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Calculus of Vector-Valued Functions

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

Derivatives of Vector-Valued Functions

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.

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.

Notice that in the calculations in , 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.

Condensed — the full section is in OpenStax Calculus Volume 3.

Tangent Vectors and Unit Tangent Vectors

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function \(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}.\) The derivative of this function is \({r}^{'}(t)=-\text{sin}\ t\ \text{i}+\text{cos}\ t\ \text{j}.\) If we substitute the value \(t=\pi \text{/}6\) into both functions we get

\[\text{r}(\frac{\pi }{6})=\frac{\sqrt{3}}{2}\ \text{i}+\frac{1}{2}\ \text{j}\ \text{and}\ {r}^{'}(\frac{\pi }{6})=-\frac{1}{2}\ \text{i}+\frac{\sqrt{3}}{2}\ \text{j}.\]

The graph of this function appears in , along with the vectors \(\text{r}(\frac{\pi }{6})\) and \({r}^{'}(\frac{\pi }{6}).\)

Notice that the vector \({r}^{'}(\frac{\pi }{6})\) is tangent to the circle at the point corresponding to \(t=\pi \text{/}6.\) This is an example of a tangent vector to the plane curve defined by \(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}.\)

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative \({r}^{'}(t).\) Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Condensed — the full section is in OpenStax Calculus Volume 3.

Integrals of Vector-Valued Functions

We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral. Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.

The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time t, then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.

Definition

Let f, g, and h be integrable real-valued functions over the closed interval \([a,b].\)

The indefinite integral of a vector-valued function \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}\) is
\[\int [f(t)\ \text{i}+g(t)\ \text{j}]dt=[\ \int f(t)dt]\ \text{i}+[\ \int g(t)dt]\ \text{j}.\]
The definite integral of a vector-valued function is
\[{\int }_{a}^{b}[f(t)\ \text{i}+g(t)\ \text{j}]dt=[\ {\int }_{a}^{b}f(t)dt]\ \text{i}+[\ {\int }_{a}^{b}g(t)dt]\ \text{j}.\]
The indefinite integral of a vector-valued function \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}\) is
\[\int [f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}]dt=[\ \int f(t)dt]\ \text{i}+[\ \int g(t)dt]\ \text{j}+[\ \int h(t)dt]\ \text{k}.\]
The definite integral of the vector-valued function is
\[{\int }_{a}^{b}[f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}]dt=[\ {\int }_{a}^{b}f(t)dt]\ \text{i}+[\ {\int }_{a}^{b}g(t)dt]\ \text{j}+[\ {\int }_{a}^{b}h(t)dt]\ \text{k}.\]

Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have

\[\int f(t)dt=F(t)+{C}_{1}\ \text{and}\ \int g(t)dt=G(t)+{C}_{2},\]

where F and G are antiderivatives of f and g, respectively. Then

\[\begin{array}{ll}\int [f(t)\ \text{i}+g(t)\ \text{j}]dt & =[\ \int f(t)dt]\ \text{i}+[\ \int g(t)dt]\ \text{j} \\ & =(F(t)+{C}_{1})\ \text{i}+(G(t)+{C}_{2})\ \text{j} \\ & =F(t)\ \text{i}+G(t)\ \text{j}+{C}_{1}\ \text{i}+{C}_{2}\ \text{j} \\ & =F(t)\ \text{i}+G(t)\ \text{j}+\ \text{C},\end{array}\]

where \(\text{C}={C}_{1}\ \text{i}+{C}_{2}\ \text{j}.\) Therefore, the integration constant becomes a constant vector.

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
Many of the properties of differentiation from the Introduction to Derivatives also apply to vector-valued functions.
The derivative of a vector-valued function \(\text{r}(t)\) is also a tangent vector to the curve. The unit tangent vector \(\text{T}(t)\) is calculated by dividing the derivative of a vector-valued function by its magnitude.
The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.

Key Equations

Derivative of a vector-valued function	\({r}^{'}(t)=\underset{\text{\Delta }t\to 0}{\text{lim}}\frac{\text{r}(t+\text{\Delta }t)-\text{r}(t)}{\text{\Delta }t}\)
Principal unit tangent vector	\(\text{T}(t)=\frac{{r}^{'}(t)}{\text{‖}\ {r}^{'}(t)\text{‖}}\)
Indefinite integral of a vector-valued function	\(\int [f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}]dt=[\ \int f(t)dt]\ \text{i}+[\ \int g(t)dt]\ \text{j}+[\ \int h(t)dt]\ \text{k}\)
Definite integral of a vector-valued function	\({\int }_{a}^{b}[f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}]dt=[\ {\int }_{a}^{b}f(t)dt]\ \text{i}+[\ {\int }_{a}^{b}g(t)dt]\ \text{j}+[\ {\int }_{a}^{b}h(t)dt]\ \text{k}\)

Calculus of Vector-Valued Functions

Compute the derivatives of the vector-valued functions.

For the following problems, find a tangent vector at the indicated value of t.

Find the unit tangent vector for the following parameterized curves.

Let \(\text{r}(t)=t\ \text{i}+{t}^{2}\ \text{j}-{t}^{4}\ \text{k}\) and \(\text{s}(t)=\text{sin}(t)\ \text{i}+{e}^{t}\ \text{j}+\text{cos}(t)\ \text{k}.\) Here is the graph of the function:

Find the following.

A particle moves on a circular path of radius b according to the function \(\text{r}(t)=b\ \text{cos}(\omega t)\ \text{i}+b\ \text{sin}(\omega t)\ \text{j},\) where \(\omega\) is the angular velocity, \(d\theta \text{/}dt.\)

The position vector for a particle is \(\text{r}(t)=t\ \text{i}+{t}^{2}\ \text{j}+{t}^{3}\ \text{k}.\) The graph is shown here:

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Use the definition to calculate the derivative of the function
\[\text{r}(t)=(3t+4)\ \text{i}+({t}^{2}-4t+3)\ \text{j}.\]

ਜਵਾਬ ਦਿਓ

Let’s use :
\[\begin{array}{ll}{r}^{'}(t) & =\underset{\text{\Delta }t\to 0}{\text{lim}}\frac{\text{r}(t+\text{\Delta }t)-\text{r}(t)}{\text{\Delta }t} \\ & =\underset{\text{\Delta }t\to 0}{\text{lim}}\frac{[(3(t+\text{\Delta }t)+4)\ \text{i}+({(t+\text{\Delta }t)}^{2}-4(t+\text{\Delta }t)+3)\ \text{j}]-[(3t+4)\ \text{i}+({t}^{2}-4t+3)\ \text{j}]}{\text{\Delta }t} \\ & =\underset{\text{\Delta }t\to 0}{\text{lim}}\frac{(3t+3\text{\Delta }t+4)\ \text{i}-(3t+4)\ \text{i}+({t}^{2}+2t\text{\Delta }t+{(\text{\Delta }t)}^{2}-4t-4\text{\Delta }t+3)\ \text{j}-({t}^{2}-4t+3)\ \text{j}}{\text{\Delta }t} \\ & =\underset{\text{\Delta }t\to 0}{\text{lim}}\frac{(3\text{\Delta }t)\ \text{i}+(2t\text{\Delta }t+{(\text{\Delta }t)}^{2}-4\text{\Delta }t)\ \text{j}}{\text{\Delta }t} \\ & =\underset{\text{\Delta }t\to 0}{\text{lim}}(3\ \text{i}+(2t+\text{\Delta }t-4)\ \text{j}) \\ & =3\ \text{i}+(2t-4)\ \text{j}.\end{array}\]
Use the definition to calculate the derivative of the function \(\text{r}(t)=(2{t}^{2}+3)\ \text{i}+(5t-6)\ \text{j}.\)

ਜਵਾਬ ਦਿਓ

\({r}^{'}(t)=4t\ \text{i}+5\ \text{j}\)
Use to calculate the derivative of each of the following functions.
1. \(\text{r}(t)=(6t+8)\ \text{i}+(4{t}^{2}+2t-3)\ \text{j}\)
2. \(\text{r}(t)=3\ \text{cos}\ t\ \text{i}+4\ \text{sin}\ t\ \text{j}\)
3. \(\text{r}(t)={e}^{t}\text{sin}\ t\ \text{i}+{e}^{t}\text{cos}\ t\ \text{j}-{e}^{2t}\ \text{k}\)
ਜਵਾਬ ਦਿਓ
We use and what we know about differentiating functions of one variable.
1. The first component of \(\text{r}(t)=(6t+8)\ \text{i}+(4{t}^{2}+2t-3)\ \text{j}\) is \(f(t)=6t+8.\) The second component is \(g(t)=4{t}^{2}+2t-3.\) We have \({f}^{'}(t)=6\) and \({g}^{'}(t)=8t+2,\) so the theorem gives \({r}^{'}(t)=6\ \text{i}+(8t+2)\ \text{j}.\)
2. The first component is \(f(t)=3\ \text{cos}\ t\) and the second component is \(g(t)=4\ \text{sin}\ t.\) We have \({f}^{'}(t)=-3\ \text{sin}\ t\) and \({g}^{'}(t)=4\ \text{cos}\ t,\) so we obtain \({r}^{'}(t)=-3\ \text{sin}\ t\ \text{i}+4\ \text{cos}\ t\ \text{j}.\)
3. The first component of \(\text{r}(t)={e}^{t}\text{sin}\ t\ \text{i}+{e}^{t}\text{cos}\ t\ \text{j}-{e}^{2t}\ \text{k}\) is \(f(t)={e}^{t}\text{sin}\ t,\) the second component is \(g(t)={e}^{t}\text{cos}\ t,\) and the third component is \(h(t)=-{e}^{2t}.\) We have \({f}^{'}(t)={e}^{t}(\text{sin}\ t+\text{cos}\ t),\) \({g}^{'}(t)={e}^{t}(\text{cos}\ t-\text{sin}\ t),\) and \({h}^{'}(t)=-2{e}^{2t},\) so the theorem gives \({r}^{'}(t)={e}^{t}(\text{sin}\ t+\text{cos}\ t)\ \text{i}+{e}^{t}(\text{cos}\ t-\text{sin}\ t)\ \text{j}-2{e}^{2t}\ \text{k}.\)
Calculate the derivative of the function
\[\text{r}(t)=(t\ \text{ln}\ t)\ \text{i}+(5{e}^{t})\ \text{j}+(\text{cos}\ t-\text{sin}\ t)\ \text{k}.\]

ਜਵਾਬ ਦਿਓ

\({r}^{'}(t)=(1+\text{ln}\ t)\ \text{i}+5{e}^{t}\ \text{j}-(\text{sin}\ t+\text{cos}\ t)\ \text{k}\)
Given the vector-valued functions
\[\text{r}(t)=(6t+8)\ \text{i}+(4{t}^{2}+2t-3)\ \text{j}+5t\ \text{k}\]
and
\[\text{u}(t)=({t}^{2}-3)\ \text{i}+(2t+4)\ \text{j}+({t}^{3}-3t)\ \text{k},\]
calculate each of the following derivatives using the properties of the derivative of vector-valued functions.
1. \(\frac{d}{dt}[\ \text{r}(t)\cdot \ \text{u}(t)]\)
2. \(\frac{d}{dt}[\ \text{u}(t)\ \times \ {u}^{'}(t)]\)
ਜਵਾਬ ਦਿਓ
1. We have \({r}^{'}(t)=6\ \text{i}+(8t+2)\ \text{j}+5\ \text{k}\) and \({u}^{'}(t)=2t\ \text{i}+2\ \text{j}+(3{t}^{2}-3)\ \text{k}.\) Therefore, according to property iv.:
  \[\begin{array}{ll}\frac{d}{dt}[\ \text{r}(t)\cdot \ \text{u}(t)] & ={r}^{'}(t)\cdot \ \text{u}(t)+\ \text{r}(t)\cdot \ {u}^{'}(t) \\ & =(6\ \text{i}+(8t+2)\ \text{j}+5\ \text{k})\cdot (({t}^{2}-3)\ \text{i}+(2t+4)\ \text{j}+({t}^{3}-3t)\ \text{k}) \\ & \ +((6t+8)\ \text{i}+(4{t}^{2}+2t-3)\ \text{j}+5t\ \text{k})\cdot (2t\ \text{i}+2\ \text{j}+(3{t}^{2}-3)\ \text{k}) \\ & =6({t}^{2}-3)+(8t+2)(2t+4)+5({t}^{3}-3t) \\ & \ +2t(6t+8)+2(4{t}^{2}+2t-3)+5t(3{t}^{2}-3) \\ & =20{t}^{3}+42{t}^{2}+26t-16.\end{array}\]
2. First, we need to adapt property v. for this problem:
  \[\frac{d}{dt}[\ \text{u}(t)\ \times \ {u}^{'}(t)]={u}^{'}(t)\ \times \ {u}^{'}(t)+\ \text{u}(t)\ \times \ \text{u″}(t).\]
  
  Recall that the cross product of any vector with itself is zero. Furthermore, \(\text{u″}(t)\) represents the second derivative of \(\text{u}(t)\text{:}\)
  
  \[\text{u″}(t)=\frac{d}{dt}[\ {u}^{'}(t)]=\frac{d}{dt}[2t\ \text{i}+2\ \text{j}+(3{t}^{2}-3)\ \text{k}]=2\ \text{i}+6t\ \text{k}.\]
  
  Therefore,
  
  \[\begin{array}{llllllll}\frac{d}{dt}[\ \text{u}(t)\ \times \ {u}^{'}(t)] & =0+(({t}^{2}-3)\ \text{i}+(2t+4)\ \text{j}+({t}^{3}-3t)\ \text{k})\ \times \ (2\ \text{i}+6t\ \text{k}) \\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ {t}^{2}-3 & 2t+4 & {t}^{3}-3t \\ 2 & 0 & 6t\end{array}| \\ & =6t(2t+4)\ \text{i}-(6t({t}^{2}-3)-2({t}^{3}-3t))\ \text{j}-2(2t+4)\ \text{k} \\ & =(12{t}^{2}+24t)\ \text{i}+(12t-4{t}^{3})\ \text{j}-(4t+8)\ \text{k}.\end{array}\]
Given the vector-valued functions \(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}-{e}^{2t}\ \text{k}\) and \(\text{u}(t)=t\ \text{i}+\text{sin}\ t\ \text{j}+\text{cos}\ t\ \text{k},\) calculate \(\frac{d}{dt}[\ \text{r}(t)\cdot \ {r}^{'}(t)]\) and \(\frac{d}{dt}[\ \text{u}(t)\ \times \ \text{r}(t)].\)

ਜਵਾਬ ਦਿਓ

\(\frac{d}{dt}[\ \text{r}(t)\cdot \ {r}^{'}(t)]=8{e}^{4t}\)
\(\begin{array}{l}\frac{d}{dt}[\ \text{u}(t)\ \times \ \text{r}(t)] \\ =-({e}^{2t}(\text{cos}\ t+2\ \text{sin}\ t)+\text{cos}\ 2t)\ \text{i}+({e}^{2t}(2t+1)-\text{sin}\ 2t)\ \text{j}+(t\ \text{cos}\ t+\text{sin}\ t-\text{cos}\ 2t)\ \text{k}\end{array}\)
Find the unit tangent vector for each of the following vector-valued functions:
1. \(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}\)
2. \(\text{u}(t)=(3{t}^{2}+2t)\ \text{i}+(2-4{t}^{3})\ \text{j}+(6t+5)\ \text{k}\)
ਜਵਾਬ ਦਿਓ
1. \(\begin{array}{llll}\text{First step:} & {r}^{'}(t) & = & -sin\ t\ \text{i}+\text{cos}\ t\ \text{j} \\ \text{Second step:} & \text{‖}\ {r}^{'}(t)\text{‖} & = & \sqrt{{(\text{-}\text{sin}\ t)}^{2}+{(\text{cos}\ t)}^{2}}=1 \\ \text{Third step:} & \text{T}(t) & = & \frac{{r}^{'}(t)}{\text{‖}\ {r}^{'}(t)\text{‖}}=\frac{\text{-sin}\ t\ \text{i}+\text{cos}\ t\ \text{j}}{1}=\text{-sin}\ t\ \text{i}+\text{cos}\ t\ \text{j}\end{array}\)
2. \(\begin{array}{llll}\text{First step:} & {u}^{'}(t) & = & (6t+2)\ \text{i}-12{t}^{2}\ \text{j}+6\ \text{k} \\ \text{Second step:} & \text{‖}\ {u}^{'}(t)\text{‖} & = & \sqrt{{(6t+2)}^{2}+{(-12{t}^{2})}^{2}+{6}^{2}} \\ & & = & \sqrt{144{t}^{4}+36{t}^{2}+24t+40} \\ & & = & 2\sqrt{36{t}^{4}+9{t}^{2}+6t+10} \\ \text{Third step:} & \text{T}(t) & = & \frac{{u}^{'}(t)}{\text{‖}\ {u}^{'}(t)\text{‖}}=\frac{(6t+2)\ \text{i}-12{t}^{2}\ \text{j}+6\ \text{k}}{2\sqrt{36{t}^{4}+9{t}^{2}+6t+10}} \\ & & = & \frac{3t+1}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\ \text{i}-\frac{6{t}^{2}}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\ \text{j}+\frac{3}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\ \text{k}\end{array}\)
Find the unit tangent vector for the vector-valued function
\[\text{r}(t)=({t}^{2}-3)\ \text{i}+(2t+1)\ \text{j}+(t-2)\ \text{k}.\]

ਜਵਾਬ ਦਿਓ

\(\text{T}(t)=\frac{2t}{\sqrt{4{t}^{2}+5}}\ \text{i}+\frac{2}{\sqrt{4{t}^{2}+5}}\ \text{j}+\frac{1}{\sqrt{4{t}^{2}+5}}\ \text{k}\)
Calculate each of the following integrals:
1. \(\int [(3{t}^{2}+2t)\ \text{i}+(3t-6)\ \text{j}+(6{t}^{3}+5{t}^{2}-4)\ \text{k}]\ dt\)
2. \(\int [〈t,{t}^{2},{t}^{3}〉\ \times \ 〈{t}^{3},{t}^{2},t〉]\ dt\)
3. \({\int }_{0}^{\pi \text{/}3}[\text{sin}\ 2t\ \text{i}+\text{tan}\ t\ \text{j}+{e}^{-2t}\ \text{k}]\ dt\)
ਜਵਾਬ ਦਿਓ
1. We use the first part of the definition of the integral of a space curve:
  \[\begin{array}{l}\int [(3{t}^{2}+2t)\ \text{i}+(3t-6)\ \text{j}+(6{t}^{3}+5{t}^{2}-4)\ \text{k}]dt \\ \\ =[\ \int 3{t}^{2}+2t\ dt]\ \text{i}+[\ \int 3t-6dt]\ \text{j}+[\ \int 6{t}^{3}+5{t}^{2}-4dt]\ \text{k} \\ =({t}^{3}+{t}^{2})\ \text{i}+(\frac{3}{2}{t}^{2}-6t)\ \text{j}+(\frac{3}{2}{t}^{4}+\frac{5}{3}{t}^{3}-4t)\ \text{k}+\ \text{C}.\end{array}\]
2. First calculate \(〈t,{t}^{2},{t}^{3}〉\ \times \ 〈{t}^{3},{t}^{2},t〉\text{:}\)
  \[\begin{array}{llllllll}〈t,{t}^{2},{t}^{3}〉\ \times \ 〈{t}^{3},{t}^{2},t〉 & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ t & {t}^{2} & {t}^{3} \\ {t}^{3} & {t}^{2} & t\end{array}| \\ & =({t}^{2}(t)-{t}^{3}({t}^{2}))\ \text{i}-({t}^{2}-{t}^{3}({t}^{3}))\ \text{j}+(t({t}^{2})-{t}^{2}({t}^{3}))\ \text{k} \\ & =({t}^{3}-{t}^{5})\ \text{i}+({t}^{6}-{t}^{2})\ \text{j}+({t}^{3}-{t}^{5})\ \text{k}.\end{array}\]
  Next, substitute this back into the integral and integrate:
  \[\begin{array}{ll}\int [〈t,{t}^{2},{t}^{3}〉\ \times \ 〈{t}^{3},{t}^{2},t〉]dt & =\int ({t}^{3}-{t}^{5})\ \text{i}+({t}^{6}-{t}^{2})\ \text{j}+({t}^{3}-{t}^{5})\ \text{k}\ dt \\ & =(\frac{{t}^{4}}{4}-\frac{{t}^{6}}{6})\ \text{i}+(\frac{{t}^{7}}{7}-\frac{{t}^{3}}{3})\ \text{j}+(\frac{{t}^{4}}{4}-\frac{{t}^{6}}{6})\ \text{k}+\ \text{C}.\end{array}\]
3. Use the second part of the definition of the integral of a space curve:
  \[\begin{array}{l}{\int }_{0}^{\pi \text{/}3}[\text{sin}\ 2t\ \text{i}+\text{tan}\ t\ \text{j}+{e}^{-2t}\ \text{k}]dt \\ \\ =[\ {\int }_{0}^{\pi \text{/}3}\text{sin}\ 2tdt]\ \text{i}+[\ {\int }_{0}^{\pi \text{/}3}\text{tan}\ tdt]\ \text{j}+[\ {\int }_{0}^{\pi \text{/}3}{e}^{-2t}dt]\ \text{k} \\ ={(-\frac{1}{2}\text{cos}\ 2t)|}_{0}^{\pi \text{/}3}\ \text{i}-{(\text{ln}(\text{cos}\ t))|}_{0}^{\pi \text{/}3}\ \text{j}-{(\frac{1}{2}{e}^{-2t})|}_{0}^{\pi \text{/}3}\ \text{k} \\ =(-\frac{1}{2}\text{cos}\ \frac{2\pi }{3}+\frac{1}{2}\text{cos}\ 0)\ \text{i}-(\text{ln}(\text{cos}\ \frac{\pi }{3})-\text{ln}(\text{cos}\ 0))\ \text{j}-(\frac{1}{2}{e}^{-2\pi \text{/}3}-\frac{1}{2}{e}^{-2(0)})\ \text{k} \\ =(\frac{1}{4}+\frac{1}{2})\ \text{i}-(\text{-}\text{ln}\ 2)\ \text{j}-(\frac{1}{2}{e}^{-2\pi \text{/}3}-\frac{1}{2})\ \text{k} \\ =\frac{3}{4}\ \text{i}+(\text{ln}\ 2)\ \text{j}+(\frac{1}{2}-\frac{1}{2}{e}^{-2\pi \text{/}3})\ \text{k}.\end{array}\]
Calculate the following integral:
\[{\int }_{1}^{3}[(2t+4)\ \text{i}+(3{t}^{2}-4t)\ \text{j}]dt.\]

ਜਵਾਬ ਦਿਓ

\({\int }_{1}^{3}[(2t+4)\ \text{i}+(3{t}^{2}-4t)\ \text{j}]dt=16\ \text{i}+10\ \text{j}\)
\(\text{r}(t)={t}^{3}\ \text{i}+3{t}^{2}\ \text{j}+\frac{{t}^{3}}{6}\ \text{k}\)

ਜਵਾਬ ਦਿਓ

\(〈3{t}^{2},6t,\frac{1}{2}{t}^{2}〉\)
\(\text{r}(t)=\text{sin}(t)\ \text{i}+\text{cos}(t)\ \text{j}+{e}^{t}\ \text{k}\)
\(\text{r}(t)={e}^{\text{-}t}\ \text{i}+\text{sin}(3t)\text{}\ \text{j}+10\sqrt{t}\ \text{k}.\) A sketch of the graph is shown here. Notice the varying periodic nature of the graph.

ਜਵਾਬ ਦਿਓ

\(〈\text{-}{e}^{\text{-}t},3\ \text{cos}(3t),\frac{5}{\sqrt{t}}〉\)
\(\text{r}(t)={e}^{t}\ \text{i}+2{e}^{t}\ \text{j}+\ \text{k}\)
\(\text{r}(t)=\text{i}+\ \text{j}+\ \text{k}\)

ਜਵਾਬ ਦਿਓ

\(〈0,0,0〉\)
\(\text{r}(t)=t{e}^{t}\ \text{i}+t\ \text{ln}(t)\ \text{j}+\text{sin}(3t)\ \text{k}\)
\(\text{r}(t)=\frac{1}{t+1}\ \text{i}+\text{arctan}(t)\ \text{j}+\text{ln}\ {t}^{3}\ \text{k}\)

ਜਵਾਬ ਦਿਓ

\(〈\frac{-1}{{(t+1)}^{2}},\frac{1}{1+{t}^{2}},\frac{3}{t}〉\)
\(\text{r}(t)=\text{tan}(2t)\ \text{i}+\text{sec}(2t)\ \text{j}+{\text{sin}}^{2}(t)\ \text{k}\)
\(\text{r}(t)=3\ \text{i}+4\ \text{sin}(3t)\ \text{j}+t\ \text{cos}(t)\ \text{k}\)

ਜਵਾਬ ਦਿਓ

\(〈0,12\ \text{cos}(3t),\text{cos}\ t-t\ \text{sin}\ t〉\)
\(\text{r}(t)={t}^{2}\ \text{i}+t{e}^{-2t}\ \text{j}-5{e}^{-4t}\ \text{k}\)
\(\text{r}(t)=t\ \text{i}+\text{sin}(2t)\ \text{j}+\text{cos}(3t)\ \text{k};t=\frac{\pi }{3}\)

ਜਵਾਬ ਦਿਓ

\(〈1,-1,0〉\)
\(\text{r}(t)=3{t}^{3}\ \text{i}+2{t}^{2}\ \text{j}+\frac{1}{t}\ \text{k};t=1\)
\(\text{r}(t)=3{e}^{t}\ \text{i}+2{e}^{-3t}\ \text{j}+4{e}^{2t}\ \text{k};\) \(t=\text{ln}(2)\)

ਜਵਾਬ ਦਿਓ

\(〈6,-\frac{3}{4},32〉\)
\(\text{r}(t)=\text{cos}(2t)\ \text{i}+2\ \text{sin}\ t\ \text{j}+{t}^{2}\ \text{k};t=\frac{\pi }{2}\)
\(\text{r}(t)=6\ \text{i}+\text{cos}(3t)\ \text{j}+3\ \text{sin}(4t)\ \text{k},\) \(0\le t<2\pi\) . Two views of this curve are presented here:

ਜਵਾਬ ਦਿਓ

\(\frac{1}{\sqrt{9\ {\text{sin}}^{2}(3t)+144\ {\text{cos}}^{2}(4t)}}〈0,-3\ \text{sin}(3t),12\ \text{cos}(4t)〉\)
\(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}+\text{sin}\ t\ \text{k},\) \(0\le t<2\pi .\)
\(\text{r}(t)=3\ \text{cos}(4t)\ \text{i}+3\ \text{sin}(4t)\ \text{j}+5t\ \text{k},1\le t\le 2\)

ਜਵਾਬ ਦਿਓ

\(\text{T}(t)=\frac{-12}{13}\text{sin}(4t)\ \text{i}+\frac{12}{13}\text{cos}(4t)\ \text{j}+\frac{5}{13}\ \text{k}\)
\(\text{r}(t)=t\ \text{i}+3t\ \text{j}+{t}^{2}\ \text{k}\)
\(\frac{d}{dt}[\text{r}({t}^{2})]\)

ਜਵਾਬ ਦਿਓ

\(〈2t,4{t}^{3},-8{t}^{7}〉\)
\(\frac{d}{dt}[{t}^{2}\cdot \text{s}(t)]\)
\(\frac{d}{dt}[\text{r}(t)\cdot \text{s}(t)]\)

ਜਵਾਬ ਦਿਓ

\(\text{sin}(t)+2t{e}^{t}-4{t}^{3}\text{cos}(t)+t\ \text{cos}(t)+{t}^{2}{e}^{t}+{t}^{4}\text{sin}(t)\)
Compute the first, second, and third derivatives of \(\text{r}(t)=3t\ \text{i}+6\ \text{ln}(t)\ \text{j}+5{e}^{-3t}\ \text{k}.\)
Find \(\text{r}'(t)\cdot \ \text{r}\text{″}(t)\ \text{for}\ \text{r}(t)=-3{t}^{5}\ \text{i}+5t\ \text{j}+2{t}^{2}\ \text{k}.\)

ਜਵਾਬ ਦਿਓ

\(900{t}^{7}+16t\)
The acceleration function, initial velocity, and initial position of a particle are
\(\text{a}(t)=-5\ \text{cos}\ t\ \text{i}-5\ \text{sin}\ t\ \text{j},\ \text{v}(0)=9\ \text{i}+2\ \text{j},\ \text{and}\ \text{r}(0)=5\ \text{i}.\)
Find \(\text{v}(t)\ \text{and}\ \text{r}(t).\)
The position vector of a particle is \(\text{r}(t)=5\ \text{sec}(2t)\ \text{i}-4\ \text{tan}(t)\ \text{j}+7{t}^{2}\ \text{k}.\)
1. Graph the position function and display a view of the graph that illustrates the asymptotic behavior of the function.
2. Find the velocity as t approaches but is not equal to \(\pi \text{/}4\) (if it exists).
ਜਵਾਬ ਦਿਓ
2. Undefined or infinite
Find the velocity and the speed of a particle with the position function \(\text{r}(t)=(\frac{2t-1}{2t+1})\ \text{i}+\text{ln}(1-4{t}^{2})\ \text{j}.\) The speed of a particle is the magnitude of the velocity and is represented by \(\text{‖}{r}^{'}(t)\text{‖}.\)
Find the velocity function and show that \(\text{v}(t)\) is always orthogonal to \(\text{r}(t).\)

ਜਵਾਬ ਦਿਓ

\(\text{r}'(t)=-b\omega \ \text{sin}(\omega t)\ \text{i}+b\omega \ \text{cos}(\omega t)\ \text{j}.\) To show orthogonality, note that \(\text{r}'(t)\cdot \ \text{r}(t)=0.\)
Show that the speed of the particle is proportional to the angular velocity.
Evaluate \(\frac{d}{dt}[\ \text{u}(t)\ \times \ {u}^{'}(t)]\) given \(\text{u}(t)={t}^{2}\ \text{i}-2t\ \text{j}+\ \text{k}.\)

ਜਵਾਬ ਦਿਓ

\(0\ \text{i}+2\ \text{j}+4t\ \text{k}\)
Find the antiderivative of \(\text{r}'(t)=\text{cos}(2t)\ \text{i}-2\ \text{sin}\ t\ \text{j}+\frac{1}{1+{t}^{2}}\ \text{k}\) that satisfies the initial condition \(\text{r}(0)=3\ \text{i}-2\ \text{j}+\ \text{k}.\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

imaginary unit
i² = −1.

\neq

not equal
The two sides are different.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Calculus of Vector-Valued Functions

Write an expression for the derivative of a vector-valued function.
Find the tangent vector at a point for a given position vector.
Find the unit tangent vector at a point for a given position vector and explain its significance.
Calculate the definite integral of a vector-valued function.
If
If
The first component of
The first component is

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

ਆਪਣਾ ਹੀ ਕੋਸ਼ਿਸ਼ ਕਰੋ

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ਹੋਰ ਵਿੱਚ Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems