Volume of a pond

Question

A pond with vertical sides has a depth of 3 ft and a surface area of 10 ft2. If the pond is full of water and evaporation causes the water level to drop at the rate of 0.3 inches/day, write an expression that represents the volume of water in the pond after $d$ days.

I honestly have no idea how to do this.

Answer 1

The height after $d$ days is $$3 \text{ ft} - \frac{0.3}{12} \text{ ft/day} \times d \text{ days}$$ So the volume is $$V(d)= 10 \text{ ft}^2 \times\left(3 \text{ ft} - \frac{0.3}{12} \text{ ft/day} \times d \text{ days}\right)$$

Answer 2

Lets model the task step by step:

A pond with vertical sides has a depth of 3 ft and a surface area of 10 ft2.

So the pond can be modeled as the volume which can be generated by extruding some area $A=10\,\text{ft}^2$ vertically by the distance $D = 3\,\text{ft}$. This initial volume is $V_0 = A \cdot D = 30\,\text{ft}^3$.

If the pond is full of water and evaporation causes the water level to drop at the rate of 0.3 inches/day

The rate means a change over time of the $D$ variable: $$ \frac{d}{dt} D(t) = \dot{D}(t) = -0.3 \frac{\text{inch}}{\text{day}} $$

write an expression that represents the volume of water in the pond after d days.

For this we need to know how the depth/height of the water level develops over time, thus $D(t)$.

We know the initial condition $D_0 = D(0\,\text{day}) = 3\,\text{ft}$ and obtain $D(t)$ by integrating both sides of the above equation: $$ \int\limits_{0\,\text{day}}^T \dot{D}(t) \,dt = \int\limits_{0\,\text{day}}^T -0.3 \frac{\text{inch}}{\text{day}} dt = -0.3 \frac{\text{inch}}{\text{day}} \int\limits_{0\,\text{day}}^T dt \iff \\ D(T) - D_0 = -0.3 \frac{\text{inch}}{\text{day}}(T-0\,\text{day}) \iff \\ D(T) = D_0 - 0.3 \frac{\text{inch}}{\text{day}} T $$ We multiply both sides by $A$ and get the expression for the volume: \begin{align} V(T) &= A\cdot D(T) \\ &= A\cdot D_0 - 0.3 \frac{\text{inch}}{\text{day}} A\cdot T \\ &= V_0 - 0.3 \frac{\text{inch}}{\text{day}} A\cdot T \end{align} we finally rename the time variable from $T$ to $d$ and get \begin{align} V(d) &= V_0 - 0.3 \frac{\text{inch}}{\text{day}} A\cdot d \\ &= 30\,\text{ft}^3 - 0.3 \frac{\text{inch}}{\text{day}} \frac{1\,\text{ft}}{12\,\text{inch}} 10\,\text{ft}^2 \cdot d \\ &= 30\,\text{ft}^3 - 0.25 \frac{\text{ft}^3}{\text{day}} \cdot d \\ \end{align}