Posted by Karina on .
Consider a hemispherical bowl of radius 10cm.
(4 points) If the water level in the bowl is h cm deep, find the radius of the surface of the water as a function of h.
(6 points) The water level drops at a rate of 0.1 cm per hour. At what rate is the radius of the water decreasing when the depth is 5 cm?
draw a vertical cross-section of the bowl. Pick a point H on the vertical axis at height h from the bottom. If P is a point on the rim of the bowl, HP = 10. The radius at H is
(10-h)^2 + r^2 = 10^2
r = √(100-(10-h)^2)