Solve the problem.
-The intensity $\mathrm { L } ( \mathrm { x } )$ of light $x \mathrm { ft }$ beneath the surface of a lake satisfies the differential equation $\frac { \mathrm { dL } } { \mathrm { dx } } = - 0.07 \mathrm {~L} .$ At what depth, to the nearest foot, is the intensity one tenth the intensity at the surface?
A) 22 ft
B) 49 ft
C) 16 ft
D) 33 ft

Question

Quizplus · Accepted Answer

We can solve this separable differential equation by dividing both sides by L and integrating:
$$\ln L = -0.07x + C$$
where C is the constant of integration. Exponentiating both sides, we get:
$$L = e^{-0.07x+C} = Ce^{-0.07x}$$
where we have combined the constant of integration C into a single constant.

To find the depth at which the intensity is one tenth the surface intensity, we can use the fact that:
$$\frac{L(d)}{L(0)} = \frac{1}{10}$$
where d is the depth we want to find. Substituting our expression for L, we get:
$$\frac{Ce^{-0.07d}}{C} = \frac{1}{10}$$
Simplifying, we get:
$$e^{-0.07d} = \frac{1}{10}$$
Taking the natural logarithm of both sides, we get:
$$-0.07d = \ln\frac{1}{10} = -\ln 10$$
Solving for d, we get:
$$d = \frac{\ln 10}{0.07} \approx 33$$
to the nearest foot. Therefore, the answer is D, 33 ft.

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