Calculus of Vector-Valued Functions: Learn It 1

  • Find derivatives of vector-valued functions
  • Find the tangent vector at a point for a given position vector
  • Find and interpret unit tangent vectors
  • Calculate the definite integral of a vector-valued function

Derivatives of Vector-Valued Functions

To study the calculus of vector-valued functions, we follow a path similar to real-valued functions: define the derivative, examine applications, then move on to integrals. The vector nature of these functions brings some interesting new ideas along the way.

Now that you know what a vector-valued function is and how to take its limit, the next step is learning to differentiate it. The definition closely parallels the derivative of a real-valued function, but because the range consists of vectors, the derivative is also a vector.

derivative of a vector-valued function

The derivative of a vector-valued function [latex]{\bf{r}}(t)[/latex] is

[latex]{\bf{r}}'(t)=\underset{\Delta t\to 0}{\text{lim}}\frac{{\bf{r}}(t+\Delta t)-{\bf{r}}(t)}{\Delta t}[/latex]

provided the limit exists. If [latex]{\bf{r}}'(t)[/latex] exists, then [latex]{\bf{r}}[/latex] is differentiable at [latex]t[/latex].

[latex]\\[/latex]

For [latex]{\bf{r}}[/latex] to be differentiable over a closed interval [latex][a, b][/latex], these one-sided limits must also exist:

[latex]{\bf{r}}'(a)=\underset{\Delta t\to 0^{+}}{\text{lim}}\frac{{\bf{r}}(a+\Delta t)-{\bf{r}}(a)}{\Delta t}[/latex] and [latex]{\bf{r}}'(b)=\underset{\Delta t\to 0^{-}}{\text{lim}}\frac{{\bf{r}}(b+\Delta t)-{\bf{r}}(b)}{\Delta t}[/latex]

Just as with real-valued functions, the derivative of a vector-valued function represents an instantaneous rate of change. When the function represents the position of an object at time [latex]t[/latex], the derivative represents its velocity at that same moment.

Use the definition to calculate the derivative of the function

[latex]{\bf{r}}\,(t)=(3t+4)\,{\bf{i}}+(t^{2}-4t+3){\bf{j}}[/latex]

Calculating derivatives using the limit definition works, but it’s tedious. Fortunately, we can differentiate component by component—just like we handled limits.

differentiation of vector-valued functions theorem

Let [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex] be differentiable functions of [latex]t[/latex].

  • If [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}[/latex], then [latex]{\bf{r}}'\,(t)=f'\,(t)\,{\bf{i}}+g'\,(t)\,{\bf{j}}[/latex].
  • If [latex]{\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}[/latex], then [latex]{\bf{r}}'\,(t)=f'\,(t)\,{\bf{i}}+g'\,(t)\,{\bf{j}}+h'\,(t)\,{\bf{k}}[/latex].
Differentiate each component separately using the derivative rules you already know from Calculus 1. The vector structure doesn’t change—only the components do.

Since we will be using derivatives of common single-variable functions throughout this course, we briefly recall the derivatives of common functions, along with frequently-encountered derivative rules below.

Derivatives of Common Functions

  • [latex]\frac{d}{dx} (x^n) = nx^{n-1}[/latex] (Power Rule)
  • [latex]\frac{d}{dx} (\sin x) = \cos x[/latex]
  • [latex]\frac{d}{dx} (\cos x) = -\sin x[/latex]
  • [latex]\frac{d}{dx} (\tan x) = \sec^2 x[/latex]
  • [latex]\frac{d}{dx} (\sec x) = \sec x \tan x[/latex]
  • [latex]\frac{d}{dx} (\csc x) = -\csc x \cot x[/latex]
  • [latex]\frac{d}{dx} (\cot x) = -\csc^2 x[/latex]
  • [latex]\frac{d}{dx} (e^x) = e^x[/latex]
  • [latex]\frac{d}{dx} (\ln x) = \frac{1}{x}[/latex]
  • [latex]\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}[/latex]

Derivative Rules

  • Constant Multiple: [latex]\frac{d}{dx} (cf(x)) = cf'(x)[/latex]
  • Sum and Difference: [latex]\frac{d}{dx} (f(x) \pm g(x)) = f'(x) \pm g'(x)[/latex]
  • Product Rule: [latex]\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x)[/latex]
  • Quotient Rule: [latex]\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}[/latex]
  • Chain Rule: [latex]\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)[/latex]
Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.

  1. [latex]{\bf{r}}\,(t)=(6t+8)\,{\bf{i}}+(4t^{2}+2t-3){\bf{j}}[/latex]
  2. [latex]{\bf{r}}\,(t)=3\cos{t\,{\bf{i}}}+4\sin{t\,{\bf{j}}}[/latex]
  3. [latex]{\bf{r}}\,(t)=e^{t}\sin{t\,{\bf{i}}}+e^{t}\cos{t{\bf{j}}}-e^{2t}\,{\bf{k}}[/latex]

Properties of Derivatives

The derivative properties you learned in Calculus 1 extend to vector-valued functions. The constant multiple rule, sum and difference rules, and chain rule all work the same way. The product rule has three versions because vectors can be multiplied in three different ways: scalar multiplication, dot product, and cross product.

differentiation rules for vector-valued functions

Let [latex]{\bf{r}}[/latex] and [latex]{\bf{u}}[/latex] be differentiable vector-valued functions of [latex]t[/latex], and let [latex]f[/latex] be a differentiable real-valued function of [latex]t[/latex]. Let [latex]c[/latex] be a scalar.

  1. Scalar multiple: [latex]\frac{d}{dt}\big[c{\bf{r}},(t)\big]=c{\bf{r}}',(t)[/latex]
  2. Sum and difference: [latex]\frac{d}{dt}\big[{\bf{r}},(t)\pm{\bf{u}},(t)\big]={\bf{r}}',(t)\pm{\bf{u}}',(t)[/latex]
  3. Scalar product: [latex]\frac{d}{dt}\big[f,(t),{\bf{u}},(t)\big]=f',(t),{\bf{u}},(t)+f,(t),{\bf{u}}',(t)[/latex]
  4. Dot product: [latex]\frac{d}{dt}\big[{\bf{r}},(t)\cdot{\bf{u}},(t)\big]={\bf{r}}',(t)\cdot{\bf{u}},(t)+{\bf{r}},(t)\cdot{\bf{u}}',(t)[/latex]
  5. Cross product: [latex]\frac{d}{dt}\big[{\bf{r}},(t)\times{\bf{u}},(t)\big]={\bf{r}}',(t)\times{\bf{u}},(t)+{\bf{r}},(t)\times{\bf{u}}',(t)[/latex]
  6. Chain rule: [latex]\frac{d}{dt}\big[{\bf{r}}(f,(t))\big]={\bf{r}}',(f,(t))\cdot{f}',(t)[/latex]
  7. If [latex]{\bf{r}},(t)\cdot{\bf{r}},(t)=c[/latex], then [latex]{\bf{r}},(t)\cdot{\bf{r}}',(t)=0[/latex]

Proof

The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule from the Introduction to Derivatives. Let [latex]{\bf{u}}\,(t)=g\,(t)\,{\bf{i}}+h\,(t)\,{\bf{j}}[/latex]. Then

[latex]\begin{array}{ccc}\hfill {\frac{d}{dt}\big[f\,(t)\,{\bf{u}}\,(t)\big]} & =\hfill & {\frac{d}{dt}\big[f\,(t)(g\,(t)\,{\bf{i}}+h\,(t)\,{\bf{j}})\big]}\hfill \\ \hfill & =\hfill & {\frac{d}{dt}\big[f\,(t)\,g\,(t)\,{\bf{i}}+f\,(t)\,h\,(t)\,{\bf{j}}\big]}\hfill\\\hfill & =\hfill & {\frac{d}{dt}\big[f\,(t)\,g\,(t)\big]\,{\bf{i}}+\frac{d}{dt}\big[f\,(t)\,h\,(t)\big]\,{\bf{j}}}\hfill\\\hfill& =\hfill & {\big(f'\,(t)\,g\,(t)+f\,(t)\,g'\,(t)\big)\,{\bf{i}}+\big(f'\,(t)\,h\,(t)+f\,(t)\,h'\,(t)\big)\,{\bf{j}}} \hfill \\ \hfill & =\hfill & {f'\,(t)\,{\bf{u}}\,(t)+f\,(t)\,{\bf{u}}'\,(t)}\hfill \\ \hfill \end{array}[/latex].

 

To prove property 4. Let [latex]{\bf{r}}\,(t)=f_{1}\,(t)\,{\bf{i}}+g_{1}\,(t)\,{\bf{j}}[/latex] and [latex]{\bf{u}}\,(t)=f_{2}\,(t)\,{\bf{i}}+g_{2}\,(t)\,{\bf{j}}[/latex]. Then

[latex]\begin{array}{ccc}\hfill {\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big]} & =\hfill & {\frac{d}{dt}\big[f_{1}\,(t)\,f_{2}\,(t)+g_{1}\,(t)\,g_{2}\,(t)\big]}\hfill \\ \hfill & =\hfill & {f_{1}'\,(t)f_{2}\,(t)+f_{1}\,(t)\,f_{2}'\,(t)+g_{1}'\,(t)\,g_{2}\,(t)+g_{1}\,(t)\,g_{2}'\,(t)}\hfill\\\hfill & =\hfill & {f_{1}'\,(t)f_{2}\,(t)+g_{1}'\,(t)\,g_{2}\,(t)+f_{1}\,(t)\,f_{2}'\,(t)+g_{1}\,(t)\,g_{2}'\,(t)}\hfill\\\hfill& =\hfill & {\big(f_{1}'{\bf{i}}+g_{1}'{\bf{j}}\big)\cdot(f_{2}{\bf{i}}+g_{2}{\bf{j}})+(f_{1}{\bf{i}}+g_{1}{\bf{j}})\cdot(f_{2}'{\bf{i}}+g_{2}'{\bf{j}})} \hfill \\ \hfill & =\hfill & {{\bf{r}}'\,(t)\cdot{\bf{u}}\,(t)+{\bf{r}}\,(t)\cdot{\bf{u}}'\,(t)}\hfill \\ \hfill \end{array}[/latex].

 

The proof of property 5 is similar to that of property 4. Property 6 can be proved using the chain rule. Last, property 7 follows from property 4:

[latex]\begin{array}{ccc}\hfill {\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{r}}\,(t)\big]} & =\hfill & {\frac{d}{dt}[c]}\hfill \\ \hfill {{\bf{r}}'\,(t)\cdot{\bf{r}}\,(t)+{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)} & =\hfill & {0} \hfill\\\hfill {2{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)} & =\hfill & {0} \hfill\\\hfill {{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)} & =\hfill & {0}.\hfill \\ \hfill \end{array}[/latex]

[latex]_\blacksquare[/latex]

Given the vector-valued functions

[latex]{\bf{r}}\,(t)=(6t+8)\,{\bf{i}}+(4t^{2}+2t-3)\,{\bf{j}}+5t\,{\bf{k}}[/latex]

and

[latex]{\bf{u}}\,(t)=(t^{2}-3)\,{\bf{i}}+(2t+4)\,{\bf{j}}+(t^{3}-3t)\,{\bf{k}}[/latex],

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

  1. [latex]\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big][/latex]
  2. [latex]\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{u}}'\,(t)\big][/latex]