Answer:
0.358 ft/min
Step-by-step explanation:
Let's say r and h are the radius and height of the water, and R and H are the radius and height of the tank.
The volume of the water is:
V = π/3 r² h
Using similar triangles:
r / h = R / H
r = h R/H
Substitute:
V = π/3 (h R/H)² h
V = π/3 (R/H)² h³
Take derivative with respect to time:
dV/dt = π (R/H)² h² dh/dt
Plug in values:
2 = π (2/3)² (2)² dh/dt
2 = 16π/9 dh/dt
1 = 8π/9 dh/dt
dh/dt = 9/(8π)
dh/dt ≈ 0.358
The water is rising at approximately 0.358 ft/min.
The water is rising at approximately 0.358 ft/min.
Here we assume r and h are the radius and height of the water, and R and H are the radius and height of the tank.
We know that
The volume of the water is:
[tex]V = \pi \div 3 r^2 h[/tex]
Now here we used similar triangles:
[tex]r \div h = R \div H\\\\r = h R\div H[/tex]
Now
[tex]V = \pi \div 3 (h R\div H)^2 h\\\\V = \pi \div 3 (R\div H)^2 h^3[/tex]
Now take the derivative with respect to time:
So,
[tex]dV\div dt = \pi (R\div H)^2 h^2 \\\\2 = \pi (2\div 3)^3 (2)^3\\\\2 = 16\pi \div 9\\\\1 = 8\pi \div 9\\\\dh/dt = 9\div (8\pi )[/tex]
≈ 0.358
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