(b) The water level drops at a rate of 0.1 cm per hour. At what rate is the radius of the surface of the water decreasing when the depth is 8 cm? Round your

Question

(a) A hemispherical bowl of radius 12 cm contains water to a depth of h cm. Find the radius of the surface of the water as a function of h.

 

r= ???

 

(b) The water level drops at a rate of 0.1 cm per hour. At what rate is the radius of the surface of the water decreasing when the depth is 8 cm?

Round your answer to four decimal places.

Answer 1

Think of a circular cross section of a hemisphere with center of the circle at (x, y) = (0, r) where x and y are the horizontal and vertical axes respectively, and the radius of the circle is r (constant). The height is along the y axis here. The area of the circle is ( x - 0)² + (y - r)² = r² . Solve for x: x = [r² - (y - r)²]^0.5 . 

 

In order to find the rate of change of the  radius or horizontal component as a function of the change of the height (vertical component), take the derivative of x and y with respect to time (t).   So the change in the radius dx/dt = 0.5[r² - (y - r)²]^-0.5[-2*(y - r)*dy/dt].  Substitute and solve this equation for r = 12 cm, y = 8 cm, dy/dt = -0.1 cm/hr. The result for the rate the radius is decreasing is dx/dt = -0.0354 cm/hr.  Check the math!