AP Precalculus -3.3 Sine and Cosine Values- FRQ Exam Style Questions - Effective Fall 2023

(A) Exponential model \(P(t) = ae^{bt}\)

• Given: \(P(1) = 145\), \(P(2) = 115\).
• Divide the equations: \(\frac{P(2)}{P(1)} = \frac{115}{145} = \frac{23}{29} \approx 0.7931 = e^{b(2-1)} = e^b\).
• So \(b = \ln\left(\frac{23}{29}\right) \approx \ln(0.7931) \approx -0.2318\).
• Use \(P(1) = ae^{b\cdot1} = 145\): \(a = 145 \cdot e^{-b} = 145 \cdot \frac{29}{23} \approx 145 \cdot 1.2609 \approx 182.83\).
• Alternatively, compute directly: \(a = P(1) e^{-b} = 145 \cdot \frac{29}{23} = \frac{145 \cdot 29}{23} = \frac{4205}{23} \approx 182.826\).
• Thus, \(P(t) \approx 182.826 e^{-0.2318 t}\).

(B) i. Average rate of change from \(t = 0\) to \(t = 4\)

• \(P(0) = a \approx 182.826\).
• \(P(4) = 182.826 e^{-0.2318 \cdot 4} \approx 182.826 e^{-0.9272} \approx 182.826 \cdot 0.3956 \approx 72.34\).
• Average rate = \(\frac{P(4) – P(0)}{4 – 0} \approx \frac{72.34 – 182.826}{4} = \frac{-110.486}{4} \approx -27.622\) pellets per minute.
• Meaning: On average over the first 4 minutes, the dog ate about 27.6 pellets per minute.

(B) ii. Comparison of average rates

• For an exponential decay \(P(t) = ae^{bt}\) with \(b < 0\), the function is decreasing and concave up (since \(P”(t) = ab^2 e^{bt} > 0\)).
• In a concave up decreasing function, the average rate of change over an interval decreases in magnitude (becomes less negative) as the interval moves to the right.
• Thus, the average rate from \(t = 4\) to \(t = t_a\) will be greater (less negative) than that from \(t = 0\) to \(t = 4\).
• Reason: The instantaneous rate (derivative) is negative but increasing (becoming less negative), so later intervals have a smaller rate of loss.

(C) Limit as \(t \to \infty\)

• \(\lim_{t \to \infty} P(t) = \lim_{t \to \infty} ae^{bt} = 0\) because \(b < 0\).
• Context: The model predicts the pellets approach zero but never actually reach it. In reality, the dog will finish all pellets in finite time, so the model is not perfectly accurate for very large \(t\). However, it is a good approximation for the decay process while pellets remain.
• Thus, the model does not fully make sense as \(t \to \infty\) because it predicts the dog never finishes, while in reality the bowl becomes empty at some finite time.