AP Calculus BC 3.5 Selecting Procedures for Calculating Derivatives Exam Style Questions - FRQ
▶️ Answer/Explanation
Solution
A. Velocity of particle H at t=1:
\[ v_H(t) = \frac{d}{dt} e^{t^2 – 4t} = e^{t^2 – 4t}(2t – 4) \] \[ v_H(1) = e^{1 – 4}(2 – 4) = e^{-3}(-2) = -\frac{2}{e^3} \]
B. Opposite direction intervals:
First find when each particle changes direction:
- Particle H: \( v_H(t) = 0 \) when \( 2t-4=0 \) ⇒ \( t=2 \)
- Particle J: \( v_J(t) = 0 \) when \( t=0,1 \) (since \( t^2-1=0 \) ⇒ \( t=±1 \), but domain is 0≤t≤5)
Test intervals:
| Interval | \( v_H(t) \) | \( v_J(t) \) | Opposite? |
|---|---|---|---|
| (0,1) | Negative | Negative | No |
| (1,2) | Negative | Positive | Yes |
| (2,5) | Positive | Positive | No |
Particles move in opposite directions during \( (1,2) \).
C. Speed of particle J at t=2:
Since \( v_J(2) = 2(2)(4-1)^3 = 4(27) = 108 > 0 \) and \( v_J'(2) > 0 \), both velocity and acceleration are positive ⇒ speed is increasing.
D. Position of particle J at t=2:
\[ x_J(2) = 7 + \int_0^2 2t(t^2 – 1)^3 dt \] Let \( u = t^2 – 1 \), \( du = 2t dt \): \[ = 7 + \int_{-1}^3 u^3 du = 7 + \left[ \frac{u^4}{4} \right]_{-1}^3 = 7 + \left( \frac{81}{4} – \frac{1}{4} \right) = 7 + 20 = 27 \]