Calculus I

Lesson 19: Applications of Inverse Trig Functions

1) A lighthouse is on a small island 3 km away from the nearest point O

on a straight shoreline and its light makes 4 revolutions per minute.

a) How fast is the beam of light moving along the shoreline when it is 1 km from O? See diagram.

b) Plot the rate of change of x (from the diagram) with respect to time.

c) What is ?

d) What is ?

> restart; with(plottools): with(plots):

Warning, the name changecoords has been redefined

> PO := line([0,0],[0,3], linestyle=3, color=brown):
PX := line([0,3], [-1, 3], linestyle=1, thickness=2, color=blue):
XP := line([-1, 3], [0,0], thickness=3, color=yellow):
shoreline := line([-3,3],[3,3], color=coral):
text:=textplot({[.1,0,`P`],[0,3.05,`O`],[-1.1,3.05,`x`],[-.4,.9,`alpha`],[.1,1.5,`3`],[-.5,3.05,`1`]},color=black,align={ABOVE, RIGHT}):
a := arc([0,0], .8, Pi/2..Pi/1.65, thickness=2):
display(PO, PX, XP, shoreline, text, a, axes=none);

In the above diagram, the length of segment OP is given to be 3km.

Let be the angle xPO.

We are given that = 4 rev/min = 4 ( 2 ) 60 = 480 radians/hour.

We are asked to find when x = 1 km. As tan ( ) = we see

that = .

Thus,

= 480 =

and therefore

= 480 3 ( 1 + ( ) .

> dxt:= x -> 480 * Pi * 3 * ( 1 + (x/3)^2);

> evalf(dxt(1));

Thus when x = 1 is about km/hour.

b)

> plot(dxt(x), x = 0..100);

c) and d) We see that

and that

2) Where should the point P be chosen on the line segment AB

so as to maximize the angle theta = ? What is the maximal value of ?

Assume that the segment AB has length 3 and the vertical line segments

have lengths 5 and 2. See Diagram.

> X := 1:
ground := line([0,0],[3,0], color=brown, thickness=3):
stake1 := line([0,0],[0,5], color=blue, thickness=2):
stake2 := line([3,0],[3,2], color=blue, thickness=2):
line1 := line([X,0],[0,5], color=red, linestyle=2, thickness=2):
line2 := line([X,0],[3,2], color=red, linestyle=2, thickness=2):
text := textplot({[-.1,-.1,`A`],[X,-.1,`P`],[3,-.1,`B`],[X+.4,1.3,`theta`], [-.1,2.5,`5`],[3.1,1,`2`]}):
a := arc([1,0], 1, arccot((3-X)/2)..Pi-arccot(X/5)):
display(ground, stake1, stake2, line1, line2, text,a,axes=none,scaling=constrained);

Let x denote the length of the line segment AP.

Determine in terms of x. Plot this function.;

we are looking for what value of x makes maximal.

What happens to as x varies between 0 and 3?

We have:

= - - .

Thus, = .

Setting = 0 we have:

iff

.

Hence we have two critical values; as x must be between 0 and 3 we

have as our critical value.

> evalf( 5 - 2* sqrt(5));

> th:= x -> Pi - arccot(x/5) - arccot((3-x)/2);

> th(x);

> plot(th(x), x = 0..3);

> evalf(th(0));

> evalf(th(.527864044));

From the plot of , we see that has one maximum which occurs

at which is about units.

You can also use the first derivative test to see that indeed this

critical value leads to a maximal value of . Make sure you can

do this analysis.

From the plot of , we see that as x increases from 0 to 3,

increases from approximately to

and then strictly decreases to 0 .