46. If a particle moves in the \( xy \)-plane so that at time \( t > 0 \) its position vector is \( \langle \sin(3t - \frac{\pi}{2}),\; 3t^2 \rangle \), then at time \( t = \frac{\pi}{2} \) the velocity vector is






Answer is: option1

\( \langle -3,\; 3\pi \rangle \)

Solution:

The curve is a four-leaved rose.

The area is determined by evaluating

\[ \int_{0}^{2\pi} \frac{1}{2}[r(\theta)]^2 \, d\theta = \int_{0}^{2\pi} \frac{1}{2}[\sin(2\theta)]^2 \, d\theta \]

With a calculator, we determine the value of this definite integral to be approximately \( 1.571 \).

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