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Calculus Help?

Y

yosuke188

Platinum Member
I have an issue with the following Calculus problem; I have no idea how to do it. Here it is....


At time t=0, a bacterial culture weighs 1 gram. Two hours later, the culture weighs 2 grams. The maximum weight of the culture is 10 grams.

1. Write a non-differential logistics equation that models the weight of the bacterial culture. Be sure to define each constant and the variables in terms of this problem.
2. Find the culture's weight after 5 hours.
3. When will the culture's weight reach 8 grams?
4. Write a logistics differential equation that models the growth rate of the culture's weight. Then repeat parts 2. and 3. using Euler's Method with a step size of h=1. Compare the approximation with the exact answers.
5. At what time is the culture's weight increasing most rapidly?

Thanks in advance for any help anyone can give me!
 
Also, if anyone can help me integrate the following first order differential equation, I'd appreciate it!...

dy=(y(tan x) + 2e^x)dx

I got it this far....

y = (C/abs(cos x))*(integral of e^x*abs(cos x)*dx)
 
I'll help, I usually ask for help here. Let me get some paper and read the thing.

This involves carrying capacity. Let me get my notes.
 
1. y = (1)e^(kt) where k = (ln2 over 2) and t = time in hours (ex. t = 2 --> 2 hours);

2. 4 root 2 grams, or about 5.657 grams.

3. at t = 4 root 2 hours, or about 5.657 hours.

4. dG/dt = kG(10-G) 10 is carrying capacity, G is weight in grams.

 
Shit, I just read my notes wrong. I was looking at exponential growth and decay, not the logistics stuff.

let me redo it. all 🙁
 
5. I recall my teacher saying something about the half-way point being the fastest growing point in logistical equations. So where G = 5, that's the time where it's the most rapidly increasing.

 
I'm working on doing the problems myself with your answers as a guide, and I'm getting there. But please stick around, cuz I haven't gotten to #4 yet.

Can you take a look at that first order linear differential equation? Here's the FOLDE

dy=(y(tan x) + 2e^x)dx

I got it this far....

y = (C/abs(cos x))*(integral of e^x*abs(cos x)*dx)
 
Originally posted by: yosuke188
Did you use an integrating factor?

http://en.wikipedia.org/wiki/Integrating_factor
Click to expand...

The integrating factor, M[x] = Exp[Integral(-Tan[x],dx)]. That would simplify to Cos(x). Then solve for y(x) as it is defined in the wiki page provided.

So y(x)=(Integral(2*Exp(x)*Cos(x)dx)+C)/Cos(x). When all said and done, you should get something like.

y(x) = Constant*Sec(x) + Exp(x)*Sec(x)*(Cos(x) + Sin(x)), where Sec(x) = 1/Cos(x).

Hope that helps.
 
Thanks, yes that helped a lot!

Does anybody know what the solution to this equation is?

dt/dv = 1/(kv - 9.8) , where k is a constant,
 
Originally posted by: yosuke188
Thanks, yes that helped a lot!

Does anybody know what the solution to this equation is?

dt/dv = 1/(kv - 9.8) , where k is a constant,
Click to expand...

That's just your standard 1/x integral 🙂
 
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