# Related rates problem

• Nov 16th 2011, 09:12 AM
Kjcube
Related rates problem
One of the practice problems in my book is giving me trouble.

"A hemispherical bowl of radius 10 cm contains water to a depth of h cm. Find the radius of the surface of the water as a function of h."

I can see that the radius gets smaller as h gets smaller but I'm not how to go about finding an equation. The back of the book says that the answer is r=(20h -h^2)^(1/2) and I think they used the circle equation but I'm not sure. Does anyone have any hints?
• Nov 16th 2011, 09:30 AM
Quacky
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)