AP Precalculus -3.8 The Tangent Function- FRQ Exam Style Questions - Effective Fall 2023

Step 1: Identify parameters
– Clock diameter = 24 inches ⇒ radius \( r = 12 \) inches.
– Clock center height above ground: 5 feet = \( 5 \times 12 = 60 \) inches.
– Bottom of clock: \( 60 – 12 = 48 \) inches above ground.
– Top of clock: \( 60 + 12 = 72 \) inches above ground.
– Cycle period = 30 seconds ⇒ \( b = \frac{2\pi}{30} = \frac{\pi}{15} \).
– At \( t = 0 \), second hand points straight up ⇒ height is maximum: \( h(0) = 72 \).

Step 2:
Form of \( h(t) \) General form: \( h(t) = a\cos(b(t + c)) + d \).
Maximum height = \( d + |a| = 72 \)
Minimum height = \( d – |a| = 48 \)
Solve: \( d = \frac{72+48}{2} = 60 \), \( |a| = 72 – 60 = 12 \).
Since at \( t = 0 \) we have max height, choose cosine with no horizontal shift that gives max at \( t=0 \) ⇒ \( a > 0 \), \( c = 0 \) (since \(\cos(0)=1\) gives max).
Thus \( a = 12 \), \( c = 0 \), \( b = \frac{\pi}{15} \), \( d = 60 \).

Step 3: Final function \[ h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 \]

Step 1:
Set up equation \[ 12\cos\left( \frac{\pi}{15} t \right) + 60 = 72 – 6\sqrt{2} \] \[ 12\cos\left( \frac{\pi}{15} t \right) = 12 – 6\sqrt{2} \] \[ \cos\left( \frac{\pi}{15} t \right) = 1 – \frac{\sqrt{2}}{2} \]
Note: \( 1 – \frac{\sqrt{2}}{2} \approx 0.2929 \).

Step 2: General solutions Let \( \theta = \frac{\pi}{15} t \). \(\cos \theta = 1 – \frac{\sqrt{2}}{2} \).
Since cosine is positive, solutions are in QI and QIV (for principal values \( 0 \leq \theta < 2\pi \)).
\(\theta_1 = \cos^{-1}\left(1 – \frac{\sqrt{2}}{2}\right)\) \(\theta_2 = 2\pi – \theta_1\).

Step 3: Find \( t \)
\( t = \frac{15}{\pi} \theta \).
Numerically: \( 1 – \frac{\sqrt{2}}{2} \approx 0.292893 \) \(\theta_1 \approx \cos^{-1}(0.292893) \approx 1.2735 \) rad.
\( t_1 \approx \frac{15}{\pi} \times 1.2735 \approx 6.08 \) s. \(\theta_2 \approx 2\pi – 1.2735 \approx 5.0097 \) rad. \( t_2 \approx \frac{15}{\pi} \times 5.0097 \approx 23.92 \) s.
Both are in \( 0 < t \leq 30 \).

Step 4: Exact form Exact \( \theta_1 = \cos^{-1}\left( 1 – \frac{\sqrt{2}}{2} \right) \), but simpler:
Note \( 1 – \frac{\sqrt{2}}{2} = \frac{2 – \sqrt{2}}{2} \).
Thus: \[ t_1 = \frac{15}{\pi} \cos^{-1}\left( \frac{2 – \sqrt{2}}{2} \right), \quad t_2 = 30 – t_1 \]
because cosine is symmetric about \( \theta = \pi \). Final times: \( t \approx 6.08 \) s and \( t \approx 23.92 \) s.

Step 1: Second derivative
\( h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 \) \( h'(t) = -12 \cdot \frac{\pi}{15} \sin\left( \frac{\pi}{15} t \right) \)
\( h”(t) = -12 \cdot \frac{\pi^2}{225} \cos\left( \frac{\pi}{15} t \right) = -\frac{4\pi^2}{75} \cos\left( \frac{\pi}{15} t \right) \).

Step 2: Inflection points occur when \( h”(t) = 0 \) \[ \cos\left( \frac{\pi}{15} t \right) = 0 \] \[ \frac{\pi}{15} t = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] \[ t = 7.5 + 15n \]

Step 3: In \( 0 \leq t < 30 \) \( n = 0 \): \( t = 7.5 \) s \( n = 1 \): \( t = 22.5 \) s
So yes, there are two points of inflection at \( t = 7.5 \) s and \( t = 22.5 \) s.

At a point of inflection, the rate of change of height \( h'(t) \) is at a local extremum (maximum or minimum magnitude of vertical speed). Equivalently, the acceleration \( h”(t) \) changes sign, meaning the rate at which the height is changing (speed) stops increasing and starts decreasing, or vice versa.
In this problem:
– At \( t = 7.5 \) s, second hand is horizontal (3 o’clock position), vertical speed is maximum in downward direction, and acceleration changes from negative to positive (decelerating downward motion before starting to slow and reverse).
– At \( t = 22.5 \) s, second hand is horizontal (9 o’clock position), vertical speed is maximum in upward direction, and acceleration changes from positive to negative.
Thus, these are the moments when the vertical motion switches from speeding up to slowing down (or vice versa) while moving toward the extreme top or bottom of its path.