calculus problem 8-17: A pipe with a ten-inch radius....

[leslie2]

A pipe with a 10" radius runs along the floor and wall of a room.

a) What is the radius of the largest pipe that will fit between this pipe and the corner?

b) If the diameter of the larger pipe is given as D, what is the diameter of the smaller pipe?

 

[stapel]

By "corner", do you mean "where the wall meets the floor"?

Thank you.

Eliz.

 

[soroban]

Re: calculus problem 8-17

Hello, leslie2!

A pipe with a 10" radius runs along the floor and wall of a room.
a) What is the radius of the largest pipe that will fit between this pipe and the corner?
b) If the diameter of the larger pipe is given as D, what is the diameter of the smaller pipe?

      |           * * *
      |      *             *
      |   *                   *
      | *                       *
      |*                         *
      |
      *             O             *
    A * - - - - - - *             * 
      *           / :             *
      |       R /   :
      |*      /     :            *
      | *   /       :R          *
      |   *D        :         *
      | /    *      :      *
      * - - - - - * * * - - - - - -
. . . B - - - - - - C

The center of the pipe is \(\displaystyle O\).
Its radius is: .\(\displaystyle R\,=\,OA\,=\,OD\,=\,OC.\)

The smaller pipe has diameter \(\displaystyle DB\).

In square \(\displaystyle ABCO\) with side \(\displaystyle R\)
\(\displaystyle \;\;\)the diagonal is: \(\displaystyle \,OB\,=\,R\sqrt{2}\)

Hence: \(\displaystyle \text{ diameter }DB\:=\:OB\,-\,OD\:=\:R\sqrt{2}\,-\,R\:=\:R\left(\sqrt{2}\,-\,1\right)\)

Therefore: \(\displaystyle \,\text{radius}\;=\;\frac{1}{2}R\left(\sqrt{2}\,-\,1\right)\)