ANSWER :
- [tex] \small \sf{Therefore \: the \: water \: level \: is \: rising \: at \: the \: rate \: of \frac{32}{25\pi}{\: m/min \: when \: the \: water \: is \: 5m deep. }}[/tex]
SOLUTION :
- Let T minutes = Be the time that has elapsed since water started to flow into the tank.
- At T minutes let H meters = Be the height of the water level.
- Let R = Be the radius of the surface of the water.
- Let V the cubic meters of the water in the tank.
The volume of water in the tank can be expressed in terms of the volume of a cone. See the attached image.
[tex] \large\bold{v \: = \frac{1}{3} \pi \: r {}^{2} \: h }[/tex]
- V, r , and h are all functions of t. Because water is flowing into the tank at the rate of 2m³/min , dV/dt = 2. We'll find the dh/dt when h = 5. To express the r terms of h we must have from similar triangles ;
[tex] \large \bold{ \frac{r}{h} = \frac{4}{16} → \: r \: = \frac{1}{4} h}[/tex]
- Substituting this value of r into (4) we obtain ;
[tex] \large \bold{V \: = \: \frac{1}{3}\pi( \frac{1}{4} h) {}^{2}(h) \: → \: \: V = \: \frac{1}{48}\pi \: h {}^{3} }[/tex]
- By differentiating both sides of this equation with respect to t,
[tex] \large \bold{ \frac{dV}{dt} = \frac{1}{16} \pi \: h {}^{2} \: \frac{dh}{dt} }[/tex]
- Substituting 2 for dV/dt and solving for dh/dt we get ;
[tex] \large \bold{ \frac{dh}{dt} = \frac{32}{\pi \: h {}^{2} } }[/tex]
[tex]\small \sf{Therefore \: the \: water \: level \: is \: rising \: at \: the \: rate \: of \frac{32}{25\pi}{\: m/min \: when \: the \: water \: is \: 5m deep. }}[/tex]
[tex]\sf \red{\overline{ \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}[/tex]
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