Skip to main content
Mathematics LibreTexts

4.9E: Exercises for Section 4.8

  • Page ID
    123919

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    In exercises 1 - 4, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

    1) \(x=3+t,\quad y=1−t\)

    2) \( x=8+2t, \quad y=1\)

    Answer:
    \(m=0\)

    3) \( x=4−3t, \quad y=−2+6t\)

    4) \( x=−5t+7, \quad y=3t−1\)

    Answer:
    \(m= -\frac{3}{5}\)

     

    In exercises 5 - 9, determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter.

    5) \( x=3\sin t,\quad y=3\cos t, \quad \text{for }t=\frac{π}{4}\)

    6) \( x=\cos t, \quad y=8\sin t, \quad \text{for }t=\frac{π}{2}\)

    Answer:
    Slope\(=0; y=8.\)

    7) \( x=2t, \quad y=t^3, \quad \text{for } t=−1\)

    8) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t}, \quad \text{for }t=1\)

    Answer:
    Slope is undefined; \( x=2\).

    9) \( x=\sqrt{t}, \quad y=2t, \quad \text{for }t=4\)

     

    In exercises 10 - 13, find all points on the curve that have the given slope.

    10) \( x=4\cos t, \quad y=4\sin t,\) slope = \(0.5\)

    Answer:
    \( t=\arctan(−2); \left(\frac{4\sqrt{5}}{5},\frac{−8\sqrt{5}}{5}\right)\).

    11) \( x=2\cos t, \quad y=8\sin t,\) slope= \(−1\)

    12) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t},\) slope= \(1\)

    Answer:
    No points possible; undefined expression.

    13) \( x=2+\sqrt{t}, \quad y=2−4t,\) slope= \(0\)

     

    In exercises 14 - 16, write the equation of the tangent line in Cartesian coordinates for the given parameter \(t\).

    14) \( x=e^{\sqrt{t}}, \quad y=1−\ln t^2, \quad \text{for }t=1\)

    Answer:
    \( y=−(\frac{2}{e})x+3\)

    15) \( x=t\ln t, \quad y=\sin^2t, \quad \text{for }t=\frac{π}{4}\)

    16) \( x=e^t, \quad y=(t−1)^2,\) at \((1,1)\)

    Answer:
    \( y=2x−7\)

     

    17) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π.\) Find all values of \(t\) at which a horizontal tangent line exists.

    18) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π\). Find all values of \(t\) at which a vertical tangent line exists.

    Answer:
    A vertical tangent line exists at \(t = \frac{π}{4},\frac{5π}{4},\frac{3π}{4},\frac{7π}{4}\)

     

    19) Find all points on the curve \( x=4\cos(t), \quad y=4\sin(t)\) that have the slope of \( \frac{1}{2}\).

    20) Find \( \dfrac{dy}{dx}\) for \( x=\sin(t), \quad y=\cos(t)\).

    Answer:
    \( \dfrac{dy}{dx}=−\tan(t)\)

    21) Find the equation of the tangent line to \( x=\sin(t), \quad y=\cos(t)\) at \( t=\frac{π}{4}\).

    22) For the curve \( x=4t, \quad y=3t−2,\) find the slope and concavity of the curve at \( t=3\).

    Answer:
    \( \dfrac{dy}{dx}=\dfrac{3}{4}\) and \( \dfrac{d^2y}{dx^2}=0\), so the curve is neither concave up nor concave down at \( t=3\). Therefore the graph is linear and has a constant slope but no concavity.

    23) For the parametric curve whose equation is \( x=4\cos θ, \quad y=4\sin θ\), find the slope and concavity of the curve at \( θ=\frac{π}{4}\).

    24) Find the slope and concavity for the curve whose equation is \( x=2+\sec θ, \quad y=1+2\tan θ\) at \( θ=\frac{π}{6}\).

    Answer:
    \( \dfrac{dy}{dx}=4, \quad \dfrac{d^2y}{dx^2}=−6\sqrt{3};\) the curve is concave down at \( θ=\frac{π}{6}\).

    25) Find all points on the curve \( x=t+4, \quad y=t^3−3t\) at which there are vertical and horizontal tangents.

    26) Find all points on the curve \( x=\sec θ, \quad y=\tan θ\) at which horizontal and vertical tangents exist.

    Answer:
    No horizontal tangents. Vertical tangents at \( (1,0)\) and \((−1,0)\).

     

    In exercises 27 - 29, find \( d^2y/dx^2\).

    27) \( x=t^4−1, \quad y=t−t^2\)

    28) \( x=\sin(πt), \quad y=\cos(πt)\)

    Answer:
    \( d^2y/dx^2 = −\sec^3(πt)\)

    29) \( x=e^{−t}, \quad y=te^{2t}\)

     

    In exercises 30 - 31, find points on the curve at which tangent line is horizontal or vertical.

    30) \( x=t(t^2−3), \quad y=3(t^2−3)\)

    Answer:
    Horizontal \( (0,−9)\);
    Vertical \( (±2,−6).\)

    31) \( x=\dfrac{3t}{1+t^3}, \quad y=\dfrac{3t^2}{1+t^3}\)

     

    In exercises 32 - 34, find \( dy/dx\) at the value of the parameter.

    32) \( x=\cos t,y=\sin t, \quad \text{for }t=\frac{3π}{4}\)

    Answer:
    \(dy/dx = 1\)

    33) \( x=\sqrt{t}, \quad y=2t+4,t=9\)

    34) \( x=4\cos(2πs), \quad y=3\sin(2πs), \quad \text{for }s=−\frac{1}{4}\)

    Answer:
    \(dy/dx = 0\)

     

    In exercises 35 - 36, find \( d^2y/dx^2\) at the given point without eliminating the parameter.

    35) \( x=\frac{1}{2}t^2, \quad y=\frac{1}{3}t^3, \quad \text{for }t=2\)

    36) \( x=\sqrt{t}, \quad y=2t+4, \quad \text{for }t=1\)

    Answer:
    \(d^2y/dx^2 = 4\)

     

    37) Find intervals for \(t\) on which the curve \( x=3t^2, \quad y=t^3−t\) is concave up as well as concave down.

    38) Determine the concavity of the curve \( x=2t+\ln t, \quad y=2t−\ln t\).

    Answer:
    Concave up on \( t>0\).

     

    Contributors and Attributions

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


    This page titled 4.9E: Exercises for Section 4.8 is shared under a not declared license and was authored, remixed, and/or curated by Chau D Tran.

    • Was this article helpful?