9.4 Defining and Differentiating Vector-Valued Functions

• What is a vector-valued function?

  Function mapping scalar to vector
• A vector-valued function can be represented as $\vec{r}(t) =$$< x(t), y(t) >$, where $x(t)$ and $y(t)$ are the component functions.
• The vector-valued function $\vec{r}(t) =$$< t^{2}, 2t >$ has component functions $x(t) =$$t^{2}$ and $y(t) =$$2t$.
• How are vector-valued functions expressed in terms of component functions?

  Inside angle brackets
• The component functions of a vector-valued function describe the x and $y$ coordinates as functions of t</latex>.
• Match the term with its definition:

  Vector-valued function ↔️ Function that maps a scalar $t$ to a vector $\vec{r}(t)$

  Component functions ↔️ Scalar functions that define $x$ and $y$ coordinates of $\vec{r}(t)$

  Parametric equations ↔️ Equations expressing $x$ and $y$ in terms of $t$
• Vector-valued functions use component functions to represent parametric curves.
• What is the process for differentiating a vector-valued function \vec{r}(t) = < x(t), y(t) ></latex>?

  Differentiate each component
• Steps to differentiate a vector-valued function $\vec{r}(t) =$$< x(t), y(t) >$


  1️⃣ Find the derivative of $x(t)$

  2️⃣ Find the derivative of $y(t)$

  3️⃣ Combine the derivatives into a new vector-valued function
• The derivative of $\vec{r}(t) =$$< x(t), y(t) >$ is $\frac{d}{dt}\vec{r}(t) =$$< x'(t), y'(t) >$, where $x'(t)$ and $y'(t)$ are the derivatives of the component functions.
• What is the formula for the derivative vector $\vec{r}'(t)$ of a vector-valued function $\vec{r}(t) =$$< x(t), y(t) >$?


  $\vec{r}'(t) =$$< x'(t), y'(t) >$
• The derivative vector of \vec{r}(t) = < t^{3}, 2t^{2} ></latex> is $\vec{r}'(t) =$$< 3t^{2}, 4t >$.
• How do you combine the derivatives of the components of a vector-valued function into a new vector $\vec{r}'(t)$?


  $\vec{r}'(t) =$$< x'(t), y'(t) >$
• To find the derivative vector of a vector-valued function, you differentiate each component
• Steps to find the derivative vector of a vector-valued function $\vec{r}(t) =$$< x(t), y(t) >$


  1️⃣ Differentiate the x-component: $x'(t) =$$\frac{d}{dt}(x(t))$

  2️⃣ Differentiate the y-component: $y'(t) =$$\frac{d}{dt}(y(t))$

  3️⃣ Combine derivatives into a new vector: $\vec{r}'(t) =$$< x'(t), y'(t) >$
• What is the derivative of the y-component of $\vec{r}(t) =$$< t^{3}, 2t^{2} >$?


  $4t$
• The derivative vector of $\vec{r}(t) =$$< t^{3}, 2t^{2} >$ is $\vec{r}'(t) =$$< 3t^{2}, 4t >$
• What is a vector-valued function?

  A function that returns a vector
• Vector-valued functions are similar to parametric equations.
• What are component functions in a vector-valued function?

  Scalar functions defining vector coordinates
• Vector-valued functions use component functions to represent parametric curves
• Steps to differentiate a vector-valued function $\vec{r}(t) =$$< x(t), y(t) >$


  1️⃣ Find the derivative of the first component function: $x'(t) =$$\frac{dx}{dt}$

  2️⃣ Find the derivative of the second component function: $y'(t) =$$\frac{dy}{dt}$

  3️⃣ Combine the derivatives into a new vector-valued function: $\frac{d}{dt}\vec{r}(t) =$$< x'(t), y'(t) >$
• What is the derivative of the first component of $\vec{r}(t) =$$< t^{2}, 2t >$?


  $2t$
• The derivative vector of a vector-valued function represents the direction of the tangent line at any point.
• Steps to find the tangent vector to a vector-valued function \vec{r}(t)</latex>

  1️⃣ Find the derivative $x'(t)$ and $y'(t)$ for the component functions $x(t)$ and $y(t)$.

  2️⃣ Construct the derivative vector $\vec{r}'(t) =$$< x'(t), y'(t) >$.

  3️⃣ The derivative vector $\vec{r}'(t)$ is the tangent vector at $t$.
• What is the tangent vector to $\vec{r}(t) =$$< t^{2}, 2t >$?


  $\vec{r}'(t) =$$< 2t, 2 >$
• How is the velocity vector calculated from a position vector \vec{r}(t)</latex>?

  First derivative
• The velocity vector $\vec{v}(t)$ is given by $\vec{v}(t) =$$< x'(t), y'(t) >$, where $x'(t) =$$\frac{dx}{dt}$ and $y'(t) =$$\frac{dy}{dt}$, representing the rates of change of the components
• How is the acceleration vector calculated from the velocity vector $\vec{v}(t)$?


  First derivative
• The acceleration vector $\vec{a}(t)$ is given by $\vec{a}(t) =$$< x''(t), y''(t) >$, where $x''(t) =$$\frac{d^{2}x}{dt^{2}}$ and $y''(t) =$$\frac{d^{2}y}{dt^{2}}$, representing the rates of change of the velocity vector components
• What is the velocity vector for $\vec{r}(t) =$$< t^{2}, 2t^{3} >$?


  $\vec{v}(t) =$$< 2t, 6t^{2} >$
• What is the acceleration vector for \vec{r}(t) = < t^{2},2t^{3} ></latex>?

  $\vec{a}(t) =$$< 2, 12t >$
• Match the property of differentiable vector-valued functions with its formula:

  Linearity ↔️ $\frac{d}{dt} (a\vec{r}(t) +$$b\vec{s}(t)) =$$a\vec{r}'(t) +$$b\vec{s}'(t)$

  Chain Rule ↔️ $\frac{d}{dt} \vec{r}(f(t)) =$$\vec{r}'(f(t)) \cdot f'(t)$

  Dot Product Rule ↔️ $\frac{d}{dt} (\vec{r}(t) \cdot \vec{s}(t)) =$$\vec{r}'(t) \cdot \vec{s}(t) +$$\vec{r}(t) \cdot \vec{s}'(t)$

  Cross Product Rule ↔️ $\frac{d}{dt} (\vec{r}(t) \times \vec{s}(t)) =$$\vec{r}'(t) \times \vec{s}(t) +$$\vec{r}(t) \times \vec{s}'(t)$
• In the linearity property, scalars can be factored out of the derivative.
• The chain rule for vector-valued functions involves the derivative of the composite function multiplied by the derivative of the inner
• What is the derivative of the dot product of \vec{r}(t) = < t^{2}, t ></latex> and $\vec{s}(t) =$$< 2t, t^{3} >$?


  $6t^{2} +$$4t^{3}$