Find a function called temp of seawater with latitude (as a number in the interval [0, 90]) and depth?

Question

Just as the atmosphere may be divided into layers characterized by how the temperature changes as altitude increases, the oceans may be divided into zones characterized by how the temperature changes as depth increases. We shall divide the oceans into three zones: the surface zone comprises the water at depths between 0 m and 200 m; the thermocline comprises the water at depths between 200 m and 1000 m; and the deep zone comprises water at depths exceeding 1000 m

We shall assume that at all latitudes and longitudes, temperature is a continuous function of depth. Informally, this means that you can sketch the graph of temperature versus depth (at a particular latitude and longitude) without lifting your pencil from the page

Further, we shall assume that the temperature in the surface zone is 2 ◦C at the poles, and 24 ◦C at the equator

Answer 1

starting temp:

2 C at the poles

24 C at the equator

0-200 m is level 1

200 - 1000 is level 2

below 1000 is level 3

[0,90] is the latitude where 45 is the equator

starting temp range = 22 while latitute ranges 45.

slope is 22/45

(45,24) is the max point

point slope form is:

(y - 24) = (22/45)(x-45)

y = (22/45)(x-45) + 24

(22/45)(x)- (22/45)(45) +24

(22/45)x + 2

is the slope intercept form...

next [0,90] needs to be mapped to [0,45]

with (0,0) , (45,45) and (90,0)

it is {

x if 0<=x<=45

90-x if 45<=x<=90

}

So the function is:

temp(x, L) =

{

(22/45)x +2 - (L-1)*m if 0<=x<=45

(22/45)(90-x) + 2 - (L-1)*m if 45 <=x <= 90

}

where x=[0,90] is the latitude and L is the level={1,2,3}

Note that m is the parameter that specifies how much

the temperature drops for each level.