\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.4in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Instantaneous Velocity} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, 2015} \rfoot{\small Last revised: September 15, 2015} \textwidth 7.4in \textheight 9.3in \setlength{\headheight}{27pt} \setlength{\parindent}{0in} \begin{document} Suppose that an object is moving along a vertical line, and its vertical position is given by $L\left( t\right) $. \ The average velocity of the object between $t_{1}$ and $t_{2}$ is \begin{equation*} v_{\text{av}}=\dfrac{L\left( t_{2}\right) -L\left( t_{1}\right) }{t_{2}-t_{1}% } \end{equation*}% We define the instantenous velocity at $t$ as the limit of the average velocities, where the time interval around $t$ is getting smaller and smaller. \ In short, the instantaneous velocity at time $t$ is the following limit (if this limit exists)% \begin{equation*} v\left( t\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{t+h-t}=\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h} \end{equation*} \bigskip \bigskip \begin{center} {\LARGE Sample Problems} \end{center} \begin{enumerate} \item The location function of an object is $L\left( t\right) =t^{2}-3t$. $\ $Compute the instantaneous velocity of the object a) \ at $t=7$ second \ \ \ \ \ \ b) \ at $t=10$ second \ \ \ \ \ \ \ c) \ at $t$. \item The location function of an object is $L\left( t\right) =t^{3}$. $\ $% Compute the instantaneous velocity of the object a) \ at $t=4$ second \ \ \ \ \ \ b) \ at $t$. \item The location function of an object is $L\left( t\right) =\sqrt{t}$. $\ $Compute the instantaneous velocity of the object a) \ at $t=49$ second \ \ \ \ \ \ \ \ \ \ \ b) \ at $t$ \item The location function of an object is $L\left( t\right) =\dfrac{1}{t}$% . $\ $Compute the instantaneous velocity of the object a) \ at $t=5$ second \ \ \ \ \ \ \ \ \ \ \ b) \ at $t$ \end{enumerate} \bigskip \begin{center} {\Large Practice Problems} {\Large \bigskip } \end{center} \begin{enumerate} \item The location function of an object is $L\left( t\right) =-t^{2}+t$. $\ $Compute the instantaneous velocity of the object a) at\ $t=3$ second \ \ \ \ \ \ b) \ at\ $t=4$ second \ \ \ \ \ \ c) \ at $t $ \item The location function of an object is $L\left( t\right) =t^{4}$. $\ $% Compute the instantaneous velocity of the object a) \ at\ $t=3$ second \ \ \ \ \ \ \ b) \ at $t$ (Hint: you may need the following formula: \ $\left( x+y\right) ^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}$) \item The location function of an object is $L\left( t\right) =\sqrt{2t+1}$. $\ $Compute the instantaneous velocity of the object a) \ at $t=12$ second \ \ \ \ \ \ \ \ \ b) at $t$ \item The location function of an object is $L\left( t\right) =\dfrac{1}{3t+5% }$. $\ $Compute the instantaneous velocity of the object a) \ at $t=2$ second \ \ \ \ \ \ \ \ b) \ at $t$ \item (Enrichment) \ The location function of an object is given by $L\left( t\right) =2t^{3}-15t^{2}$. \ When is the object moving upward? \ \end{enumerate} \bigskip \begin{center} {\LARGE Answers - Sample Problems}\bigskip \end{center} \begin{enumerate} \item a) \ $v\left( 7\right) =11$ \ \ \ \ \ \ b) \ $v\left( 10\right) =17$ \ \ \ \ \ \ c) \ $v\left( t\right) =2t-3$ \item a) \ $v\left( 4\right) =48$ \ \ \ \ \ \ b) \ $v\left( t\right) =3t^{2}$ \item a) \ $v\left( 49\right) =\dfrac{1}{14}$ \ \ \ \ b) \ $v\left( t\right) =\dfrac{1}{2\sqrt{t}}=\dfrac{\sqrt{t}}{2t}$ \item a) \ $v\left( 5\right) =-\dfrac{1}{25}$ \ \ \ \ \ \ \ b) \ $v\left( t\right) =-\dfrac{1}{t^{2}}$\bigskip \end{enumerate} \bigskip \begin{center} {\LARGE Answers - Practice Problems}\bigskip \end{center} \begin{enumerate} \item a) \ $v\left( 3\right) =-5$ \ \ \ \ b) \ $v\left( 4\right) =-7$ \ \ \ \ \ \ \ c) \ $v\left( t\right) =-2t+1$ \item a) \ $v\left( 3\right) =108$ \ \ \ \ \ b) \ $v\left( t\right) =4t^{3}$ \item a) \ $v\left( 12\right) =\dfrac{1}{5}$ \ \ \ \ \ \ b) \ $L^{\prime }\left( t\right) =\dfrac{1}{\sqrt{2t+1}}$ \item a) \ $v\left( 2\right) =-\dfrac{3}{121}$ \ \ \ \ \ \ \ \ b) \ $v\left( t\right) =-\dfrac{3}{\left( 3t+5\right) ^{2}}$\bigskip \end{enumerate} \bigskip \begin{center} {\LARGE Sample Problems - Solutions \bigskip } \end{center} \begin{enumerate} \item The location function of an object is $L\left( t\right) =t^{2}-3t$. $\ $ a) \ Compute the instantaneous velocity of the object at $t=7$ second. Solution: \ \begin{equation*} v_{7}=\lim\limits_{h\rightarrow 0}\dfrac{L\left( 7+h\right) -L\left( 7\right) }{h} \end{equation*}% We compute first $L\left( 7+h\right) $% \begin{equation*} L\left( 7+h\right) =\left( 7+h\right) ^{2}-3\left( 7+h\right) =h^{2}+14h+49-21-3h=h^{2}+11h+38 \end{equation*}% We also compute $L\left( 7\right) $% \begin{equation*} L\left( 7\right) =7^{2}-3\cdot 7=49-21=38 \end{equation*}% So now the velocity: \ \begin{eqnarray*} v_{7} &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( 7+h\right) -L\left( 7\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+11h+38-38}{h}% =\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+11h}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}\left( h+11\right) }{\NEG{h}} \\ &=&\lim\limits_{h\rightarrow 0}\left( h+11\right) =11 \end{eqnarray*}% So at $t=7$, the velocity of the object is $11$. \ In short, $v\left( 7\right) =11.$ b) \ Compute the instantaneous velocity of the object at $t=10$ second. Solution: \ \begin{equation*} v\left( 10\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( 10+h\right) -L\left( 10\right) }{h} \end{equation*}% We compute first $L\left( 10+h\right) $% \begin{equation*} L\left( 10+h\right) =\left( 10+h\right) ^{2}-3\left( 10+h\right) =h^{2}+20h+100-30-3h=h^{2}+17h+70 \end{equation*}% We also compute $L\left( 10\right) $% \begin{equation*} L\left( 10\right) =10^{2}-3\cdot 10=100-30=70 \end{equation*}% So now the velocity: \ \begin{eqnarray*} v\left( 10\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( 10+h\right) -L\left( 10\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+17h+70-70}{h% }=\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+17h}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}\left( h+17\right) }{\NEG{h}} \\ &=&\lim\limits_{h\rightarrow 0}\left( h+17\right) =17 \end{eqnarray*}% So at $t=10$, the velocity of the object is $17$. \ In short, $v\left( 10\right) =17$.\bigskip c) \ Compute the instantaneous velocity of the object at $t$. Solution: \ If we do that and we obtain an expression in terms of $t,$ then we created a new function, the velocity function.% \begin{equation*} v\left( t\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h} \end{equation*}% We compute first $L\left( t+h\right) $% \begin{equation*} L\left( t+h\right) =\left( t+h\right) ^{2}-3\left( t+h\right) =h^{2}+2th+t^{2}-3t-3h \end{equation*}% So now the velocity: \ \begin{eqnarray*} v\left( 10\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( 10+h\right) -L\left( 10\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\left( h^{2}+2th+t^{2}-3t-3h\right) -\left( t^{2}-3t\right) }{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+2th+t^{2}-3t-3h-t^{2}+3t}{h}% =\lim\limits_{h\rightarrow 0}\dfrac{h^{2}+2th-3h}{h}=\lim\limits_{h% \rightarrow 0}\dfrac{\NEG{h}\left( h+2t-3\right) }{\NEG{h}} \\ &=&\lim\limits_{h\rightarrow 0}\left( h+2t-3\right) =2t-3 \end{eqnarray*}% So if an object's location is given by $L\left( t\right) =t^{2}-3t$, then its velocity at time $t$ is $v\left( t\right) =2t-3$. \ If we look at this formula, $v\left( 7\right) =2\cdot 7-3=11$ and $v\left( 10\right) =2\cdot 10-3=17$ agrees with previous findings. \pagebreak \item The location function of an object is $L\left( t\right) =t^{3}$. $\ $ a) \ Compute the instantaneous velocity of the object at $t=4$ second. Solution: \ \begin{equation*} v\left( 4\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( 4+h\right) -L\left( 4\right) }{h} \end{equation*}% We compute first $L\left( 4+h\right) $% \begin{equation*} L\left( 4+h\right) =\left( 4+h\right) ^{3}=4^{3}+3\cdot 4^{2}h+3\cdot 4\cdot h^{2}+h^{3}=h^{3}+12h^{2}+48h+64 \end{equation*}% We also compute $L\left( 4\right) =64$. \ So now the velocity: \ \begin{eqnarray*} v\left( 4\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( 4+h\right) -L\left( 4\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{% h^{3}+12h^{2}+48h+64-64}{h}=\lim\limits_{h\rightarrow 0}\dfrac{% h^{3}+12h^{2}+48h}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}\left( h^{2}+12h+48\right) }{\NEG{h}} \\ &=&\lim\limits_{h\rightarrow 0}\left( h^{2}+12h+48\right) =48 \end{eqnarray*}% So at $t=4$, the velocity of the object is $48$. \ In short, $v\left( 4\right) =48.$ b) \ Compute the instantaneous velocity of the object at $t$. Solution: \ If we do that and we obtain an expression in terms of $t,$ then we created a new function, the velocity function.% \begin{equation*} v\left( t\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h} \end{equation*}% We compute first $L\left( t+h\right) $% \begin{equation*} L\left( t+h\right) =\left( t+h\right) ^{3}=t^{3}+3t^{2}h+3th^{2}+h^{3} \end{equation*}% So now the velocity: \ \begin{eqnarray*} v\left( t\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{% t^{3}+3t^{2}h+3th^{2}+h^{3}-t^{3}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{% 3t^{2}h+3th^{2}+h^{3}}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}\left( 3t^{2}+3th+h^{2}\right) }{\NEG{h}}=\lim\limits_{h\rightarrow 0}\left( 3t^{2}+3th+h^{2}\right) =3t^{2} \end{eqnarray*}% So if an object's location is given by $L\left( t\right) =t^{3}$, then its velocity at time $t$ is $v\left( t\right) =3t^{2}$. \ If we look at this formula, $v\left( 4\right) =3\cdot 4^{2}=48$ agrees with previous findings. \item The location function of an object is $L\left( t\right) =\sqrt{t}$. $\ $ a) \ Compute the instantaneous velocity of the object at $t=49$ second. Solution: \ \begin{equation*} v\left( 49\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( 49+h\right) -L\left( 49\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{49+h}-\sqrt{% 49}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{49+h}-7}{h} \end{equation*}% Since this is an indeterminate with radicals, we will use the conjugate of $% \sqrt{49+h}-7$.% \begin{eqnarray*} v\left( 49\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{49+h}-7}{h}% =\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{49+h}-7}{h}\cdot \dfrac{\sqrt{49+h}% +7}{\sqrt{49+h}+7} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{49+h-49}{h\left( \sqrt{49+h}+7\right) }% =\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}}{\NEG{h}\left( \sqrt{49+h}% +7\right) }=\lim\limits_{h\rightarrow 0}\dfrac{1}{\sqrt{49+h}+7}=\dfrac{1}{14% } \end{eqnarray*}% \ So at $t=49$, the velocity of the object is $\dfrac{1}{14}$. \ In short, $% v\left( 49\right) =\dfrac{1}{14}$.\pagebreak b) \ Compute the instantaneous velocity of the object at $t$. Solution: \ \begin{equation*} v\left( t\right) =\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{t+h}-\sqrt{t}% }{h} \end{equation*}% Since this is an indeterminate with radicals, we will use the conjugate of $% \sqrt{t+h}-\sqrt{t}$.% \begin{eqnarray*} v\left( 49\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{t+h}-\sqrt{t}}{% h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{t+h}-\sqrt{t}}{h}\cdot \dfrac{% \sqrt{t+h}+\sqrt{t}}{\sqrt{t+h}+\sqrt{t}} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{t+h-t}{h\left( \sqrt{t+h}+\sqrt{t}% \right) }=\lim\limits_{h\rightarrow 0}\dfrac{\NEG{h}}{\NEG{h}\left( \sqrt{t+h% }+\sqrt{t}\right) }=\lim\limits_{h\rightarrow 0}\dfrac{1}{\sqrt{t+h}+\sqrt{t}% }=\dfrac{1}{2\sqrt{t}} \end{eqnarray*}% So if an object's location is given by $L\left( t\right) =\sqrt{t}$, then its velocity at time $t$ is $v\left( t\right) =\dfrac{1}{2\sqrt{t}}$. \ If we look at this formula, $v\left( 49\right) =\dfrac{1}{2\sqrt{49}}=\dfrac{1}{% 14}$ agrees with previous findings. \item The location function of an object is $L\left( t\right) =\dfrac{1}{t}$% . $\ $ a) \ Compute the instantaneous velocity of the object at $t=5$ second. Solution: \ \begin{eqnarray*} v\left( 5\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( 5+h\right) -L\left( 5\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{1}{5+h}-% \dfrac{1}{5}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{~~\dfrac{5-\left( 5+h\right) }{5\left( 5+h\right) }~~}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left( \dfrac{1}{h}\cdot \dfrac{5-5-h}{% 5\left( 5+h\right) }\right) =\lim\limits_{h\rightarrow 0}\dfrac{-h}{5h\left( 5+h\right) }=\lim\limits_{h\rightarrow 0}\dfrac{-1}{5\left( 5+h\right) }=-% \dfrac{1}{25} \end{eqnarray*}% \ So at $t=5$, the velocity of the object is $-\dfrac{1}{25}$. \ In short, $% v\left( 5\right) =-\dfrac{1}{25}$. \ The negative sign here indicates that the object is moving downward at $t=5$ second. b) \ Compute the instantaneous velocity of the object at $t$. Solution: \ \ \begin{eqnarray*} v\left( t\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{L\left( t+h\right) -L\left( t\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{1}{t+h}-% \dfrac{1}{t}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{~~\dfrac{t-\left( t+h\right) }{t\left( t+h\right) }~~}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left( \dfrac{1}{h}\cdot \dfrac{t-t-h}{% t\left( t+h\right) }\right) =\lim\limits_{h\rightarrow 0}\dfrac{-h}{th\left( t+h\right) }=\lim\limits_{h\rightarrow 0}\dfrac{-1}{t\left( t+h\right) }=-% \dfrac{1}{t^{2}} \end{eqnarray*}% So if an object's location is given by $L\left( t\right) =\dfrac{1}{t}$, then its velocity at time $t$ is $v\left( t\right) =-\dfrac{1}{t^{2}}$. \ If we look at this formula, $v\left( 5\right) =-\dfrac{1}{25}$ agrees with previous findings. \end{enumerate} \vspace{1.3in} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}