C Roettger, S15 – Math 165, Project 3C
You are Function Detectives! Evil Genius Dr Moriarty’s secret is a function f (t), about which you have discovered the following clues.
1. It is a polynomial in sin(t), so there is a polynomial g(x) such that
f (t) = g(sin t).
2. It has horizontal tangent lines whenever sin t = 1/3 or sin t = 1 or
sin t = −1.
3. Its graph passes through the point (0, 1).
4. Its graph passes through the point (π/2, 4).
Your ultimate goal is to find the function f (t) itself. Fortunately, the clues
are already in the right order to solve the puzzle. In particular, use 2. to
determine the degree of g(x). Then use it to determine f � (x). This leaves
you with two parameters in f (x) undetermined. The last two clues nail these
down (use antiderivatives again). Have some fun by plotting several (eg ten)
different functions satisfying all clues 1. - 3., including the correct one, over
the interval [−π, π]. Include the plot in your writeup.
Solution.
First, f � (t) = g � (sin t) cos t, so clue # 2 tells you g � (x) = 0 for x = 1/3.
Whenever sin t = 1 or sin t = −1, automatically cos t = 0, so these two
t-values for horizontal tangent lines are already taken care of. We can take
g(x) of degree One,
a
g � (x) = a(x − 1/3) = (3x − 1),
3
and we integrate:
g(x) =
�
a
g (x) dx =
3
�
�
�
3x2
− x + C.
2
We use clue 3: for t = 0, x = sin t = 0, so 1 = f (0) = g(0) and therefore
C = 1.
Our function f (t) at this point has the shape
�
�
a 3 sin2 t
f (t) =
− sin t + 1.
3
2
Finally, clue 4 says f (π/2) = 4, and since sin π/2 = 1, g(1) = 4. We use this
to determine a:
�
�
a 3
a
4 = g(1) =
−1 +1= +1
3 2
6
and so a = 18 which completely determines the function f (t) = g(sin t), since
we know at this point
�
� 2
3x
− x + 1.
g(x) = 6
2
The graph below shows several functions that satisfy all clues # 1 - 3, with
the one plotted in bold red being Dr Moriarty’s secret. Note how they all
share the x-coordinates of points with horizontal tangent lines (these are
local maxima / minima).
I saw many projects which wrote that clue # 2 tells you g � (x) = 0 for
x = 1/3, 1, −1 – no points taken off for that.
With that as given,
a
a
g � (x) = a(x − 1/3)(x − 1)(x + 1) = (3x − 1)(x2 − 1) = (3x3 − x2 − 3x + 1),
3
3
and we integrate:
�
�
�
a 3x4 x3 3x2
�
g(x) = g (x) dx =
−
−
+ x + C.
3
4
3
2
We use clue 3: for t = 0, x = sin t = 0, so 1 = f (0) = g(0) and therefore
C = 1.
Our function f (t) at this point has the shape
�
�
a 3 sin4 t sin3 t 3 sin2 t
f (t) =
−
−
+ sin t + 1.
3
4
3
2
Finally, clue 4 says f (π/2) = 4, and since sin π/2 = 1, g(1) = 4. We use this
to determine a:
�
�
a 3 1 3
a
4 = g(1) =
− − +1 +1=− +1
3 4 3 2
36
and with this reading of the clues, a = −108 which completely determines
the function f (t) = g(sin t), giving
� 4
�
3x
x3 3x2
g(x) = −36
−
−
+ x + 1.
4
3
2
The graph below shows these functions plotted for several values of a, with
the one plotted in bold red being Dr Moriarty’s secret. Note how they all
share the x-coordinates of points with horizontal tangent lines.