Respuesta :
Answer:
The water level rising when the water is 4 inches deep is [tex]\frac{3}{8\times \pi} inch/s[/tex].
Step-by-step explanation:
Rate of water pouring out in the cone = R=[tex]\frac{3}{2} inch^3/s[/tex]
Height of the cup = h = 6 inches
Radius of the cup = r = 3 inches
[tex]\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}[/tex]
r = h/2
Volume of the cone = [tex]V=\frac{1}{3}\pi r^2h[/tex]
[tex]V=\frac{1}{3}\pi r^2h[/tex]
[tex]\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}[/tex]
[tex]\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}[/tex]
h = 4 inches
[tex]\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}[/tex]
[tex]\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2[/tex]
[tex]\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s[/tex]
The water level rising when the water is 4 inches deep is [tex]\frac{3}{8\times \pi} inch/s[/tex].
