Home / AP Calculus BC 9.1 Defining and Differentiating Parametric Equations- Exam Style Questions – MCQs

AP Calculus BC 9.1 Defining and Differentiating Parametric Equations- Exam Style Questions - MCQs - New Syllabus

AP Calculus BC-MCQs and Answers - All Topics

Question

For \(t>0\), a parametric curve in the \(xy\)-plane is defined by \(x(t)=t^2-4t+5\) and \(y(t)=t^2-6t\). For what value of \(t\) does the curve have a vertical tangent?
A. \(2\)
B. \(3\)
C. \(5\)
D. There is no such value of \(t\).
▶️ Answer/Explanation
Detailed solution
Vertical tangent for a parametric curve occurs when \( \dfrac{dx}{dt}=0 \) and \( \dfrac{dy}{dt}\neq 0 \).
\(x'(t)=2t-4,\quad y'(t)=2t-6\).
Set \(x'(t)=0 \Rightarrow 2t-4=0 \Rightarrow t=2\). Then \(y'(2)=4-6=-2\neq 0\).
Therefore the curve has a vertical tangent at \( \boxed{t=2} \).
✅ Correct: A

Question

Let \( f(x)=\sin(x^{2}) \). What are the first three nonzero terms of the Maclaurin series for \( f'(x) \)?
(A) \( -2x^{3}+\dfrac{7}{3}x^{7}-\dfrac{x^{11}}{60} \)
(B) \( 1-\dfrac{x^{4}}{2}+\dfrac{x^{8}}{24} \)
(C) \( 2x-x^{5}+\dfrac{x^{9}}{12} \)
(D) \( 2x+2x^{3}+x^{5} \)
▶️ Answer/Explanation
Detailed solution

\(\sin u = u-\dfrac{u^{3}}{3!}+\dfrac{u^{5}}{5!}-\cdots\). Substitute \(u=x^{2}\): \[ f(x)=x^{2}-\frac{x^{6}}{6}+\frac{x^{10}}{120}-\cdots. \]
Differentiate term-by-term: \[ f'(x)=2x-x^{5}+\frac{x^{9}}{12}-\cdots. \]

✅ Answer: (C)

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