# Triangular Prism

1. Nov 26, 2006

### stunner5000pt

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!!

2. Nov 27, 2006

### HallsofIvy

Staff Emeritus
h is the height of the triangle so it is the depth of the water- you triangle has its vertex downward, remember?

3. Nov 27, 2006

### stunner5000pt

ahhh true

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

but that wouldnt make a difference??

4. Jul 23, 2011

### math4math

Yes, the depth of water is the height of the prism and which is root 3 as given.

5. Jul 23, 2011

### 1MileCrash

I doubt stunner5000pt is still working on this problem after 5 years.