Question

This question is about derivateves. Please help

Answer 1

how would it look like in 2d

Answer 2

Well this is 3d not 2d and i think you can't project it onto 2d?

Sorry I can't help you at the moment my brain is messed up now. Maybe others can help you.

Answer 3

The cross-sectional area of the water in the trough is an equilateral triangle. When the triangle has height h, its area would be h^2 / sqrt(3).

The volume of water would be:
V = (h^2 / sqrt(3))(200)

dV/dt = (dV/dh)(dh/dt)
355 cm^3/min = {[400/sqrt(3)]h}(dh/dt)

For h = 12:
355 cm^3/min = {[400/sqrt(3)](12)}(dh/dt)
dh/dt = 0.128 cm/min

Answer 4

???

Answer 5

how did you get the volume of the water

Answer 6

isn't that the volume of a cone?

Answer 7

my bad im thinking of something else

Answer 8

sorry give me a min xD