from another answer but I think it works in general). dg/dt= Kg, dg/g = Kdt, g^-1*dg = Kdt, integral g^-1dg = integral Kdt... ln(g) + C = Kt + C I am still working on the rest of the problem. This is for you first problem about the dye.

Answer 41

C is a constant and the C's on either side should be labelled C subscript 1 and C subscript 2 because they aren't necessarily the same constant. Anyway you can group them together and make them into another constant and just call that C and then we can solve for C. lng = Kt + C

Answer 42

Okay we know that g=0.5 when t=0 because that is how much dye is initially injected. So we have ln(0.5) = K*0 + C so C = ln(0.5). Now we know need to solve for K the proportionality constant. We can do this because we know that when t = 30min g=0.1 so we plug in to get ln(0.1) = K*30 + ln(0.5), (ln(0.1)-ln(0.5))/30 = K, K = -0.0536, Now we plug in all the values we know to solve for g at t=1.5 or 90minutes since we used minutes earlier. ln(g) = (-0.0536)*90 + ln(0.5), ln(g) = 0.004g which is hilarious because it is the same answer I got using my original method lol guess what I did for that first problem was fine.

Answer 43

sorry I meant g = 0.004g not ln(g) = 0.004g but yeah awesome =)