Rate of Change of Water Level in Triangular Prism

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Homework Help Overview

The problem involves a triangular prism with a base that is an equilateral triangle, where water is being added at a specified rate. The objective is to determine the rate of change of the water level when the water reaches a certain depth.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the height of the water and the dimensions of the triangular prism, questioning the definitions of height and depth in the context of the problem.

Discussion Status

Some participants have provided clarifications regarding the interpretation of height in relation to the water level, while others express uncertainty about the implications of the prism's orientation. There is an ongoing exploration of the definitions and relationships involved in the problem.

Contextual Notes

Participants note potential confusion stemming from the diagram in the textbook, which may have influenced their understanding of the problem setup.

stunner5000pt
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A trinagular prism is 10 m long and has an equilaterlal trinagle for its base. WAter is added at a rate of 2m^3/min. Determine the rate of change of the water level when the water is [itex]\sqrt{3}[/itex] m deep.

ok so the volume of a prism is
[tex]V = \frac{1}{2} lwh[/tex] ... (1)
l is the length
w is the width
h is the height

now dl/dt = 0 because the length of the column of water is constant

to find a relation between h and w i got this because the triangle is an equilaterla triangle

[tex]h = \frac{\sqrt{3}}{2} w[/tex] ... (2)

and it follos that
[tex]\frac{dh}{dt} = \frac{\sqrt{3}}{2}\frac{dx}{dt}[/tex] ... (3)

now subbing 1 into 2

[tex]V = \frac{1}{\sqrt{3}} lh^2[/tex]

[tex]\frac{dV}{dt} = \frac{l}{\sqrt{3}} 2h \frac{dh}{dt}[/tex]

now here's the problem ... what is h??
h does not represent the depth of the water, does it??
it reprsnts the height of the remainder of the prism that has not been filled iwht water. so really

[tex]\frac{dh_{water}}{dt} = \frac{dh_{empty part}}{dt}[/tex]

is it reasonable to say that??

thank you for all your input!
 
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stunner5000pt said:
A trinagular prism is 10 m long and has an equilaterlal trinagle for its base. WAter is added at a rate of 2m^3/min. Determine the rate of change of the water level when the water is [itex]\sqrt{3}[/itex] m deep.

ok so the volume of a prism is
[tex]V = \frac{1}{2} lwh[/tex] ... (1)
l is the length
w is the width
h is the height

now dl/dt = 0 because the length of the column of water is constant

to find a relation between h and w i got this because the triangle is an equilaterla triangle

[tex]h = \frac{\sqrt{3}}{2} w[/tex] ... (2)

and it follos that
[tex]\frac{dh}{dt} = \frac{\sqrt{3}}{2}\frac{dx}{dt}[/tex] ... (3)

now subbing 1 into 2

[tex]V = \frac{1}{\sqrt{3}} lh^2[/tex]

[tex]\frac{dV}{dt} = \frac{l}{\sqrt{3}} 2h \frac{dh}{dt}[/tex]

now here's the problem ... what is h??
h does not represent the depth of the water, does it??
it reprsnts the height of the remainder of the prism that has not been filled iwht water. so really

[tex]\frac{dh_{water}}{dt} = \frac{dh_{empty part}}{dt}[/tex]

is it reasonable to say that??

thank you for all your input!
h is the height of the triangle so it is the depth of the water- you triangle has its vertex downward, remember?
 
HallsofIvy said:
h is the height of the triangle so it is the depth of the water- you triangle has its vertex downward, remember?

ahhh true

the book was rather deceptive in taht it drew the triangle right side up

but that wouldn't make a difference??
 
Yes, the depth of water is the height of the prism and which is root 3 as given.
 
I doubt stunner5000pt is still working on this problem after 5 years.
 

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