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 \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!
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!