Chapter 13: Vector-Valued Functions and Motion in Space - Practice Exercises - Page 777: 7

$v(1,0)=-j$ (Clockwise Motion)

$v=\dfrac{dx}{dt}i+\dfrac{dy}{dt}j$ The equation of the circle is $x^2+y^2=1$ $2 x x_t+2y y_t=0$ $\implies \dfrac{dy}{dt}=\dfrac{-x}{y} \dfrac{dx}{dt}$ Since, $\dfrac{dx}{dt}=y$ $\dfrac{dy}{dt}=\dfrac{-x}{y} \times y$ So, $\dfrac{dy}{dt}=-x$ Now, $v(1,0)=\dfrac{dx}{dt}i+\dfrac{dy}{dt}j=0 i -(1) j $ So, $v(1,0)=-j$ (Clockwise Motion)