5.2E:
- Page ID
- 26283
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
Compute the derivatives of the vector-valued functions.
1) \(\mathrm{r(t)=t^3 \mathbf{i}+3t^2 \mathbf{j}+\frac{t^3}{6}\mathbf{k}}\)
2) \(\mathrm{r(t)=\sin(t) \mathbf{i}+\cos(t) \mathbf{j}+e^t \mathbf{k}}\)
3) \(\mathrm{r(t)=e^{−t} \mathbf{i}+\sin(3t) \mathbf{j}+10 \sqrt{t} \mathbf{k}}\). A sketch of the graph is shown here. Notice the varying periodic nature of the graph.
4) \(\mathrm{r(t)=e^t \mathbf{i}+2e^t \mathbf{j}+\mathbf{k}}\)
5) \(\mathrm{r(t)=\mathbf{i}+\mathbf{j}+\mathbf{k}}\)
6) \(\mathrm{r(t)=te^t \mathbf{i}+t \ln(t) \mathbf{j}+\sin(3t)\mathbf{k}}\)
7) \(\mathrm{r(t)=\frac{1}{t+1} \mathbf{i}+\arctan(t) \mathbf{j}+\ln t^3 \mathbf{k}}\)
8) \(\mathrm{r(t)= \tan(2t) \mathbf{i}+\sec(2t) \mathbf{j}+\sin ^2 (t) \mathbf{k}}\)
9) \(\mathrm{r(t)=3 \mathbf{i}+4 \sin (3t) \mathbf{j}+ t \cos(t) \mathbf{k}}\)
10) \(\mathrm{r(t)=t^2 \mathbf{i}+te^{−2t} \mathbf{j}−5e^{−4t} \mathbf{k}}\)
- Answer
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1) \(\mathrm{⟨3t^2,6t,\frac{1}{2}t^2⟩}\)
3) \(\mathrm{⟨−e^{−t},3\cos (3t),5t⟩}\)
5) \(\mathrm{⟨0,0,0⟩}\)
7) \(\mathrm{⟨\frac{−1}{(t+1)^2},\frac{1}{1+t^2},\frac{3}{t}⟩}\)
9) \(\mathrm{⟨0,12 \cos(3t), \cos t−t \sin t⟩}\)
Exercise \(\PageIndex{2}\)
For the following problems, find a tangent vector at the indicated value of t.
1) \(\mathrm{r(t)=t \mathbf{i}+\sin(2t) \mathbf{j}+\cos(3t) \mathbf{k}}\); \(\mathrm{t=\frac{π}{3}}\)
2) \(\mathrm{r(t)=3t^3 \mathbf{i}+2t^2 \mathbf{j}+\frac{1}{t} \mathbf{k};t=1}\)
3) \(\mathrm{r(t)=3e^t \mathbf{i}+2e^{−3t} \mathbf{j}+4e^{2t} \mathbf{k}; t= \ln(2)}\)
4) \(\mathrm{r(t)=\cos(2t) \mathbf{i}+2 \sin t \mathbf{j}+t^2 \mathbf{k};t=\frac{π}{2}}\)
- Answer
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1) \(\mathrm{\frac{1}{\sqrt{2}}⟨1,−1,0⟩}\)
3) \(\mathrm{\frac{1}{\sqrt{1060.5625}}⟨6,−34,32⟩}\)
Exercise \(\PageIndex{3}\)
Find the unit tangent vector for the following parameterized curves.
1) \(\mathrm{r(t)=6 \mathbf{i}+\cos(3t) \mathbf{j}+3\sin(4t) \mathbf{k}, 0≤t<2π}\)
2) \(\mathrm{r(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\sin t \mathbf{k}, 0≤t<2π}\). Two views of this curve are presented here:

3) \(\mathrm{r(t)=3 \cos(4t) \mathbf{i}+3 \sin(4t) \mathbf{j}+5t \mathbf{k},1≤t≤2}\)
4) \(\mathrm{r(t)=t \mathbf{i}+3t \mathbf{j}+t^2 \mathbf{k}}\)
- Answer
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1) \(\mathrm{\frac{1}{\sqrt{9sin ^2 (3t)+144\cos ^2 (4t)}}⟨0,−3\sin(3t),12\cos(4t)⟩}\)
3) \(\mathrm{T(t)=−\frac{12}{13} \sin(4t) \mathbf{i}+ \frac{12}{13}\cos (4t) \mathbf{j}+\frac{5}{13} \mathbf{k}}\)
Exercise \(\PageIndex{4}\)
Let \(\mathrm{r(t)=t \mathbf{i}+t^2 \mathbf{j}−t^4 \mathbf{k}}\) and \(\mathrm{s(t)=\sin(t) \mathbf{i}+e^t \mathbf{j}+ \cos(t) \mathbf{k}}\) Here is the graph of the function:
Find the following.
1) \(\mathrm{\frac{d}{dt}[r(t^2)]}\)
2) \(\mathrm{\frac{d}{dt}[t^2⋅s(t)]}\)
3) \(\mathrm{\frac{d}{dt}[r(t)⋅s(t)]}\)
- Answer
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1) \(\mathrm{⟨2t,4t^3,−8t^7⟩}\)
3) \(\mathrm{\sin(t)+2te^t−4t^3 \cos(t)+tcos(t)+t^2e^t+t^4sin(t)}\)
Exercise \(\PageIndex{5}\)
1) Compute the first, second, and third derivatives of \(\mathrm{r(t)=3t \mathbf{i}+6\ln(t) \mathbf{j}+5e^{−3t}\mathbf{k}}\).
2) Find \(\mathrm{r'(t)⋅r''(t) \; for \; r(t)=−3t^5 \mathbf{i}+5t \mathbf{j}+2t^2 \mathbf{k}.}\)
- Answer
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\(\mathrm{900t^7+16t}\)
Exercise \(\PageIndex{6}\)
1) The acceleration function, initial velocity, and initial position of a particle are
\(\mathrm{a(t)=−5 \cos t \mathbf{i}−5\sin t \mathbf{j},v(0)=9 \mathbf{i}+2 \mathbf{j},and \; r(0)=5 \mathbf{i}.}\) Find \(\mathrm{v(t) \; and \; r(t)}\).
2) The position vector of a particle is \(\mathrm{r(t)=5 \sec(2t) \mathrm{i}−4tan(t) \mathrm{j}+7t^2 \mathrm{k}}\).
a. Graph the position function and display a view of the graph that illustrates the asymptotic behavior of the function.
b. Find the velocity as t approaches but is not equal to \(\mathrm{\frac{π}{4}}\) (if it exists)
3) Find the velocity and the speed of a particle with the position function \(\mathrm{r(t)=(\frac{2t−1}{2t+1}) \mathbf{i}+\ln(1−4t^2) \mathbf{j}}\). The speed of a particle is the magnitude of the velocity and is represented by \(\mathrm{‖r′(t)‖}\).
4) A particle moves on a circular path of radius b according to the function \(\mathrm{r(t)=b \cos(\omega t) \mathbf{i}+b\sin(\omega) \mathbf{j},}\) where \(\mathrm{\omega}\) is the angular velocity, \(\mathrm{\frac{d \theta}{dt}}\).
Find the velocity function and show that \(\mathrm{v(t)}\) is always orthogonal to \(\mathrm{r(t)}\).
5) Show that the speed of the particle is proportional to the angular velocity.
6) Evaluate \(\mathrm{\frac{d}{dt}[u(t) \times u′(t)]}\) given \(\mathrm{u(t)=t^2 \mathbf{i}−2t \mathbf{j}+\mathbf{k}}\).
7) Find the antiderivative of \(\mathrm{r'(t)=\cos(2t) \mathbf{i}−2\sin t \mathbf{j}+\frac{1}{1+t^2} \mathbf{k}}\) that satisfies the initial condition \(\mathrm{r(0)=3 \mathbf{i}−2 \mathbf{j}+\mathbf{k}}\).
8) Evaluate \(\mathrm{\int_0^3‖ti+t^2j‖dt}\).
- Answer
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2)
, Undefined or infinite.
4) \(\mathrm{r'(t)=−b \omega \sin( \omega t) \mathbf{i}+b \omega cos(\omega t)\mathbf{j}}\). To show orthogonality, note that \(\mathrm{r'(t)⋅r(t)=0}\).
6) \(\mathrm{0 \mathbf{i} +2 \mathbf{j}+4t \mathbf{j}}\)
8) \(\mathrm{\frac{1}{3}(10^{\frac{3}{2}}−1)}\)
Exercise \(\PageIndex{7}\)
1) An object starts from rest at point \(\mathrm{P(1,2,0)}\) and moves with an acceleration of\(\mathrm{ a(t)=\mathbf{j}+2 \mathbf{k},}\) where \(\mathrm{‖a(t)‖}\) is measured in feet per second per second. Find the location of the object after \(\mathrm{t=2}\) sec.
2) Show that if the speed of a particle travelling along a curve represented by a vector-valued function is constant, then the velocity function is always perpendicular to the acceleration function.
3) Given \(\mathrm{r(t)=t \mathbf{i}+3t \mathbf{j}+t^2 \mathbf{k}}\) and \(\mathrm{u(t)=4t \mathbf{i}+t^2 \mathbf{j}+t^3 \mathbf{k}}\), find \(\mathrm{\frac{d}{dt}(r(t) \times u(t))}\).
4) Given \(\mathrm{r(t)=⟨t+ \cos t,t− \sin t⟩}\), find the velocity and the speed at any time.
5) Find the velocity vector for the function \(\mathrm{r(t)=⟨e^t,e^{−t},0⟩}\).
6) Find the equation of the tangent line to the curve \(\mathrm{r(t)=⟨e^t,e^{−t},0⟩}\) at \(\mathrm{t=0}\).
7) Describe and sketch the curve represented by the vector-valued function \(\mathrm{r(t)=⟨6t,6t−t^2⟩}\).
8) Locate the highest point on the curve \(\mathrm{r(t)=⟨6t,6t−t^2⟩}\) and give the value of the function at this point.
- Answer
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2) \(\begin{align} \mathrm{‖v(t)‖ \;} & \mathrm{= k} \\ \mathrm{v(t)·v(t) \; } & \mathrm{= k} \\ \mathrm{ddt(v(t)·v(t)) \; } & \mathrm{=\frac{d}{dt}k=0} \\ \mathrm{v(t)·v′(t)+v′(t)·v(t) \;} & \mathrm{= 0} \\ \mathrm{2v(t)·v′(t) \;} & \mathrm{= 0} \\ \mathrm{v(t)·v′(t) \;} & \mathrm{= 0}\end{align}\)
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.
4) \(\mathrm{v(t)=⟨1− \sin t,1−\cos t⟩, speed=−v(t)‖=\sqrt{4−2( \sin t+\cos t)}}\)
6) \(\mathrm{x−1=t,y−1=−t,z−0=0}\)
8) \(\mathrm{r(t)=⟨18,9⟩}\) at \(\mathrm{t=3}\)
Exercise \(\PageIndex{8}\)
The position vector for a particle is \(\mathrm{r(t)=t \mathbf{i}+t^2 \mathbf{j}+t^3 \mathbf{k}}\).The graph is shown here:
1) Find the velocity vector at any time.
2) Find the speed of the particle at time \(\mathrm{t=2}\) sec.
3) Find the acceleration at time \(\mathrm{t=2}\) sec.
- Answer
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2) \(\mathrm{\sqrt{593}}\)
Exercise \(\PageIndex{9}\)
A particle travels along the path of a helix with the equation \(\mathrm{r(t)= \cos(t) \mathbf{i}+\sin(t) \mathbf{j}+t \mathbf{k}}\). See the graph presented here:
Find the following:
1) Velocity of the particle at any time
2) Speed of the particle at any time
3) Acceleration of the particle at any time
4) Find the unit tangent vector for the helix.
- Answer
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1) \(\mathrm{v(t)=⟨−\sin t,\cos t,1⟩}\)
3) \(\mathrm{a(t)=−\cos t \mathbf{i}− \sin t \mathbf{j}+0 \mathbf{j}}\)
Exercise \(\PageIndex{10}\)
A particle travels along the path of an ellipse with the equation \(\mathrm{r(t)=\cos t \mathbf{i}+2 \sin t \mathbf{j}+0 \mathbf{k}}\). Find the following:
1) Velocity of the particle
2) Speed of the particle at \(\mathrm{t=\frac{π}{4}}\)
3) Acceleration of the particle at \(\mathrm{t=\frac{π}{4}}\)
- Answer
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1) \(\mathrm{v(t)=⟨−\sin t,2 \cos t,0⟩}\)
3) \(\mathrm{a(t)=⟨−\frac{\sqrt{2}}{2},−\sqrt{2},0⟩}\)
Exercise \(\PageIndex{11}\)
Given the vector-valued function \(\mathrm{r(t)=⟨\tan t,\sec t,0⟩}\) (graph is shown here), find the following:
1) Velocity
2) Speed
3) Acceleration
4) Find the minimum speed of a particle traveling along the curve \(\mathrm{r(t)=⟨t+\cos t,t−\sin t⟩}\) \mathrm{t∈[0,2π)}\).
- Answer
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2) \(\mathrm{‖v(t)‖=\sqrt{\sec ^4 t+\sec ^2 t \tan ^2 t}=\sqrt{\sec ^2 t(\sec ^2 t+\tan ^2 t)}}\)
4) 2
Exercise \(\PageIndex{12}\)
Given \(\mathrm{r(t)=t \mathbf{i}+2\sin t \mathbf{j}+2 \cos t \mathbf{k}}\) and \(\mathrm{u(t)=\frac{1}{t} \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k}}\), find the following:
1) \(\mathrm{r(t) \times u(t)}\)
2) \(\mathrm{\frac{d}{dt}(r(t) \times u(t))}\)
3) Now, use the product rule for the derivative of the cross product of two vectors and show this result is the same as the answer for the preceding problem.
- Answer
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2) \(\mathrm{⟨0,2 \sin t(t− \frac{1}{t})−2 \cos t(1+ \frac{1}{t^2}),2 \sin t(1+ \frac{1}{t^2})+2 \cos t(t−\frac{2}{t})⟩}\)
Exercise \(\PageIndex{13}\)
Find the unit tangent vector T(t) for the following vector-valued functions.
1) \(\mathrm{r(t)=⟨t,\frac{1}{t}⟩}\). The graph is shown here:

2) \(\mathrm{r(t)=⟨t \cos t,t sin t⟩}\)
3) \(\mathrm{r(t)=⟨t+1,2t+1,2t+2⟩}\)
- Answer
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1) \(\mathrm{T(t)=⟨\frac{t^2}{\sqrt{t^4+1}},\frac{-1}{\sqrt{t^4+1}⟩}}\)
3) \(\mathrm{T(t)=\frac{1}{3} ⟨1,2,2⟩}\)
Exercise \(\PageIndex{14}\)
Evaluate the following integrals:
1) \(\mathrm{\int (e^t \mathbf{i}+\sin t \mathbf{j}+ \frac{1}{2t−1} \mathbf{k})dt}\)
2) \(\mathrm{\int_0^1 r(t)dt}\), where \(\mathrm{r(t)=⟨\sqrt[3]{t},\frac{1}{t+1},e^{−t}⟩}\)
- Answer
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2) \(\mathrm{\frac{3}{4}\mathbf{i}+\ln(2) \mathbf{j}+(1−\frac{1}{e}) \mathbf{j}}\)