Derivatives of Vector-Valued Functions

Learning Outcomes

Write an expression for the derivative of a vector-valued function

Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.

Definition

A derivative of a vector-valued function ${\bf{r}}\,(t)$ is

r^{'} (t) = lim Δ t \to 0 \frac{r (t + Δ t) - r (t)}{Δ t},

${\bf{r}}'\,(t)=\displaystyle\lim_{\Delta{t}{\to}0}\frac{{\bf{r}}(t+\Delta{t})-{\bf{r}}\,(t)}{\Delta{t}},$

provided the limit exists. If ${\bf{r}}'\,(t)$ exists, then ${\bf{r}}$ is differentiable at $t$ . If ${\bf{r}}'\,(t)$ exists for all $t$ in an open interval $(a,\ b)$ , then ${\bf{r}}$ is differentiable over the interval $(a,\ b)$ . For the function to be differentiable over the closed interval $[a,\ b]$ , the following two limits must exist as well:

r^{'} (a) = lim Δ t \to 0^{+} \frac{r (a + Δ t) - r (a)}{Δ t}

${\bf{r}}'\,(a)=\displaystyle\lim_{\Delta{t}{\to}0^{+}}\frac{{\bf{r}}\,(a+\Delta{t})-{\bf{r}}\,(a)}{\Delta{t}}$ and

r^{'} (b) = lim Δ t \to 0^{-} \frac{r (b + Δ t) - r (b)}{Δ t}

${\bf{r}}'\,(b)=\displaystyle\lim_{\Delta{t}{\to}0^{-}}\frac{{\bf{r}}\,(b+\Delta{t})-{\bf{r}}\,(b)}{\Delta{t}}$

Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.

We now demonstrate taking the derivative of a vector-valued function.

Example: Finding the derivative of a vector-valued function

Use the definition to calculate the derivative of the function

${\bf{r}}\,(t)=(3t+4)\,{\bf{i}}+(t^{2}-4t+3){\bf{j}}$

Show Solution

Let’s use the equation from the above definition:

\begin{matrix} r^{'} (t) & = & lim Δ t \to 0 \frac{r (t + Δ t) - r (t)}{Δ t} = & lim Δ t \to 0 \frac{[(3 (t + Δ t) + 4) i + ({(t + Δ t)}^{2} - 4 (t + Δ t) + 3) j] - [(3 t + 4) i + (t^{2} - 4 t + 3) j]}{Δ t} = & lim Δ t \to 0 \frac{(3 t + 3 Δ t + 4) i - (3 t + 4) i + (t^{2} + 2 t Δ t + (Δ t)^{2} - 4 t - 4 Δ t + 3) j - (t^{2} - 4 t + 3) j}{Δ t} = & lim Δ t \to 0 \frac{(3 Δ t) i + (2 t Δ t + (Δ t)^{2} - 4 Δ t) j}{Δ t} = & lim Δ t \to 0 (3 i + (2 t + Δ t - 4) j = & 3 i + (2 t - 4) j \end{matrix}

$\begin{array}{ccc}\hfill {\bf{r}}'\,(t) & =\hfill & {\displaystyle\lim_{\Delta{t}{\to}0}\frac{{\bf{r}}(t+\Delta{t})-{\bf{r}}(t)}{\Delta{t}}} \hfill \\ \hfill & =\hfill & {\displaystyle\lim_{\Delta{t}{\to}0}\frac{\big[(3(t+\Delta{t})+4){\bf{i}}+\big({(t+\Delta{t})}^{2}-4(t+\Delta{t})+3\big){\bf{j}}\big]-\big[(3t+4){\bf{i}}+(t^{2}-4t+3){\bf{j}}\big]}{\Delta{t}}}\hfill\\\hfill & =\hfill & {\displaystyle\lim_{\Delta{t}{\to}0}\frac{(3t+3\Delta{t}+4){\bf{i}}-(3t+4){\bf{i}}+(t^{2}+2t\Delta{t}+(\Delta{t})^{2}-4t-4\Delta{t}+3){\bf{j}}-(t^{2}-4t+3){\bf{j}}}{\Delta{t}}}\hfill\\\hfill& =\hfill & {\displaystyle\lim_{\Delta{t}{\to}0}\frac{(3\Delta{t}){\bf{i}}+\big(2t\Delta{t}+(\Delta{t})^{2}-4\Delta{t}\big){\bf{j}}}{\Delta{t}}} \hfill \\ \hfill & =\hfill & {\displaystyle\lim_{\Delta{t}{\to}0}{(3\,{\bf{i}}+(2t+{\Delta}t-4)\,{\bf{j}}}} \hfill \\ \hfill & =\hfill & {3\,{\bf{i}}+(2t-4)\,{\bf{j}}}\hfill \\ \hfill \end{array}$

TRY IT

Use the definition to calculate the derivative of the function ${\bf{r}}\,(t)=(2t^{2}+3)\,{\bf{i}}+(5t-6)\,{\bf{j}}.$

Show Solution

${\bf{r}}'\,(t)=4t\,{\bf{i}}+5\,{\bf{j}}$

Notice that in the calculations in the example above, we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.

Differentiation of Vector-Valued functions theorem

Let $f$ , $g$ , and $h$ be differentiable functions of $t$ .

If ${\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}$ , then ${\bf{r}}'\,(t)=f'\,(t)\,{\bf{i}}+g'\,(t)\,{\bf{j}}$ .
If ${\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}$ , then ${\bf{r}}'\,(t)=f'\,(t)\,{\bf{i}}+g'\,(t)\,{\bf{j}}+h'\,(t)\,{\bf{k}}$ .

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.

Recall: Derivatives of common functions

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

Recall: Derivative Rules

Constant Multiple Rule: $\frac{d}{dx} (cf(x)) = cf'(x)$

Sum and Difference Rule: $\frac{d}{dx} f(x) \pm g(x)) = f'(x) \pm g'(x)$

Product Rule: $\frac{d}{dx} (f(x)g(x)) = g(x)f'(x) + f(x)g'(x)$

Quotient Rule: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

Chain Rule: $\frac{d}{dx} f(g(x) = f'(g(x)) g'(x)$

Example: calculating the derivative of vector-valued functions

Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.

${\bf{r}}\,(t)=(6t+8)\,{\bf{i}}+(4t^{2}+2t-3){\bf{j}}$
${\bf{r}}\,(t)=3\cos{t\,{\bf{i}}}+4\sin{t\,{\bf{j}}}$
${\bf{r}}\,(t)=e^{t}\sin{t\,{\bf{i}}}+e^{t}\cos{t{\bf{j}}}-e^{2t}\,{\bf{k}}$

Show Solution

We use Differentiation of Vector-Valued Functions and what we know about differentiating functions of one variable.

The first component of ${\bf{r}}\,(t)=(6t+8)\,{\bf{i}}+(4t^{2}+2t-3)\,{\bf{j}}$ is $f\,(t)=6t+8$ . The second component is $g\,(t)=4t^{2}+2t-3$ . We have $f'\,(t)=6$ and $g'\,(t)=8t+2$ , so the theorem gives ${\bf{r}}'\,(t)=6\,{\bf{i}}+(8t+2)\,{\bf{j}}$
The first component is $f\,(t)=3\cos{t}$ and the second component is $g\,(t)=4\sin{t}$ . We have $f'\,(t)=-3\sin{t}$ and $g'\,(t)=4\cos{t}$ , so we obtain ${\bf{r}}'\,(t)=-3\sin{t\,{\bf{i}}}+4\cos{t\,{\bf{j}}}$
The first component of ${\bf{r}}\,(t)=e^{t}\sin{t\,{\bf{i}}}+e^{t}\cos{t{\bf{j}}}-e^{2t}\,{\bf{k}}$ is $f\,(t)=e^{t}\sin{t}$ , the second component is $g\,(t)=e^{t}\cos{t}$ , and the third component is $h\,(t)=-e^{2t}$ . We have $f'\,(t)=e^{t}(\sin{t}+\cos{t}),\ g'\,(t)=e^{t}(\cos{t}-\sin{t})$ , and $h'\,(t)=-2e^{2t}$ , so the theorem gives ${\bf{r}}'\,(t)=e^{t}(\sin{t}+\cos{t})\,{\bf{i}}+e^{t}(\cos{t}-\sin{t})\,{\bf{j}}-2e^{2t}\,{\bf{k}}$

TRY IT

Calculate the derivative of the function

${\bf{r}}\,(t)=(t\ln{t})\,{\bf{i}}+(5e^{t})\,{\bf{j}}+(\cos{t}-\sin{t})\,{\bf{k}}$

Show Solution

${\bf{r}}'\,(t)=(1+\ln{t})\,{\bf{i}}+5e^{t}\,{\bf{j}}-(\sin{t}+\cos{t})\,{\bf{k}}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.5” here (opens in new window).

We can extend to vector-valued functions the properties of the derivative. In particular, the constant multiple rule, the sum and difference rules, the product rule, and the chain rule all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dot product of two vector-valued functions, and (3) for the cross product of two vector-valued functions.

Differentiation of Vector-Valued functions theorem

Let $\bf{r}$ and $\bf{u}$ be differentiable vector-valued functions of $t$ , let $f$ be a differentiable real-valued function of

$\begin{alignat}{2} \hspace{5cm}\text{1.}& &\quad \frac{d}{dt}\big[c{\bf{r}}\,(t)\big]&=c{\bf{r}}'\,(t) &\quad &\text{Scalar multiple}\\ \text{2.}& &\quad \frac{d}{dt}\big[{\bf{r}}\,(t)\pm{\bf{u}}\,(t)\big]&={\bf{r}}'\,(t)\pm{\bf{u}}'\,(t) &\quad &\text{Sum and difference}\\ \text{3.}& &\quad \frac{d}{dt}\big[f\,(t)\,{\bf{u}}\,(t)\big]&=f'\,(t)\,{\bf{u}}\,(t)+f\,(t)\,{\bf{u}}'\,(t) &\quad &\text{Scalar product}\\ \text{4.}& &\quad \frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big]&={\bf{r}}'\,(t)\cdot{\bf{u}}\,(t)+{\bf{r}}\,(t)\cdot{\bf{u}}'\,(t) &\quad &\text{Dot product}\\ \text{5.}& &\quad \frac{d}{dt}\big[{\bf{r}}\,(t)\times{\bf{u}}\,(t)\big] &={\bf{r}}'\,(t)\times{\bf{u}}\,(t)+{\bf{r}}\,(t)\times{\bf{u}}'\,(t) &\quad &\text{Cross product}\\ \text{6.}& &\quad \frac{d}{dt}\big[{\bf{r}}(f\,(t))\big]&={\bf{r}}'\,(f\,(t))\cdot{f}'\,(t) &\quad &\text{Chain rule}\\ \text{7.}& &\quad {\bf{r}}\,(t)\cdot{\bf{r}}\,(t) &=c, \text{ then }{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)=0. \end{alignat}$

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 ${\bf{u}}\,(t)=g\,(t)\,{\bf{i}}+h\,(t)\,{\bf{j}}$ . Then

\begin{matrix} \frac{d}{d t} [f (t) u (t)] & = & \frac{d}{d t} [f (t) (g (t) i + h (t) j)] = & \frac{d}{d t} [f (t) g (t) i + f (t) h (t) j] = & \frac{d}{d t} [f (t) g (t)] i + \frac{d}{d t} [f (t) h (t)] j = & (f^{'} (t) g (t) + f (t) g^{'} (t)) i + (f^{'} (t) h (t) + f (t) h^{'} (t)) j = & f^{'} (t) u (t) + f (t) u^{'} (t) \end{matrix}

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

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

\begin{matrix} \frac{d}{d t} [r (t) \cdot u (t)] & = & \frac{d}{d t} [f_{1} (t) f_{2} (t) + g_{1} (t) g_{2} (t)] = & f_{1}^{'} (t) f_{2} (t) + f_{1} (t) f_{2}^{'} (t) + g_{1}^{'} (t) g_{2} (t) + g_{1} (t) g_{2}^{'} (t) = & f_{1}^{'} (t) f_{2} (t) + g_{1}^{'} (t) g_{2} (t) + f_{1} (t) f_{2}^{'} (t) + g_{1} (t) g_{2}^{'} (t) = & (f_{1}^{'} i + g_{1}^{'} j) \cdot (f_{2} i + g_{2} j) + (f_{1} i + g_{1} j) \cdot (f_{2}^{'} i + g_{2}^{'} j) = & r^{'} (t) \cdot u (t) + r (t) \cdot u^{'} (t) \end{matrix}

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

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:

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

$_\blacksquare$

Now for some examples using these properties.

Example: Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

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

and

${\bf{u}}\,(t)=(t^{2}-3)\,{\bf{i}}+(2t+4)\,{\bf{j}}+(t^{3}-3t)\,{\bf{k}}$ ,

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

$\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big]$
$\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{u}}'\,(t)\big]$

Show Solution

We have ${\bf{r}}'\,(t)=6\,{\bf{i}}+(8t+2)\,{\bf{j}}+5\,{\bf{k}}$ and ${\bf{u}}'\,(t)=2t\,{\bf{i}}+2\,{\bf{j}}+(3t^{2}-3)\,{\bf{k}}$ . Therefore, according to property 4:
$\begin{array}{ccc}\hfill {\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big]} & =\hfill & {{\bf{r}}'\,(t)\cdot{\bf{u}}\,(t)+{\bf{r}}\cdot{\bf{u}}'\,(t)}\hfill \\ \hfill & =\hfill & {(6\,{\bf{i}}+(8t+2)\,{\bf{j}}+5\,{\bf{k}})\cdot\big((t^{2}-3)\,{\bf{i}}+(2t+4)\,{\bf{j}}+(t^{3}-3t)\,{\bf{k}}\big) \\ +\big((6t+8)\,{\bf{i}}+(4t^{2}+2t-3)\,{\bf{j}}+5t\,{\bf{k}}\big)\cdot\big(2t\,{\bf{i}}+2\,{\bf{j}}+(3t^{2}-3)\,{\bf{k}}\big)}\hfill\\\hfill & =\hfill & {6(t^{2}-3)+(8t+2)(2t+4)+5(t^{3}-3t)+2t(6t+8)+2(4t^{2}+2t-3)+5t(3t^{2}-3)}\hfill\\\hfill& =\hfill & {20t^{3}+42t^{2}+26t-16}\hfill \\ \hfill \end{array}$
First, we need to adapt property 5 for this problem:
$\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{u}}'\,(t)\big]={\bf{u}}'\,(t)\times{\bf{u}}'\,(t)+{\bf{u}}\,(t)\times{\bf{u}}''\,(t).$

Recall that the cross product of any vector with itself is zero. Furthermore, ${\bf{u}}''\,(t)$ represents the second derivative of ${\bf{u}}\,(t)$ :

${\bf{u}}''\,(t)=\frac{d}{dt}[{\bf{u}}'\,(t)]=\frac{d}{dt}[2t\,{\bf{i}}+2\,{\bf{j}}+(3t^{2}-3)\,{\bf{k}}]=2\,{\bf{i}}+6t\,{\bf{k}}.$

Therefore,

$\begin{array}{ccc}\hfill {\frac{d}{dt}\big[{\bf{u}}\,(t)\cdot{\bf{u}}'\,(t)\big]} & =\hfill & {0+\big((t^{2}-3)\,{\bf{I}}+(2t+4)\,{\bf{j}}+(t^{3}-3t)\,{\bf{k}}\big)\times(2\,{\bf{I}}+6t\,{\bf{k}})}\hfill \\ \hfill & =\hfill & {\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}}\\t^{2}-3&2t+4&t^{3}-3t\\2&0&6t\end{vmatrix}}\hfill\\\hfill & =\hfill & {6t\,(2t+4)\,{\bf{i}}-\big(6t\,(t^{2}-3)-2(t^{3}-3t)\big)\,{\bf{j}}-2\,(2t-4)\,{\bf{k}}}\hfill\\\hfill& =\hfill & {(12t^{2}+24t)\,{\bf{i}}+(12t-4t^{3})\,{\bf{j}}-(4t+8)\,{\bf{k}}}\hfill \\ \hfill \end{array}$

TRY IT

Given the vector-valued functions ${\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}-e^{2t}\,{\bf{k}}$ and ${\bf{u}}\,(t)=t\,{\bf{i}}+\sin{t\,{\bf{j}}}+\cos{t\,{\bf{k}}}$ , calculate $\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)\big]$ and $\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{r}}\,(t)\big]$ .

Show Solution

$\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)\big]=8e^{4t}$

$\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{r}}\,(t)\big]=-\big(e^{2t}(\cos{t}+2\sin{t})+\cos{2t}\big)\,{\bf{i}}+\big(e^{2t}(2t+1)-\sin2t\big)\,{\bf{j}}+\big(t\cos{t}+\sin{t}-\cos{2t}\big)\,{\bf{k}}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.6” here (opens in new window).

Module 3: Vector-Valued Functions