Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only)

(5\pi, 1/2)

(-5, $\pi/2$)

(5, $\pi/4$)

(5, $\pi/2$)

Inward circular spiral decreasing in radius towards origin.

Linear trajectory moving diagonally outward from origin.

Circular motion with constant radius equal to initial value at time zero.

Outward spiral movement increasing in radius over time.

($\pi, \pi$)

($1, 0$)

($0, \pi$)

($0, 0$)

$\langle 2, 9 \rangle$

$\langle 4, 32 \rangle$

$\langle 5, 20 \rangle$

$\langle 3, 12 \rangle$

Scalar projection

y-component

Direction

Magnitude

(-4, $\pi$)

(4, $-\pi$)

(-4, $-\frac{\pi}{2}$)

(-4, $\frac{\pi}{2}$)

$t = -1$

No such value exists

$t = 0$ or $t = 1$

$t = \frac{1}{3}$

$\left( -\frac{b}{a}, \frac{c-b}{a} \right)$

$\left( \frac{b}{a}, \frac{c-b}{a} \right)$

$\left( -\frac{b}{a}, \frac{c-b}{-a} \right)$

$\left( -\frac{b}{a}, \frac{c+b}{a} \right)$

Three Instantaneous Speed calculated by finding antiderivative $S(T)$.

Instantaneous Speed equals derivative $s'(t)$.

Two Instantaneous Speed equals average rate of change between two points on $s(t)$.

One Instantaneous Speed equals integral under $v(t) , dt$.

Switching to $\ln(\sqrt{t+1})$ would moderately increase acceleration because of slightly increased curvature.

Adjusting it to $(\ln(t+1))^2$ minimally affects acceleration as it does not change directionality.

Modifying it to $\ln((t+1)^2)$ would lead to an increase in acceleration because of compound effects on growth rate.

Changing it to $\ln(5(t+1))$ significantly increases acceleration due to a steeper slope at that point.