A quantity with both direction and magnitude.

The length of the vector.

A function that returns a vector, often written as $r(t) = <f(t), g(t)>$.

The tangent vector to the curve at a given point, representing the velocity vector.

The point at which the vector originates.

The final point of the vector, represented with an arrowhead.

The horizontal and vertical components that define the vector's direction and magnitude.

‖𝑣‖

r'(t) = <f'(t), g'(t)> where r(t) = <f(t), g(t)>

A vector that represents the location of a point in space relative to an origin.

Step 1: Differentiate the horizontal component: $d/dt (4t^2) = 8t$. Step 2: Differentiate the vertical component: $d/dt (7t^7) = 49t^6$. Step 3: Combine the derivatives: $r'(t) = <8t, 49t^6>$.

Step 1: Find the derivative of the position vector, $r'(t)$, which gives the velocity vector $v(t)$. Step 2: Substitute the given time $t$ into the velocity vector $v(t)$ to find the velocity at that specific time.

Step 1: Find the derivative of the vector-valued function, $r'(t)$. Step 2: Evaluate $r'(t)$ at the value of $t$ corresponding to the given point. This gives the tangent vector at that point.

Step 1: Find the derivatives of both vector-valued functions. Step 2: Evaluate the derivatives at the given point. Step 3: Check if one vector is a scalar multiple of the other. If so, they are parallel.

Step 1: Find the derivatives of both vector-valued functions. Step 2: Evaluate the derivatives at the given point to obtain the tangent vectors. Step 3: Use the dot product formula to find the angle between the tangent vectors: $\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.

Step 1: Find the derivative of the vector-valued function, $r'(t) = <f'(t), g'(t)>$. Step 2: Set the vertical component of the derivative equal to zero, $g'(t) = 0$, and solve for $t$. Step 3: Substitute the values of $t$ back into the original vector-valued function $r(t)$ to find the corresponding points.

Step 1: Find the derivative of the vector-valued function, $r'(t) = <f'(t), g'(t)>$. Step 2: Set the horizontal component of the derivative equal to zero, $f'(t) = 0$, and solve for $t$. Step 3: Substitute the values of $t$ back into the original vector-valued function $r(t)$ to find the corresponding points.

Step 1: Find the derivative of the vector-valued function, $r'(t)$. Step 2: Find the magnitude of the derivative, $|r'(t)|$. Step 3: Integrate the magnitude of the derivative over the interval $[a, b]$: $\int_{a}^{b} |r'(t)| dt$.

Step 1: Find the derivative of the vector-valued function, $r'(t)$. Step 2: Find the magnitude of the derivative, $|r'(t)|$. Step 3: Divide the derivative by its magnitude to obtain the unit tangent vector: $T(t) = \frac{r'(t)}{|r'(t)|}$.

Step 1: Find the first derivative of the position vector, $r'(t)$, which gives the velocity vector $v(t)$. Step 2: Find the derivative of the velocity vector, $v'(t)$, which gives the acceleration vector $a(t)$. Step 3: Substitute the given time $t$ into the acceleration vector $a(t)$ to find the acceleration at that specific time.

‖𝑣‖ = $\sqrt{a^2 + b^2}$ where $v = <a, b>$.

$tan(θ)= \frac{\text{vertical component}}{\text{horizontal component}}$

$r'(t) = <f'(t), g'(t)>$

$v(t) = r'(t)$

$a^2 + b^2 = c^2$

horizontal component = magnitude * cos(θ)

vertical component = magnitude * sin(θ)

$\hat{u} = \frac{\vec{v}}{|\vec{v}|}$

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos(\theta)$

$\theta = cos^{-1}(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|})$