. At a given time
t
, the tangent line to the unit circle at the position
P(t)
will determine a right triangle
in the first quadrant. (Connect the origin with the
y
-intercept and
x
-intercept of the tangent line.)
The identity
might be useful in some parts of this question.
(a) The area of the right triangle is
a(t)
=
12
sin
(
t
)
cos
(
t
)
.
(b)
If the tangent line to
y
=
f
(
x
) at (
5
,
4
) passes through the point (0,
3
), find
f
(
5
) and
f'
(
5
).
f
(
5
) =
4
f'
(
5
)
=
15
sin(2t)=2sin(t)cost(t)

∞
(c)
(d)
1
(e) With our restriction on
t
, the smallest
t
so that
a(t)=2
(f) With our restriction on
t
, the largest
t
so that
a(t)=2
(g) The average rate of change of the area of the triangle on the time interval [
π
/6,
π
/4] is
.
(g) The average rate of change of the area of the triangle on the time interval [
π
/4,
π
/3] is
a
(
t
)=
lim
t
→
pi
/2
−
a
(
t
)=
lim
t
→
0
+
a
(
t
)=
lim
t
→
pi
/4
is
is
.

6.
12/12 points |
Previous Answers
The solutions
(x,y)
of the equation
x
2
+ 16
y
2
= 16 form an ellipse as pictured below. Consider the point
P
as pictured, with
x
-coordinate
2
.
(a) Let
h
be a small non-zero number and form the point
Q
with
x
-coordinate
2
+h
, as pictured. The slope of the secant line
through
PQ
, denoted
s(h)
, is given by the formula
$$(
√
(
−
h
2
−
4
h
+12)
−√
12)4
h
.
(b) Rationalize the numerator of your formula in (a) to rewrite the expression so that it looks like
f(h)/g(h)
, subject to these two
conditions: (1) the numerator
f(h)
defines a line of slope -1, (2) the function
f(h)/g(h)
is defined for
h=0
. When you do this
f(h)=
−
h
−
4
g(h)=
4
√
12
−
4
h
−
h
2+8
√
3
.
(c) The slope of the tangent line to the ellipse at the point
P
is
=
−√
312