Microsoft ponds

Calculus Level 4

Mark found a pond which is shaped exactly like an elliptic paraboloid and filled to the edge with water. He gets a little interested and measures the depth of the pond as well as its diameter. He finds out that the depth is $h = 3m$ and the diameter is $d = 10m$ . He also finds out that water evaporates at a rate of $30l$ per $m^2$ and per day. After a month (31 days) he comes back to the pond and notices, that the water level significantly dropped.

Assuming the evaporation rate didn't change and it didn't rain during that month, what is the remaining height of the water in meters?

The answer is 2.07.

1 solution

Brian Moehring
Jul 30, 2018

Note that $30 \frac{\text{l}}{\text{m}^2\times \text{ day}} = 0.03 \frac{\text{m}^3}{\text{m}^2 \times \text{ day}} = 0.03 \frac{\text{m}}{\text{day}}$

Therefore the height, in meters, after 31 days is $3 - 0.03\cdot 31 = \boxed{2.07}$

As the depth changes, so does the area of the surface. That makes the supposed answer of 2.07 a gross underestimate of how much the water level drops. You'd either need to calculate it again after each day, so this problem would probably require a computer program. A per second or minute rate would make the answer a lot more accurate too. Why is this listed under calculus again?

Mark Kessler - 2 years, 10 months ago

"As the depth changes, so does the area of the surface." - This is true. However, since the evaporation rate in liters per day is directly proportional to the area on the surface, it turns out that the actual surface area doesn't matter to the rate the depth changes. Therefore,...

"That makes the supposed answer of 2.07 a gross underestimate of how much the water level drops." - This is false, and...

"You'd either need to calculate it again after each day, so this problem would probably require a computer program. A per second or minute rate would make the answer a lot more accurate too." - This doesn't even make any sense. The initial rate given in the problem as volume evaporated per area of the surface per day is necessarily an instantaneous rate. It wouldn't make any sense physically if it were a simple daily rate (which surface area do we choose from the day?), so there's absolutely no accuracy gained by using a smaller time scale.

To answer your final question of why is it listed under calculus... Well, I may have to take some of the blame for that confusion. I did some sleight of hand in my solution when I interpreted my new rate $0.03 \frac{\text{m}}{\text{day}}$ as a rate for the depth.

In order to justify this, we use the fact that the boundary of the lake is continuous and therefore the lake itself can be approximated by cylinders whose base area is the cross-sectional area of horizontal slices of the lake. When you apply the instantaneous evaporation given in the problem to each cylinder, you'll see it becomes exactly a constant rate on the depth of the lake. As the cylinders become thinner, this doesn't change, so we see that the rate the depth changes is still the constant rate $0.03 \frac{\text{m}}{\text{day}}$ . This type of argument where we take approximations with ever increasing accuracy is exactly the concept of a limit.

Brian Moehring - 2 years, 10 months ago

So i actually intended that you'd calculate the volume of the paraboloid. And indeed i wrote a programm to calculate the depth of water using small time segments. Knowing that it could be done in such an easy fashion actually feels kind of embarassing... ._. But at least im not wrong and learned sth :)

Dominik Döner - 2 years, 10 months ago

Microsoft ponds

The answer is 2.07.

1 solution

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