ab7696…: “Despite a teal bay”

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The curve $\mathcal {C}$ traced out by $\vec {p}(t)$ is shown below. The points associated to $\vec {p}(t)$ for various $t$ values are shown on the curve.

Provide the most accurate response to the following questions.

Which of the following vectors is parallel to $\vec {p}(1)?$

$\vector {1,0}$ $\vector {0,1}$ $\vector {1,1}$ $\vector {1,-1}$ $\vector {-1,1}$ more than one of these.

Which of the following vectors is parallel to $\vec {p}'(1)?$

$\vector {1,0}$ $\vector {0,1}$ $\vector {1,1}$ $\vector {1,-1}$ $\vector {-1,1}$ more than one of these.

Which of the following vectors is orthogonal to $\vec {p}'(2)?$

$\vector {1,0}$ $\vector {0,1}$ $\vector {1,1}$ $\vector {1,-1}$ $\vector {-1,1}$ more than one of these.

Which of the following vectors is parallel to $\vector {y(3),-x(3)}?$

$\vector {1,0}$ $\vector {0,1}$ $\vector {1,1}$ $\vector {1,-1}$ $\vector {-1,1}$ more than one of these.

Make sure you understand the logic behind each response.

Note that $\vec {p}(1)$ will extend from the origin to the point associated to $\vec {p}(1)$ , so any vector with an $x$ -component of $0$ and a nonzero $y$ -component will be parallel to $\vec {p}(1)$ .
Note that $\vec {p}'(1)$ will be a tangent vector to the curve at the point associated to $\vec {p}(1)$ . From the image, the $x$ and $y$ components of such a vector should be approximately equal.
Note that $\vec {p}'(2)$ will be a tangent vector to the curve at the point associated to $\vec {p}(2)$ . From the image, the $x$ -component of such a vector can be nonzero, but the $y$ -component should be $0$ . Thus, a vector orthogonal to this will have nonzero $y$ -component, but its $x$ -component should be $0$ .
Note that $\vec {p}(3)$ will extend from the origin to the point associated to $\vec {p}(3)$ . From the image, it looks like $x(3) \approx -y(3)$ so any vector parallel to $\vector {y(3),-x(3)}$ will have approximately the same $x$ and $y$ components.