Please find the error in the following solution and proved a correct version of the solution.

A water pipe with radius 10 cm and length 20 m is filling with water at the rate of 120 cm* per minute. If both ends are sealed off and the pipe is parallel to the ground, how quickly is the water level in the pipe rising when the water evel is 15 cm? What is wrong with the following "solution"?' Find the error and produce a correct solution to this problem. Answer: Let us consider the more general problem when the pipe has radius R and length _ since this is a case when the problem is simplified by avoiding concrete numbers. The accompanying diagram shows a cross section of the pipe with the green line being vertical and perpendicular to the side of the triangle. Let t denote time. Notice that a variable / has been introduced as well to represent the angle indicated and that 0 depends on time t. Let W(@) denote the water level as a function of 0. The problems asks for d'It/ dt so the chain rule will be used. The following represents W explicitly as a function of 0: W(0) = R + Rcos(@) = R(1 + cos(0)) (1) The area of the circle below the waterline at height W is equal to: the area of the circle minus the pie-like segment subtended between two radii forming an angle of 20 and the circumference of the circle . plus the area of the triangle outlined in red. Hence the area A as a function of 0 is: A(0) = TR2 xR2 + 2((Rcos(0) Resin(0)/2) = R'(x - 0 + cos(0) sin(0)) (2) and multiplying this by the length of the pipe yields the volume V as a function of 0: V(@) = LR'(x - @ + cos(0) sin(@)) (3) RSMLo) R Water loved = W Since it is known that the change in volume is 120 cm per minute it follows that 120cm = dV (@) dt _ d LR (x - 0 + cos(0) sin(0)) = 0 LR? (x - 0 + cos(0) sin(8)) - LR' (-1 + cos? (0) - sin?(0)- = -2LR2 sin? (0) dt and hence 120 = -2LR3 sin?(0)- (4) and hence 120 60 de (5) -2LR' sin' (0) LR- sin (@) dt noting that the denominator is non-zero so long as # > 0. From (1) it follows that d w(0) = R- (1 + cos(0)) = -Rsin(0) (6) so that dw dW de -60 60 dt do dt -Rsin(0)- LRsin(0) (7) LR2 sin(0) and so all that is needed is the value of sin (0) when the water level is 15 cm. Given that R = 10cm, this happens when R cos(0) = 5crn = R/2 or, in other words, when 0 = w/4. Since sin(/4) = 1/v2 it follows from (7) that at this height dw 60 60/2 dt LR/V2 LR and at the specific values given for L = 2000 and R = 10 it follows that the rate of change of the water level is 601/2 312 x 10-3 2 x 10* or approximately 0.0042 cm per minute