Continuously Compounded Interest and Diving.

[niyati]

The intensity L(x) of ligh X feet beneath the surface of the ocean satisfies the differential equation:

dL/dx = -kL

As a diver, you know from experience that diving to 18 ft in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light? We are learning compound interest and whatnot. So, the equation we manipulate is:

A = Ao(e^(rt))

A is probably a tenth of (-kL), which is the original surface value of the ocea. I somehow should use the ratio of 18ft cutting the intensity in half, but I'm not sure how.

 

[HallsofIvy]

I was wondering what in the world "compound interest" had to do with "diving". I thought perhaps you had to take out a loan to pay for your diving equipment!

Yes, the "differential equation" dL/dx= -kL has solution L(x)= Ce^-kx as you can see by differentiating that. That, of course, contains two unknow constants, C and k. If you've done "compound interest" problems that are similar, you probably know you can find C by looking at the "initial valule". Here, L(0)= Ce⁰= C and, since x is "depth", is the amount of light at the surface. You also know " diving to 18 ft in the Caribbean Sea cuts the intensity in half. " Okay, measuring x in feet, you now know that L(18)= Ce^-18k= C/2. The C's cancel and you have e^-18k= 1/2. You can solve that for k by taking logarithms: -18k= ln(1/2)= -ln(2) so k= ln(2)/18.
Now L(x)= Ce^-ln(2)x/18. You are told "cannot work without artificial light when the intensity falls below one-tenth of the surface value. " That is, when L(x)= C/10.
Solve L(x)= Ce^-ln(2)x/18= C/10 for x.

By the way, while e is convenient for derivatives, all exponentials are equivalent. From e^-18k= 1/2, you can calculate that (e^-k)¹⁸= 1/2 or e^-k= (1/2)^1/18. You could rewrite L(x)= Ce^-kt as L(x)= C(1/2)^t/18 instead.

 

[niyati]

Thank you!

D