A little help with integration and trigonometric functions.

[Unicyclist]

I was doing my exam today and ran into a couple problems.
First one: how do you differentiate \tan^2?
I converted it into \sec^2 - 1 and used the u/v = (u`v - v`u)/v^2 method, but I would like somebody clever to do it for me, just to be sure, please.

Homework Statement

Another problem.
Rate of change of population P equals \lambda P \cos(\lambda t)
Find the formula for population P in terms of P_0,\lambda and t.

Then, find t. When P = 2P_0

The Attempt at a Solution

I had problems with finding t.

t came out to be arcsin of something. The problem is, they never said anything about degrees or radians and so t could vary quite a bit, depending on that. Did I do it wrong or am I missing something out?

I don't have the exact question. It was in the exam I did two hours ago. All help is appreciated.

Thank you.

 

[mjsd]

differentiate tan^2 x? tan x = sinx/cosx , chain rule or remember the result.. d(tanx)/dx = sec^2 x, chain rule simply gives d(tan^2x)/dx = 2 tan x sec^2 x

dP(t)/dt = L P(t) cos (Lt)
=> dP(t)/P(t) = L cos(Lt) dt
=> integrate both sides
=> Log_e P(t) = sin (Lt) +C
=> P(t) = A e^(sin Lt)
where A is some constant
I suppose P0 means initial value so A = P0
and so when P(t) = 2 P0
=> 2 P0 = P0 e^(sin L t)
=> Log_e 2 = sin Lt
=> t = 1/L arcsin (Log_e 2)

ok... I am expecting you to be able to do it next time

 

[cristo]

1. To differentiate tan^2x, one would use the chain rule. \frac{d}{dx}\tan^2x=2\tan x\frac{d}{dx}(\tan x)=2\tan x \sec^2 x where the last step is done either using the quotient rule for tangent, or from memory.

2. For this question you need to solve the differential equation \frac{dP}{dt}=\lambda P\cos(\lambda t), using the initial conditions given. Is that what you did?

 

[Unicyclist]

1. To differentiate tan^2x, one would use the chain rule. \frac{d}{dx}\tan^2x=2\tan x\frac{d}{dx}(\tan x)=2\tan x \sec^2 x where the last step is done either using the quotient rule for tangent, or from memory.

2. For this question you need to solve the differential equation \frac{dP}{dt}=\lambda P\cos(\lambda t), using the initial conditions given. Is that what you did?

1)Chain rule, of course!
I got \frac{\cos^4(x)}{\sin(2x)} for that one.<br /> <br /> 2) Yes, I did that, it&#039;s easy. But then you have to make t the subject of the formula and substitute some values to find when the population doubles. That&#039;s where I had the problem, as I wasn&#039;t sure what to use: degrees or radians. Try it yourself and see.

 

[Unicyclist]

For the second one you get P = P_0e^\sin(\lambda t) <= I think.

When P = 2P_0, t = \frac{\arcsin(ln 2)}{\lambda}. <= did it in my head, so could be wrong, but the principle is there. Arcsin can give you different values, depending on what system you use: degrees, radians, grads. Or am I getting something wrong here?

 

[cristo]

1)Chain rule, of course!
I got \frac{\cos^4(x)}{\sin(2x)} for that one.

Looks like you've got the reciprocal of the correct answer (or just made a typo here!)

2) Yes, I did that, it's easy. But then you have to make t the subject of the formula and substitute some values to find when the population doubles. That's where I had the problem, as I wasn't sure what to use: degrees or radians. Try it yourself and see.

Ok, I see your point now. It doesn't really matter which you use, but I would always use radians. They'll both give the same answer, it's just the units of time will be different.

 

[Unicyclist]

The units will be different, eh? How do I even figure out the units? They said to give your answer in minutes. And t is in days, I think.

 

[Gib Z]

When dealing with logs, its usually best to use radians as angle measures that require units give workable but difficult results, such as 2 log seconds.

 

[Unicyclist]

I used degrees, doesn't matter. I'll probably lose one mark for that or something.

And I'll know next time. I actually thought about using radians, but couldn't be bothered to change the mode on my calculator.

Thanks to everyone for their help.

 

[Unicyclist]

Could someone integrate his for me, please?

2\int\sec^4x dx

I had it in my exam today and I want to know if I did it correctly. Thank you.

 

[CompuChip]

The result is

\frac{2\,\left( 2 + {\sec (x)}^2 \right) \,\tan (x)}{3}

Did you have that?

 

[Unicyclist]

Nope.

\tan x + \tan^2 x + \sec^2 x

If I remember correctly. The sad thing is, I didn't really have to do that integration. I missed out the minus in ln, which would've given me \sec^2 x, which is considerably easier to integrate.

 

[Unicyclist]

Woops, I meant tan and sec to the power of 3, not too. Is that more correct in any way?

 

[CompuChip]

OK, my Mathematica gives something else (the expression above).
Then perhaps it's time to show your calculation :)

 

[Unicyclist]

Forget about it, I've got to revise for a different exam tomorrow. Thanks for your help anyway.

 

[CompuChip]

OK, good luck with the exam.