Question

A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. if the trough is being filled with water at the rate of 0.2 m3/min, how fast is the water level rising when the water is 30 cm deep?

Answer 1

Approximately
`0.033\; {\rm m\cdot min^(-1)}` (meters per minute.)

Step-by-step explanation:

Let
`A` and
`B` denote the bottom and top width of the trough, respectively. It is given that
`A = 30\; {\rm cm} = 0.30\; {\rm m}` and
`B = 80\; {\rm cm} = 0.80\; {\rm m}`. Let
`H` denote the height of this trough;
`H = 50\; {\rm cm} = 0.50\; {\rm m}`.

Let
`h` denote the current depth of the water in this trough.

Let
`b` denote the current width of the surface of the water. As water fills the trough, this width increases from
`A = 0.30\; {\rm m}` (width of bottom of trough) to
`B = 0.80\; {\rm m}` (width of top of trough.)

The relationship between
`b` and
`h` is linear:

`\displaystyle b = (h)/(H)\, (B - A) + A`.

Cross-section area of water in this trough:

`\begin{aligned}(\text{area}) &= (1)/(2)\, (A + b) \, h \\ &= (1)/(2)\, \left(A + (h)/(H)\, (B - A) + A\right)\, h \\ &= A\, h + (1)/(2\, H) (B - A)\, h^(2)\end{aligned}`.

Let
`L` denote the length of this trough;
`L = 10\; {\rm m}`.

Let
`v` denote the volume of water in this trough:

`\begin{aligned}v &= (\text{area})\, L \\ &= (1)/(2)\, (A + b) \, h\, L \\ &= (1)/(2)\, \left(A + (h)/(H)\, (B - A) + A\right)\, h\, L \\ &= A\, L\, h + (L)/(2\, H) (B - A)\, h^(2)\end{aligned}`.

Differentiate both sides implicitly with respect to
`v`:

`\displaystyle (d)/(dv)[v] = (d)/(dv)\left[A\, L\, h + (L)/(2\, H)\, (B - A)\, h^(2)\right]`.

`\displaystyle (d)/(dv)[v] = (d)/(dv)[A\, L\, h] + (d)/(dv)\left[(L)/(2\, H)\, (B - A)\, h^(2)\right]`.

`\displaystyle 1 = A\, L\, (dh)/(dv) + (L)/(2\, H)\, (B - A)\, 2\, h\, (dh)/(dv)`.

(Note that
`A`,
`B`,
`L`, and
`H` are constants.)

Rearrange this equation to obtain:

`\displaystyle 1 = A\, L\, (dh)/(dv) + (L)/(H)\, (B - A)\, h\, (dh)/(dv)`.

`\displaystyle 1 = \left(A\, L + (L)/(H)\, (B - A)\, h \right)\, (dh)/(dv)`.

`\displaystyle (dh)/(dv) = (1)/(\displaystyle A\, L + (L)/(H)\, (B - A)\, h)`.

Let
`t` denote time. It is given that the trough is being filled at a rate of
`0.2\; {\rm m^(3)\cdot min^(-1)}`. In other words:

`\displaystyle (dv)/(dt) = 0.2\; {\rm m^(3)\cdot min^(-1)}`.

Apply the chain rule to find the rate at which
`h` is changing with respect to time
`t`:

`\begin{aligned} (dh)/(dt) &= (dh)/(dv)\cdot (dv)/(dt) \\ &= (1)/(\displaystyle A\, L + (L)/(H)\, (B - A)\, h)\cdot (dv)/(dt) \end{aligned}`.

Substitute in
`A = 0.30\; {\rm m}`,
`L = 10\; {\rm m}`,
`H = 0.50\; {\rm m}`,
`B = 0.80\; {\rm m}`,
`h = 0.30\; {\rm m}` (converted from
`30\; {\rm cm}`), and that the rate of change in
`v` is
`0.2\; {\rm m^(3)\cdot min^(-1)}`:

`\begin{aligned} (dh)/(dt) &= (dh)/(dv)\cdot (dv)/(dt) \\ &= (1)/(\displaystyle A\, L + (L)/(H)\, (B - A)\, h)\cdot (dv)/(dt) \\ &= \frac{0.2\; {\rm m^(3)\cdot min^(-1)}}{\displaystyle 0.30\; {\rm m} * 10\; {\rm m} + \frac{10\; {\rm m}}{0.60\; {\rm m}}\, (0.80\; {\rm m} - 0.30\; {\rm m}) \, 0.30\; {\rm m}} \\ &=0.033\; {\rm m\cdot min^(-1)} \end{aligned}`.

In other words, the depth of the water in this trough increases at a rate of approximately
`0.033\; {\rm m \cdot min^(-1)}` (meters per minute.)