# 4.1: Differentiation and Integration of Vector Valued Functions

- Page ID
- 564

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The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.

Definition: The Derivative of a Vector Valued Function

Let \(r(t)\) be a vector valued function, then

\[ r'(t) = \lim_{h \rightarrow 0} \dfrac{r(t+h)-r(t)}{h}.\]

Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.

Example \(\PageIndex{1}\)

\[ \dfrac{d}{dt} (3 \hat{\text{i}} + \sin t \hat{\text{j}}) = \cos t \hat{\text{j}}\]

\[ \dfrac{d}{dt} \left(3t^2\, \hat{\text{i}} + \cos{(4t)}\, \hat{\text{j}} + te^t \, \hat{\text{k}} \right) = 6t \, \hat{\text{i}} -4\sin{(t)}\,\hat{\text{j}} + (e^t + te^t)\, \hat{\text{k}}\]

## Properties of Vector Valued Functions

All of the properties of differentiation still hold for vector values functions. Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.

Suppose that \(\text{v}(t)\) and \(\text{w}(t)\) are vector valued functions, \(f(t)\) is a scalar function, and \(c\) is a real number then

- \(\dfrac{d}{dt} \left( \text{v}(t) + \text{w}(t) \right) = \dfrac{d}{dt}\text{v}(t) + \dfrac{d}{dt} \text{w}(t)\),
- \(\dfrac{d}{dt} c\text{v}(t) = c\, \dfrac{d}{dt} \text{v}(t) \),
- \(\dfrac{d}{dt}(f(t) \text{v}(t)) = f '(t) \text{v}(t) + f(t) \text{v}(t)'\),
- \( \left( v(t) \cdot \text{w}(t) \right)' = \text{v}'(t) \cdot \text{w}(t)+ \text{v}(t) \cdot \text{w}'(t)\),
- \((v(t) \times \text{w}(t))' = \text{v}'(t) \times \text{w}(t) + \text{v}(t) \times \text{w}'(t)\),
- \(\dfrac{d}{dt} v(f(t)) = \text{v}(t)'(f(t)) f '(t)\).

Example \(\PageIndex{2}\)

Show that if \(r\) is a differentiable vector valued function with constant magnitude, then

\[ r \cdot r' = 0.\]

**Solution**

Since \(r\) has constant magnitude, call its magnitude \(k\),

\[ k^2 = |r|^2 = r \cdot r.\]

Taking derivatives of the left and right sides gives

\[ 0 = (r \cdot r)' = r' \cdot r + r \cdot r' \]

\[ = r \cdot r' + r \cdot r' = 2r \cdot r' . \]

Divide by two and the result follows

## Integration of vector valued functions

We define the *integral of a vector valued function *as the integral of each component. This definition holds for both definite and indefinite integrals.

Example \(\PageIndex{3}\)

Evaluate

\[ \int (\sin t)\, \hat{\textbf{i}} + 2t\, \hat{\textbf{j}} - 8t^3 \, \hat{\textbf{k}} \; dt. \]

**Solution**

Just take the integral of each component

\[ \int (\sin t)\,dt \, \hat{\textbf{i}} + \int 2\,t \, dt \, \hat{\textbf{j}} - \int 8\,t^3 \,dt \, \hat{\textbf{k}}. \]

\[ = (-\cos t + c_1)\, \hat{\textbf{i}} + (t^2 + c_2)\, \hat{\textbf{j}} + (2\,t^4 + c_3)\, \hat{\textbf{k}}.\]

Notice that we have introduce three different constants, one for each component.

Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.