Question

Water is being pumped into a cylindrical container at a constant rate. The depth of the water is increasing linearly. At 12 pm , the depth of the water was 1.5 feet It is now 5 pm, and the depth of the water is now 4.0 feet. What will the depth of the water be at 6 pm?

Answer 1

a rate that moves linearly means a straight line and to get the equation of any straight line, we simply need two points off of it, let's use those ones in the picture below

`(\stackrel{x_1}{12}~,~\stackrel{y_1}{1.5})\qquad (\stackrel{x_2}{17}~,~\stackrel{y_2}{4.0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4.0}-\stackrel{y1}{1.5}}}{\underset{\textit{\large run}} {\underset{x_2}{17}-\underset{x_1}{12}}}\implies \cfrac{2.5}{5}\implies \cfrac{1}{2}`

`\begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1.5}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{12})\implies y-1.5=\cfrac{1}{2}x-6 \\\\\\ y=\cfrac{1}{2}x-6+1.5\implies y=\cfrac{1}{2}x+4.5 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{at 6pm, x = 12+6=18}}{y=\cfrac{1}{2}(18)+4.5}\implies y=9+4.5\implies {\Large \begin{array}{llll} y=13.5 \end{array}}`