How to find the antiderivative of cot(x)

[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!