\nopagenumbers \magnification 1600 \hsize 6.7truein \hoffset -.3truein \vsize 9.7truein \voffset -.3truein \parskip 10pt \parindent 0pt Suppose $d(t) = 40t$ represents miles traveled after $t$ hours. \vskip 1in \vfill Average velocity is $\underline{\hskip 1cm}$ Instantaneous velocity at $t = t_0$ is $\underline{\hskip 1cm}$ \vskip 10pt \hrule Suppose $d(t) = t^2$ represents miles traveled after $t$ hours. \vskip 1in \vfill $\matrix{ t_0 & \hbox{change in time} & \hbox{change in distance} & \hbox{average velocity} \cr & \hbox{btwn } t = 0 \hbox{ and } t = t_0 & \hbox{btwn } t = 0 \hbox{ and } t = t_0 & \hbox{btwn } t = 0 \hbox{ and } t = t_0 \cr ~ & ~ & ~ & \cr 2 & 2 - 0 & 2^2 - 0^2 & { 2^2 - 0^2 \over 2 - 0} = 2 ~ & ~ & ~ & \cr 1 & 1 - 0 & 1^2 - 0^2 & { 1^2 - 0^2 \over 1 - 0} = 1 ~ & ~ & ~ & \cr .5 & .5 - 0 & (.5)^2 - 0^2 & { (.5)^2 - 0^2 \over .5 - 0} = .5 ~ & ~ & ~ & \cr .1 & .1 - 0 & (.1)^2 - 0^2 & { (.1)^2 - 0^2 \over .1 - 0} = .1 ~ & ~ & ~ & \cr .01 & .01 - 0 & (.01)^2 - 0^2 & { (.01)^2 - 0^2 \over .01 - 0} = .01 ~ & ~ & ~ & }$ \vskip 10pt Instantaneous velocity at $t = 0$ is $\underline{\hskip 1cm}$ \eject Suppose $d(t) = t^2$ represents miles traveled after $t$ hours. \vfill \vskip 1.5in \vfill $\matrix{ t_0 & \hbox{change in time} & \hbox{change in distance} & \hbox{average velocity} \cr & \hbox{btwn } t = 2 \hbox{ and } t = t_0 & \hbox{btwn } t = 2 \hbox{ and } t = t_0 & \hbox{btwn } t = 2 \hbox{ and } t = t_0 \cr ~ & ~ & ~ & \cr 4 & 4 - 2 & 4^2 - 2^2 & { 4^2 - 2^2 \over 4 - 2} = 6 ~ & ~ & ~ & \cr 3 & 3 - 2 & 3^2 - 2^2 & { 3^2 - 2^2 \over 3 - 2} = 5 ~ & ~ & ~ & \cr 2.5 & 2.5 - 2 & (2.5)^2 - 2^2 & { (2.5)^2 - 2^2 \over 2.5 - 2} = 4.5 ~ & ~ & ~ & \cr 2.1 & 2.1 - 2 & (2.1)^2 - 2^2 & { (2.1)^2 - 2^2 \over 2.1 - 2} = 4.1 ~ & ~ & ~ & \cr 1.9 & 1.9 - 2 & (1.9)^2 - 2^2 & { (1.9)^2 - 2^2 \over 1.9 - 2} = 3.9 ~ & ~ & ~ & \cr 1.5 & 1.5 - 2 & (1.5)^2 - 2^2 & { (1.5)^2 - 2^2 \over 1.5 - 2} = 3.5 ~ & ~ & ~ & \cr 1 & 1 - 2 & 1^2 - 2^2 & { 1^2 - 2^2 \over 1 - 2} = 3 ~ & ~ & ~ & \cr }$ \vskip 10pt Instantaneous velocity at $t = 2$ is $\underline{\hskip 1cm}$ \vskip 10pt \vskip 10pt SLOPE OF SECANT LINE = AVERAGE VELOCITY SLOPE OF TANGENT LINE = INSTANTANEOUS VELOCITY in general, SLOPE = RATE OF CHANGE SLOPE OF SECANT LINE = AVERAGE RATE OF CHANGE SLOPE OF TANGENT LINE = INSTANTANEOUS RATE OF CHANGE \end{document}