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 [latex]{\bf{r}}\,(t)[/latex] is
provided the limit exists. If [latex]{\bf{r}}'\,(t)[/latex] exists, then [latex]{\bf{r}}[/latex] is differentiable at [latex]t[/latex]. If [latex]{\bf{r}}'\,(t)[/latex] exists for all [latex]t[/latex] in an open interval [latex](a,\ b)[/latex], then [latex]{\bf{r}}[/latex] is differentiable over the interval [latex](a,\ b)[/latex]. For the function to be differentiable over the closed interval [latex][a,\ b][/latex], the following two limits must exist as well:
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
[latex]{\bf{r}}\,(t)=(3t+4)\,{\bf{i}}+(t^{2}-4t+3){\bf{j}}[/latex]
TRY IT
Use the definition to calculate the derivative of the function [latex]{\bf{r}}\,(t)=(2t^{2}+3)\,{\bf{i}}+(5t-6)\,{\bf{j}}.[/latex]
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 [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].
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
- [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} (\sec x) = \sec x \tan x [/latex]
- [latex] \frac{d}{dx} (\csc x) = -\csc x \cot x [/latex]
- [latex] \frac{d}{dx} (\tan x) = \sec^2 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]
Recall: Derivative Rules
Constant Multiple Rule: [latex] \frac{d}{dx} (cf(x)) = cf'(x) [/latex]
Sum and Difference Rule: [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]
Example: calculating the derivative of vector-valued functions
Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the following functions.
- [latex]{\bf{r}}\,(t)=(6t+8)\,{\bf{i}}+(4t^{2}+2t-3){\bf{j}}[/latex]
- [latex]{\bf{r}}\,(t)=3\cos{t\,{\bf{i}}}+4\sin{t\,{\bf{j}}}[/latex]
- [latex]{\bf{r}}\,(t)=e^{t}\sin{t\,{\bf{i}}}+e^{t}\cos{t{\bf{j}}}-e^{2t}\,{\bf{k}}[/latex]
TRY IT
Calculate the derivative of the function
[latex]{\bf{r}}\,(t)=(t\ln{t})\,{\bf{i}}+(5e^{t})\,{\bf{j}}+(\cos{t}-\sin{t})\,{\bf{k}}[/latex]
Watch the following video to see the worked solution to the above Try It
Differentiation of Vector-Valued functions theorem
Let [latex]\bf{r}[/latex] and [latex]\bf{u}[/latex] be differentiable vector-valued functions of [latex]t[/latex], let [latex]f[/latex] be a differentiable real-valued function of
[latex]\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}[/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
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
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]_\blacksquare[/latex]
Now for some examples using these properties.
Example: Using the properties of derivatives of vector-valued functions
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.
- [latex]\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{u}}\,(t)\big][/latex]
- [latex]\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{u}}'\,(t)\big][/latex]
TRY IT
Given the vector-valued functions [latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}-e^{2t}\,{\bf{k}}[/latex] and [latex]{\bf{u}}\,(t)=t\,{\bf{i}}+\sin{t\,{\bf{j}}}+\cos{t\,{\bf{k}}}[/latex], calculate [latex]\frac{d}{dt}\big[{\bf{r}}\,(t)\cdot{\bf{r}}'\,(t)\big][/latex] and [latex]\frac{d}{dt}\big[{\bf{u}}\,(t)\times{\bf{r}}\,(t)\big][/latex].
Watch the following video to see the worked solution to the above Try It