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The water level increases by 0.608 meters per minute when the water is 3.5 m deep
The given parameters are:
The relationship between the radius and height is:
r/h = 3/7
Make r the subject
r = 3h/7
The volume of a cone is;
[tex]V = \frac 13\pi r^2h[/tex]
This gives
[tex]V = \frac 13\pi (\frac{3h}{7})^2h[/tex]
Expand
[tex]V = \frac{3h^3}{49}\pi[/tex]
Differentiate
[tex]V' = \frac{9h^2}{49}\pi h'[/tex]
Make h' the subject
[tex]h' = \frac{49}{9\pi h^2}V'[/tex]
When the water level is 3.5.
We have:
[tex]h' = \frac{49}{9\pi * 3.5^2}V'[/tex]
Also, we have:
V' = 4.3
So, the equation becomes
[tex]h' = \frac{49}{9\pi * 3.5^2} * 4.3[/tex]
Evaluate the products
[tex]h' = \frac{210.7}{346.36}[/tex]
Evaluate the quotient
h' = 0.608
Hence, the water level increases by 0.608 meters per minute
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