How to find the antiderivative of cot(x)

[sonofjohn]

5) As you can see ln(sin(1)) +1/ln.5 does not yield one of the above answers. Can someone tell me where I went wrong?

 

[math8]

I think you made a mistake when finding C.
C should be equal to 1-ln(.5) not 1/ln(.5).
The answer should be the last one (E).

 

[sonofjohn]

Thanks for the help, didn't see that.

 

[sonofjohn]

Does anyone have an idea as to how to figure 6 out? I don't believe I have ever tried one of those problems.

 

[math8]

You should try finding out what is the second derivative of [f^2(x)].
When you're done, you just plug in the number 3 and it is easy to find out what's the answer.

 

[sonofjohn]

isn't the second derivative of f^2x = 2?

 

[math8]

No, consider [f^2(x)]=[f(x)]^2
you know that (u^m)'=m (u^(m-1)) u' right?
So it's the same thing for u=f(x) and m=2.

Can you try it and see what you get when you do it twice?

 

[NoMoreExams]

isn't the second derivative of f^2x = 2?

Certainly not. Let f(x) = x^3, then f^{2}(x) = x^6. Still think your rule works?

 

[sonofjohn]

I can't see how to take the second derivative of f^2(x). Am I supposed to use the previous f(3) = 2 and ect for this? I have never actually learned how.

 

[NoMoreExams]

Use the chain rule, if I told you to differentiate [sin(x)]^2 what would you do?

 

[sonofjohn]

so we start with

f(x)^2
2f(x)
and then what?

 

[NoMoreExams]

No, that's not it. Try again :) You are saying if we define f(x) = sin(x) then f^(2)(x) = sin^2(x) and you are saying derivative of f^(2)(x) is 2*sin(x)? You know that's not true...

 

[sonofjohn]

could I possibly multiply them and get f(2x^2)?

 

[math8]

[f(x)^2]'= 2f(x)*f '(x)
you're not using the chain rule correctly. Don't forget the f '(x) at the end :)

Now can you find out what is [2f(x)*f '(x)] ' using the product rule?

 

[NoMoreExams]

could I possibly multiply them and get f(2x^2)?

Where are you getting 2x^2 from? Think of f^2(x) as f(x)*f(x). What's the derivative of that. That uses the product rule instead of the chain rule, if you have trouble remembering that one.

 

[sonofjohn]

2f'(x) + f"(x)2f(x) would that be it and then just number plug?

 

[NoMoreExams]

No! Why is the first term 2f'(x)?

 

[sonofjohn]

it should be 2f'(x)f'(x) + f"(x)2f(x) = 42. I messed up the chain rule again :(

 

[sonofjohn]

Now 7 is another problem I have yet to face. I see that t is in seconds so I think I should multiply t by 60 to convert it to minutes. Should I then plug 1 in for y and take the derivative?

 

[NoMoreExams]

it should be 2f'(x)f'(x) + f"(x)2f(x) = 42. I messed up the chain rule again :(

Actually that's the product rule.

Now 7 is another problem I have yet to face. I see that t is in seconds so I think I should multiply t by 60 to convert it to minutes. Should I then plug 1 in for y and take the derivative?

If \frac{dy}{dt} = ky Then y = ? Solve this separable DE.

 

[sonofjohn]

Ok so using the seperable differentiation (not what it's called) I come out with dy/y = (dt)k

where do I go from there?

 

[jgens]

I would integrate it.

 

[sonofjohn]

Integrating yields y = dk ?

 

[jgens]

No. The integral of dy/y is not equal to y and the integral of (k)(dt) is not equal to dk. You may want to review your integrals.

 

[sonofjohn]

Yes I do. I need to review all of Calc 1 really. Is the integration lny = t(dk/dt)

 

[jgens]

Much closer but not quite. Your equation should read: ln(y) = kt.

 

[sonofjohn]

yes, because k is a constant! Now if I plug y in I get ln1/t but t is in seconds so it would be ln1/60 right? But ln1 is zero and that ruins it all.

 

[jgens]

Write your equation in terms of y and then solve.

Edit: try this. We know y = e^kt. If we let y designate the initial amount of the isotope then y = e^k. Now based on the information we may express the half life as: y/2 = e^60k or y = 2e^60k. Equate the two and solve.

 

[sonofjohn]

I see where the 60 seconds is coming from in the e^60k but not where the y/2 comes from on the other side of the equation.

 

[jgens]

Half-life. Half of the amount of the initial isotope has decayed at t = 60, hence, y/2.

 

[NoMoreExams]

I see where the 60 seconds is coming from in the e^60k but not where the y/2 comes from on the other side of the equation.

That's probably what HALF life means

 

[sonofjohn]

Ahhh, got it that makes sense :) So now do I plug the dy/dt = ky into the equation?

 

[jgens]

No, you just need to solve for k.

 

[sonofjohn]

If I solve for k I would need to take the natural log of both sides and get k = 30lny

 

[NoMoreExams]

Write your equation in terms of y and then solve.

Edit: try this. We know y = e^kt. If we let y designate the initial amount of the isotope then y = e^k. Now based on the information we may express the half life as: y/2 = e^60k or y = 2e^60k. Equate the two and solve.

Why does y = e^k? If y denotes the initial amount i.e. when t = 0...

 

[jgens]

Bah, good point NoMoreExams. I need to be more careful when giving homework advice.

I'm terribly sorry sonofjohn.

 

[NoMoreExams]

No problem. What the OP should understand is what half life means i.e. we are told at some point time t, half of your stuff is gone. I.e. if y is the total amount, then as you said y/2 is half of the amount we are told that that happens at time t = 60 seconds.

So y/2 = e^(60*k).

We also know that initially, i.e. at t = 0, we had the whole amount i.e.

y = e^(0*k) = 1

Now you know y = 1 and y/2 = e^(60k). I am hopeful you can solve for k

 

[sonofjohn]

haha no worries. So instead what should I do?

 

[NoMoreExams]

haha no worries. So instead what should I do?

Refresh the page and read what I said probably.

 

[sonofjohn]

Great! k = 60ln(1/2) or (a). Thanks!

 

[NoMoreExams]

Great! k = 60ln(1/2) or (a). Thanks!

Are you serious...

 

[sonofjohn]

what did I do it wrong?

 

[NoMoreExams]

...show me how you got to your answer

 

[sonofjohn]

ok,

y/2 = e^k/60
ln1/2 = k/60
60ln(1/2) = k

 

[sonofjohn]

sigh, shouldn't it be ln(1/2)/60 = k thus giving me (b)

 

[NoMoreExams]

(b) looks like the better choice

 

[sonofjohn]

haha yep thanks!