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AP Calculus BC 3.2 Implicit Differentiation - FRQs - Exam Style Questions

AP Calculus BC - FRQs and Answer - All Topics

No-Calc Question

t
(minutes)		025812
\(v_A(t)\)
(meters/minute)		010040-120-150
Train A runs back and forth on an east–west section of track. Its velocity (m/min) is given by a differentiable function \(v_A(t)\). Selected values are in the table above.
(a) Find the average acceleration of Train A over \(2 \le t \le 8\).
(b) Do the data support that \(v_A(t)=-100\) m/min for some \(t\) with \(5<t<8\)? Give a reason.
(c) At \(t=2\), Train A’s position is 300 m east of the Origin Station and the train is moving east.
Write an integral expression for the position at \(t=12\).
Then use a trapezoidal sum with the three subintervals from the table to approximate that position.
(d) A second train, Train B, travels north from the Origin Station with velocity \(v_B(t)=-5t^{2}+60t+25\). At \(t=2\), Train B is 400 m north of the station.
Find the rate (m/min) at which the distance between Train A and Train B is changing at \(t=2\).

Most-appropriate topic codes (CED):

• TOPIC 2.1: Defining Average and Instantaneous Rates of Change at a Point — part (a)
• TOPIC 1.16: Working with the Intermediate Value Theorem (IVT) — part (b)
• TOPIC 6.3: Riemann Sums, Summation Notation, and Definite Integral Notation — part (c)
• TOPIC 3.2: Implicit Differentiation — part (d)

▶️ Answer/Explanation

(a)
Average acceleration on \([2,8]\):
\(\displaystyle \frac{v_A(8)-v_A(2)}{8-2} =\frac{-120-100}{6} =-\frac{220}{6} =\boxed{-\frac{110}{3}\ \text{m/min}^2}.\)

(b)
\(v_A\) is differentiable ⇒ \(v_A\) is continuous on \((5,8)\).
\(v_A(5)=40\) and \(v_A(8)=-120\).
Since \(-100\) lies between \(40\) and \(-120\), the IVT guarantees some \(t\in(5,8)\) with \(v_A(t)=-100\).
\(\boxed{\text{Yes, by IVT on }(5,8).}\)

(c)
Position function: \(\displaystyle s_A(12)=s_A(2)+\int_{2}^{12} v_A(t)\,dt =300+\int_{2}^{12} v_A(t)\,dt.\)
Trapezoidal sum with subintervals \([2,5]\), \([5,8]\), \([8,12]\):
\(\displaystyle \int_{2}^{12}\!v_A(t)\,dt \approx 3\times\frac{100+40}{2}\;+\; 3\times\frac{40+(-120)}{2}\;+\; 4\times\frac{-120+(-150)}{2}\)
\(=3\times 70\;+\;3\times(-40)\;+\;4\times(-135)\)
\(=210-120-540=-450.\)
Therefore \(s_A(12)\approx 300+(-450)=\boxed{-150\ \text{m}}\) (150 m west of the station).

(d)
Let \(x(t)\) be Train A’s east–west position, \(y(t)\) Train B’s north–south position, and \(z(t)\) the distance between them. Then \(z^2=x^2+y^2\).
Differentiate: \(\displaystyle 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt} \Rightarrow \frac{dz}{dt}=\frac{x\,x’+y\,y’}{z}.\)
At \(t=2\): \(x=300\), \(x’=v_A(2)=100\); \(y=400\), \(y’=v_B(2)=-5(2)^2+60(2)+25=125\);
\(z=\sqrt{300^2+400^2}=500.\)
\(\displaystyle \frac{dz}{dt}=\frac{300\times 100+400\times 125}{500} =\frac{80{,}000}{500} =\boxed{160\ \text{m/min}}.\)

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