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.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Triangular Prism

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