L19inverseTrig.mws
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((3X)/2)..Piarccot(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((3x)/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 .