Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 17

Answer

$$\eqalign{ & \left( a \right){\text{Graph velocity}} \cr & \left( b \right)s\left( t \right) = 6t - {t^2},{\text{ }}t \geqslant 0 \cr & \left( c \right){\text{Graph position}} \cr} $$

Work Step by Step

$$\eqalign{ & v\left( t \right) = 6 - 2t\,{\text{on the interval }}\left[ {0,5} \right], \cr & {\text{initial position }}s\left( 0 \right) = 0 \cr & \cr & \left( a \right){\text{Graph shown below }}\left( {{\text{Use GeoGebra}}} \right) \cr & *{\text{The motion of the object is in the positive direction when }} \cr & v\left( t \right) > 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) > 0{\text{ on the interval }}\left[ {0,3} \right) \cr & *{\text{The motion of the object is in the negative direction when }} \cr & v\left( t \right) < 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) < 0{\text{ on the interval }}\left( {3, 5} \right] \cr & \cr & \left( b \right){\text{ Position for }}t \geqslant 0 \cr & *{\text{Using the antiderivative method}} \cr & {\text{Recall that the velocity of an object at time }}t{\text{ is }} \cr & v\left( t \right) = s'\left( t \right) \cr & s'\left( t \right) = v\left( t \right) \cr & \frac{{ds}}{{dt}} = 6 - 2t \cr & {\text{Finding the antiderivative}} \cr & s\left( t \right) = \int {\left( {6 - 2t} \right)} dt \cr & s\left( t \right) = 6t - {t^2} + C,{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Using the initial condition }}s\left( 0 \right) = 0 \cr & 0 = 6\left( 0 \right) - {\left( 0 \right)^2} + C \cr & C = 0 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & s\left( t \right) = 6t - {t^2},{\text{ for }}t \geqslant 0 \cr & \cr & *{\text{Using the Fundamental theorem of calculus }} \cr & {\text{Let }}v\left( x \right) = 6 - 2x,{\text{ determine for }}t \geqslant 0 \cr & {\text{Use theorem 6}}{\text{.1 }}\left( {{\text{page 401}}} \right) \cr & \underbrace {s\left( t \right)}_{\scriptstyle {\text{position}} \atop \scriptstyle {\text{at }}t} = \underbrace {s\left( 0 \right)}_{\scriptstyle {\text{initial}} \atop \scriptstyle {\text{position}}} + \underbrace {\int_0^t {v\left( x \right)} dt}_{\scriptstyle {\text{displacement}} \atop \scriptstyle {\text{over }}\left[ {0,t} \right]} \cr & {\text{Therefore}}{\text{,}} \cr & s\left( t \right) = 0 + \int_0^t {\left( {6 - 2x} \right)} dx \cr & {\text{Integrating}} \cr & s\left( t \right) = \left[ {6x - {x^2}} \right]_0^t \cr & s\left( t \right) = 6t - {t^2},{\text{ }}t \geqslant 0 \cr & \cr & \left( c \right){\text{Graph the position function shown below }}\left( {{\text{Use GeoGebra}}} \right) \cr} $$
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