# Applications of Inverse Trigonometric Functions by pharmphresh30

VIEWS: 10 PAGES: 2

• pg 1
```									    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.

1
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.

2

```
To top