Mathematics: Researchers on a boat are investigating plankton cells in a sea. at a depth of h meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by p(h) = 0.2h^2e^(-0.0025h^2) for 0 ≤ h ≤ 30 and is modeled by f(h) for h ≥ 30. the continuous function f is not explicitly given. (a) find p (25). using correct units, interpret the meaning of p (25) in the context of the problem. (b) consider a vertical column of water in this sea with horizontal cross sections of constant area

I hope this is right but the answer i got is x squared = 121...

Researchers on a boat are investigating plankton

Ответ:

brazilmade1

13.03.2022

(a)

p'(25) = -1.179 millions of cells per cubic meter per meter.

The density of plankton cells in millions of cells per cubic meter is
decreasing at a rate of 1.179 millions of cells per cubic meter per meter.

____________

(b)

The volume is 3·dh, where dh is an infinitesimally small height.
Therefore, the amount of plankton is density * volume.

$\int_0^{30} p(h) \cdot 3 \, dh = 1675 \text{ million plankton cells}$

______

(c)

The total plankton cells are given by

$\begin{aligned}
\displaystyle \underbrace{\int_0^{30} 3p(h)\, dh}_{1675.415} + \int_{30}^{K}3f(h)\, dh &= 1675.415 +\int_{30}^{K}3f(h)\, dh \\
&= 1675.415 + 3 \int_{30}^{K} f(h)\, dh
\end{aligned}$

As 0 ≤ f(h) ≤ u(h) for all h ≥ 0 (f and u are nonnegative), then:

$\displaystyle \int_{30}^{K} f(h) \, dh \le \int_{30}^{K} u(h)\, dh \le \int_{30}^{\infty} u(h)\, dh = 105$

Since ultimately $\int_{30}^{K} f(h)\, dh \le 105$ , that integral can have at most 105. Therefore, the maximum total amount is

$\displaystyle 1675.415 + 3 \int_{30}^{K} f(h)\, dh \\ \\
=1675.415 + 3 (105) = 1990 \le 2000\\ \\$

____________

(d)

$\displaystyle\int_0^1 \sqrt{ \big(x'(t)\big)^2 + \big(y'(t)\big)^2}\, dt = 757.456\text{ meters}$

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