AP Calculus BC 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled - Exam Style Questions - MCQs - New Syllabus
No-CalcQuestion
What is the perimeter of the region enclosed by the graphs of \( y=3 \) and \( y=x^{2}-2x \)?
(A) \( 4+\displaystyle\int_{-1}^{3}(x^{2}-2x)\,dx \)
(B) \( 4+\displaystyle\int_{-1}^{3}\sqrt{(2x-1)^{2}}\,dx \)
(C) \( 4+\displaystyle\int_{-1}^{3}\sqrt{1+(x^{2}-2x)^{2}}\,dx \)
(D) \( 4+\displaystyle\int_{-1}^{3}\sqrt{1+(2x-2)^{2}}\,dx \)
(B) \( 4+\displaystyle\int_{-1}^{3}\sqrt{(2x-1)^{2}}\,dx \)
(C) \( 4+\displaystyle\int_{-1}^{3}\sqrt{1+(x^{2}-2x)^{2}}\,dx \)
(D) \( 4+\displaystyle\int_{-1}^{3}\sqrt{1+(2x-2)^{2}}\,dx \)
▶️ Answer/Explanation
Detailed solution
Intersections from \( x^{2}-2x=3 \Rightarrow (x-3)(x+1)=0 \Rightarrow x=-1,\,3 \). Horizontal segment on \( y=3 \) has length \( 3-(-1)=4 \). For \( y=x^{2}-2x \), \( y’ = 2x-2 \). Arc length on \([{-}1,3]\) is \[ \int_{-1}^{3}\sqrt{1+(2x-2)^{2}}\,dx. \] Total perimeter \[ = 4 + \int_{-1}^{3}\sqrt{1+(2x-2)^{2}}\,dx. \] ✅ Correct: (D)
No-Calc Question
The function \(y=f(x)\) satisfies the differential equation \(\dfrac{dy}{dx}=2xy\) with \(f(0)=5\). What is the value of \(f(2)\)?
(A) \(\dfrac{5}{21}\)
(B) \(\sqrt{33}\)
(C) \(4+e^{4}\)
(D) \(5e^{4}\)
(B) \(\sqrt{33}\)
(C) \(4+e^{4}\)
(D) \(5e^{4}\)
▶️ Answer/Explanation
Correct answer: (D)
Separate: \(\dfrac{1}{y}dy=2x\,dx\Rightarrow \ln|y|=x^{2}+C\Rightarrow y=C\,e^{x^{2}}.\)
Use \(f(0)=5\Rightarrow C=5\). Then \(f(2)=5e^{4}\).