AP Calculus BC 1.9 Connecting Multiple Representations of Limits Exam Style Questions - FRQ
▶️ Answer/Explanation
Solution
(a) \( v(t) = \int (12t^2 – 4) \, dt = 4t^3 – 4t + C \). Using \( v(0) = 0 \), \( C = 0 \). \( \boxed{v(t) = 4t^3 – 4t} \).
(b) From table, as \( t \to 1^- \): \( v(0.999) = -1.999996 \to -2 \). As \( t \to 1^+ \): \( v(1.001) = -2.000004 \to -2 \). \( \boxed{\lim_{t \to 1} v(t) = -2} \). Compute \( v(1) = 4(1)^3 – 4(1) = 0 \), which differs from -2, indicating a removable discontinuity.
(c) \( x(t) = \int (4t^3 – 4t) \, dt = t^4 – 2t^2 + C \). With \( x(1) = 3 \), \( 1 – 2 + C = 3 \implies C = 4 \). \( \boxed{x(t) = t^4 – 2t^2 + 4} \).