Respuesta :
Answer:
[tex] V = \pi r^2 h[/tex]
For this case we know that [tex] r=5m[/tex] represent the radius, [tex] h = 10m[/tex] the height and the rate given is:
[tex] \frac{dV}{dt}= \frac{100 L}{min}[/tex]
[tex] Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}[/tex]
And replacing we got:
[tex] \frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}[/tex]
And that represent [tex] 0.127 \frac{cm}{min}[/tex]
Step-by-step explanation:
For a tank similar to a cylinder the volume is given by:
[tex] V = \pi r^2 h[/tex]
For this case we know that [tex] r=5m[/tex] represent the radius, [tex] h = 10m[/tex] the height and the rate given is:
[tex] \frac{dV}{dt}= \frac{100 L}{min}[/tex]
For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:
[tex] \frac{dV}{dt}= \pi r^2 \frac{dh}{dt}[/tex]
And solving for [tex]\frac{dh}{dt}[/tex] we got:
[tex] \frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}[/tex]
We need to convert the rate given into m^3/min and we got:
[tex] Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}[/tex]
And replacing we got:
[tex] \frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}[/tex]
And that represent [tex] 0.127 \frac{cm}{min}[/tex]
