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# We Helped With This Differential Equations Assignment: Have A Similar One?

Category | Math |
---|---|

Subject | Differential Equations |

Difficulty | Undergraduate |

Status | Solved |

More Info | Differential Equations Help |

## Short Assignment Requirements

## Assignment Description

2.

The Oceanic Phosphate (P) Cycle

a. Construct a simple box model (single ODE) of the P cycle. Follow the proper steps of model building. Use the following information:

• Average concentration of P in ocean [PO_{4}] = 2.1
* 10^{-6} mol kg–1 seawater.

• Mass of the ocean = 14 * 10^{20} kg.

• Average residence time of P in ocean (with respect to river input or sediment output) = 40,000 years.

• Assume that the sediment output is proportional to the average concentration.

b. Human activity has approximately doubled the input of phosphate to the ocean. Plot the response of the ocean’s P content to a doubling of the rate of riverine input. Be sure to carry the calculation out far enough to show the new steady state.

c. Now create a more sophisticated two-box model of the ocean’s P cycle. Divide the ocean into a surface box and a deep box. The surface box will have 2.5% of the mass of the deep box and a steady state concentration 10% of the deep box. The total phosphate content of the ocean (i.e., the average concentration of phosphate) will be identical to that of part 1a, as will the total mass of the ocean. River input will be to the surface box. Use the following additional information to complete the model:

• The removal of phosphate from the surface box with biological productivity is proportional

to the concentration of phosphate in that box.

• Ninety-nine percent of this flux is released to the deep ocean during decomposition, and 1% is removed to the sediments (balancing the river input at steady state).

• Water is mixed (upwelled and down welled) between the
surface and deep boxes at a rate of 14 * 10^{17} kg y^{-1},
creating a net transfer of phosphate from deep to surface that is proportional to
this rate, and the concentration difference between deep and surface—that is,
the rate (mol y^{-1}) of phosphate added to the surface box—is (14 * 10^{17} kg y^{-1}) * ([PO_{4}]_{d} - [PO_{4}]_{s}).