Rate of Change of Water Level in Triangular Prism

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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 \sqrt{3} m deep.

ok so the volume of a prism is
V = \frac{1}{2} lwh ... (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

h = \frac{\sqrt{3}}{2} w ... (2)

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

now subbing 1 into 2

V = \frac{1}{\sqrt{3}} lh^2

\frac{dV}{dt} = \frac{l}{\sqrt{3}} 2h \frac{dh}{dt}

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

\frac{dh_{water}}{dt} = \frac{dh_{empty part}}{dt}

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 \sqrt{3} m deep.

ok so the volume of a prism is
V = \frac{1}{2} lwh ... (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

h = \frac{\sqrt{3}}{2} w ... (2)

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

now subbing 1 into 2

V = \frac{1}{\sqrt{3}} lh^2

\frac{dV}{dt} = \frac{l}{\sqrt{3}} 2h \frac{dh}{dt}

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

\frac{dh_{water}}{dt} = \frac{dh_{empty part}}{dt}

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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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