Re: Related rates problem

The bowl has a radius of 10cm. The distance from the height of the water to the top centre of the hemisphere (from where the radius is measured) is $\displaystyle (10-h)$cm

I drew a diagram in front of me for this, and you can do it using Pythagoras.

Draw the hemisphere. Draw the level of the water at any distance $\displaystyle (10-h)$cm below the top and label the $\displaystyle (10-h)$ distance. Then draw a radius from the centre top of the hemisphere to the point on the hemisphere that touches the surface of the water. If you have understood me, you should now have a right triangle. The base is $\displaystyle r$, the height is $\displaystyle (10-h)$ and the hypotenuse is $\displaystyle 10$.

I hope you'll forgive me for my dreadful artistic skills, but I have done a diagram here:

I'll check this in a second, it's sort of a leap of faith on my part, but hopefully you should get the expected result.

Edit: Yeah, I've managed to obtain their solution this way.(Cool)