1. ## Weird rate problem.

A hemispherical bowl has a radius of 3.7m. Water is added to the bowl at the rate of $pi/3$ $m^3/min$. When the depth of water in the bowl is $h$ m, and volume of water is given by $V= (1/6)pi*h^2(9-2h)m^3$. How deep is the water after 2.5 minutes? At what rate is the depth increasing at this time?

So I'm guessing that the $m^3$ is just a unit and not part of the formula.

The volume of water in the bowl is $pi/3 * 2.5 = 5pi/6
$

So $5pi/6=pi/6* h^2(9-2h)$

simplify: $5=9h^2 - 2h^3$

If I'm right so far how do I solve for h?

2. Originally Posted by coolguy00777
A hemispherical bowl has a radius of 3.7m. Water is added to the bowl at the rate of $pi/3$ $m^3/min$. When the depth of water in the bowl is $h$ m, and volume of water is given by $V= (1/6)pi*h^2(9-2h)m^3$. How deep is the water after 2.5 minutes? At what rate is the depth increasing at this time?

So I'm guessing that the $m^3$ is just a unit and not part of the formula.

The volume of water in the bowl is $pi/3 * 2.5 = 5pi/6$
The volume of the water at t= 2.5
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So $5pi/6=pi/6* h^2(9-2h)$

simplify: $5=9h^2 - 2h^3$

If I'm right so far how do I solve for h?
That's a cubic equation: $2h^3- 9h^2+ 5= 0$.
There is a general formula but it is very complex. The "rational root theorem" says that any rational roots must be of the form a/b where a divides the constant term, 5, and b divides the leading coefficient, 2. That means that, for this equation, the only possible rational roots are $\pm 1$, $\pm 5$, $\pm \frac{1}{2}$, and $\pm \frac{5}{2}$. There is no guarentee that this equation has a rational root but I recommend trying those to see. If none of those work, there is no simpler solution than the cubic formula.

3. Wow, did I make a mistake in my conclusion. We haven't used cubic roots in my calculus 1 class.