A homeowner wants to have a water tank built. This water tank must comply with the following specifications:
- it must be located two metres from his house;
- the maximum depth must be two metres;
- it must measure five metres long;
- it must follow the natural slope of the land.
The curved part is modelled by the curve $\mathscr{C}_f$ of the function $f$ on the interval $[2; 2e]$ defined by: $$f(x) = x \ln\left(\frac{x}{2}\right) - x + 2$$
The curve $\mathscr{C}_f$ is represented in an orthonormal coordinate system with unit $1\mathrm{m}$ and constitutes a profile view of the tank. We consider the points $\mathrm{A}(2; 2)$, $\mathrm{I}(2; 0)$ and $\mathrm{B}(2\mathrm{e}; 2)$.
Part AThe objective of this part is to evaluate the volume of the tank.
- Justify that the points B and I belong to the curve $\mathscr{C}_f$ and that the x-axis is tangent to the curve $\mathscr{C}_f$ at point I.
- We denote by $\mathscr{T}$ the tangent to the curve $\mathscr{C}_f$ at point B, and D the point of intersection of the line $\mathscr{T}$ with the x-axis. a. Determine an equation of the line $\mathscr{T}$ and deduce the coordinates of D. b. We call $S$ the area of the region bounded by the curve $\mathscr{C}_f$, the lines with equations $y = 2$, $x = 2$ and $x = 2\mathrm{e}$. $S$ can be bounded by the area of triangle ABI and that of trapezoid AIDB. What bounds on the volume of the tank can we deduce?
- a. Show that, on the interval $[2; 2\mathrm{e}]$, the function $G$ defined by $$G(x) = \frac{x^2}{2} \ln\left(\frac{x}{2}\right) - \frac{x^2}{4}$$ is an antiderivative of the function $g$ defined by $g(x) = x \ln\left(\frac{x}{2}\right)$. b. Deduce an antiderivative $F$ of the function $f$ on the interval $[2; 2\mathrm{e}]$. c. Determine the exact value of the area $S$ and deduce an approximate value of the volume $V$ of the tank to the nearest $\mathrm{m}^3$.
Part BFor any real number $x$ between 2 and $2\mathrm{e}$, we denote by $v(x)$ the volume of water, expressed in $\mathrm{m}^3$, in the tank when the water level in the tank is equal to $f(x)$. We admit that, for any real number $x$ in the interval $[2; 2\mathrm{e}]$, $$v(x) = 5\left[\frac{x^2}{2}\ln\left(\frac{x}{2}\right) - 2x\ln\left(\frac{x}{2}\right) - \frac{x^2}{4} + 2x - 3\right]$$
- What volume of water, to the nearest $\mathrm{m}^3$, is in the tank when the water level in the tank is one metre?
- We recall that $V$ is the total volume of the tank, $f$ is the function defined at the beginning of the exercise and $v$ the function defined in Part B. We consider the following algorithm:
| \begin{tabular}{l} Variables: |
| Processing: |
& | $a$ is a real number |
| $b$ is a real number |
| $a$ takes the value 2 |
| $b$ takes the value $2\mathrm{e}$ |
| While $v(b) - v(a) > 10^{-3}$ do: |
| $c$ takes the value $(a + b)/2$ |
| If $v(c) < V/2$, then: |
| $a$ takes the value $c$ |
| Otherwise |
| $b$ takes the value $c$ |
| End If |
| End While |
| Display $f(c)$ |
\hline \end{tabular} Interpret the result that this algorithm allows to display.