Question

The amount of ozone, Q, in the atmosphere is decreasing at a rate proportional to the amount of ozone present. If time t is measured in years, the constant of proportionality is -0.0025. Write a differential equation for Q as a function of t.
Give the general solution for the differential equation. Let C represent the arbitrary constant from integration.
Q=C

If this rate continues, approximately what percent of the ozone in the atmosphere now will decay in the next 27 years? Round your answer to one decimal.

Answer 1

Answer:

Differential equation: $\frac{dQ}{dt} = rQ = -0.0025Q$

Solution of diff equation: $Q(t) = Ke^{-0.0025t}$

6.3% of the ozone in the atmosphere now will decay in the next 27 years.

Step-by-step explanation:

The amount of ozone in the atmosphere may be found by the following differential equation:

$\frac{dQ}{dt} = rQ$

In which r is the constant of proportionality and Q is the amount of ozone. A positive value of r means that the amount of ozone in the atmosphere is going to increase, while a negative value means it is going to decrease.

Solving the differential equation:

We integrate both sides of the differential equation and apply the exponential function. So:

$\frac{dQ}{dt} = rQ$

$\frac{dQ}{Q} = r dt$

Integrating both sides

$\ln{Q} = rt + K$

Applying the exponential:

$Q(t) = Ke^{rt}$

In which K is the initial amount of ozone.

So

$Q(t) = Ke^{-0.0025t}$

If this rate continues, approximately what percent of the ozone in the atmosphere now will decay in the next 27 years?

This K-Q(27).

$Q(27) = Ke^{-0.0025*27} = 0.9347K$

K - 0.9347K = 0.0653.

6.3% of the ozone in the atmosphere now will decay in the next 27 years.