A student evaluated for the curve and obtained a value of

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

A student evaluated ∫02πy(t) x′(t) dt\displaystyle\int_{0}^{2\pi} y(t)\,x'(t)\,dt for the curve x=cos⁡t,  y=sin⁡tx=\cos t,\;y=\sin t and obtained a value of −π-\pi. Which of the following best explains why the integral is negative? (a) Because x′(t)=−sin⁡tx'(t)=-\sin t is negative for every tt in [0,2π][0,2\pi], the product y(t)x′(t)y(t)x'(t) stays negative, so the integral is negative. (b) Since y(t)y(t) is below the xx–axis for half of the interval, the negative contribution from that lower semicircle outweighs the positive contribution from the upper semicircle, making the integral negative. (c) The parameterization traverses the circle clockwise, so the signed area is negative even though the magnitude of the enclosed area is π\pi. (d) Any time a full 2π2\pi interval is used for a circular parameterization, ∫y dx\int y\,dx is automatically negative, no matter where the interval starts.

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