Precalc question

Question

As light from the surface penetrates water, its intensity is diminished. In the clear waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of depth. Thus, the intensity will have the form of a general exponential function.

(a) If the intensity of light at the water’s surface is I◦, find a formula for I(d), the intensity of light at a depth of d meters. Your formula should depend on I◦ and d.

(b) At what depth will the light intensity be decreased to 1% of its surface intensity?

Answer 1

(a) First, remember that the general form of an exponential decay function is

f(t) = Ce^-kt

but in our case we will use d since the problem specifies distance. At the surface, d=0 and l(0) = I◦,

so we have

l(0) = Ce^-k0= C = I◦

Now our formula is

l(d) = I◦e^-kd

Because, intensity decreases by 0.15 every 3 meters, we have that at d=3, l(3) = 0.85I◦, so

l(3) = I◦e^-3k=0.85I◦

Now solve for k:

I◦e^-3k=0.85I◦

e^-3k=0.85

ln(e^-3k)=ln(0.85)

-3k = ln(0.85)

k = -ln(0.85)/3

So the final formula is

l(d) = I◦e^{(ln(0.85)d)/3}

(b) We need to solve the above equation for d, when it is equal to 0.01I◦

I◦e^{(ln(0.85)d)/3}=0.01I◦

e^{(ln(0.85)d)/3}=0.01

ln(e^{(ln(0.85)d)/3})=ln(0.01)

ln(0.85)d)/3=ln(0.01)

ln(0.85)d)=3ln(0.01)

d=3ln(0.01)/ln(0.85)

So at depth d= 3ln(0.01)/ln(0.85) the light intensity is equal to 0.01I◦.

Answer 2

.85 I₀ = I₀ e^-3k

ln .85 = -3k

k= -(1/3) ln .85

I(d) = I₀ e^{-(d/3) ln .85}

From this formula you should be able to answer part b.