Answer
(a) See explanations.
(b) See graph.
(c) 21.07 ft
(d) See explanations.
Work Step by Step
(a) Use the figure given in the Exercise, the length of the pipe can be divided into two parts $x,y$, each part is a
hypotenuse of a right triangle, one has an opposite side of 9 ft to $\theta$, another has a adjacent side of 6 ft to $\theta$. In the first triangle, $sin\theta=\frac{9}{x}, x=9csc\theta$. In the second triangle $cos\theta=\frac{6}{y}, y=6sec\theta$. Thus the total pipe length $L(\theta)=x+y=9csc\theta+6sec\theta$
(b) See graph.
(c) Base on the graph, a minimum of 21.07 ft can be found at $\theta=0.8528rad$
(d) The value of L found in part (c) is the critical length that any pipe with length larger than this value will
get stuck around the corner. Thus, this value is also the longest of a pipe that can be carried around the corner.
