Applications of Inverse Trigonometric Functions by pharmphresh30

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									    Applications of Inverse Trigonometric Functions


1. A painting in an art gallery has height h and is hung so that its lower edge is
   a distance d above the eye of an observer. How far from the painting should
   an observer stand to maximize the angle subtended by the painting at the
   observer’s eye?


                                                               h

                                        θ                      d
                                        x
   Solution. The angle subtended by the painting is

                                 θ = arctan        h+d
                                                    x    − arctan       d
                                                                        x   ,

   where 0 < x < ∞. To find critical values for θ(x) set θ (x) = 0 (since there is no place
   where θ (x) does not exist on the interval 0 < x < ∞):

                         − h+d              − x2
                                              d
                                                                         h+d            d
        θ (x) = 0 ⇒         x2
                                    −                  =0 ⇒ −                       + 2     = 0.
                            h+d 2              d 2                 x2   + (h + d) 2  x + d2
                       1+    x          1+     x


   Multiplying by (x2 + (h + d)2 )(x2 + d2 ) gives

          − (h + d)(x2 + d2 ) + d x2 + (h + d)2 = 0
           ⇒ −(h + d)x2 − (h + d)d2 + dx2 + d(h + d)2 = 0
           ⇒ x2 [−(h + d) + d] − hd2 − d3 + d(h2 + 2hd + d2 ) = 0
           ⇒ −hx2 + hd2 + h2 d = 0 ⇒ −h[x2 − d(h + d)] = 0 ⇒ x =                        d(h + d).

   We can now construct the sign table corresponding to θ (x):
                                                          −→
                    θ(x)
                    θ (x)               +                  0                    −
                 0<x<∞            d(h + d)/2             d(h + d)           2d(h + d)

   The sign table above shows that x =             d(h + d) corresponds to a maximum.




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2. The tortoise and the hare begin their race directly below the starter’s plat-
   form. Both run at a constant speed but the hare runs four times faster than
   the tortoise. At what time t in seconds is the angle θ a maximum?


                                         θ
                                   φ
                               h


                                        x(t)
                                                     4x(t)
   Solution. Denote by h the height of the starter’s platform and by x the distance that
   the tortoise has run by time t. Then the hare has run 4x up to time t. The angle φ obeys

                                 tan φ =         x
                                                 h
                                                      ⇒ φ = arctan               x
                                                                                 h

   and the sum of the angles φ + θ obeys

                              tan φ + θ =               4x
                                                         h
                                                               ⇒ arctan          4x
                                                                                 h
                                                                                          .

   Hence
                          θ = (θ + φ) − φ = tan−1                     4x
                                                                      h    − tan−1            x
                                                                                              h

   Suppose that the tortoise travels at v m/s then the distance travelled by tortoise after t
   seconds is x = vt. Thus

                               θ(t) = tan−1              4vt
                                                          h      − tan−1        vt
                                                                                h     ,

   where 0 < t < ∞. To find critical values for θ(t) set θ (t) = 0 (since there is no place
   where θ (t) does not exist on the interval 0 < t < ∞):
                                  v                   v                                v                   v
                                 4h                                                   4h
              θ (t) = 0 ⇒                    −        h
                                                                 =0 ⇒                             =        h
                                 16v 2 t2               2 2
                                                                                      16v 2 t2               2 2
                            1+     h2            1   + vh2
                                                         t
                                                                               1+       h2            1   + vh2
                                                                                                              t


                               v 2 t2                 16v 2 t2                 12v 2 t2
                    ⇒ 4 1+      h2
                                          =1+           h2
                                                                 ⇒ 3=            h2
                                                                                              ⇒ t=    h
                                                                                                      2v

   We can now construct the sign table corresponding to θ (t):
                                                                 −→
                                        θ(t)
                                        θ (t)            +       0         −
                                 0<t<∞                  h
                                                        4v
                                                                 h
                                                                 2v
                                                                           h
                                                                           v

                                                 h
   The sign table above shows that t =           2v     corresponds to a maximum.




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