## Thomas' Calculus 13th Edition

a. $2$, $2.5$, $0.5$ gal/day. b. $1.25$, $3.33$, $0.75$ gal/day. c. $4$ days, $3.33$ gal/day (Other approximations are possible.)
a. Read the following values from the graph: $A(0)=16 , A(3)=10 , A(5)=3.5 , A(7)=1.5 , A(10)=0$ gal. The average rate of gasoline consumption over the time interval $[0,3]$ is given by $C_1=\frac{A(3)-A(0)}{3-0}=\frac{10-16}{3}=-2$ gal/day (negative sign indicates a decreasing trend or consumption). For the interval $[0,5]$, it is given by $C_2=\frac{A(5)-A(0)}{5-0}=\frac{3.5-16}{5}=-2.5$ gal/day. And for the interval $[7,10]$, it is given by $C_3=\frac{A(10)-A(7)}{10-7}=\frac{0-1.5}{3}=-0.5$ gal/day. b. The instantaneous rate of gasoline consumption at time $t$ is the slope of the curve at that time, and it can be estimated by drawing a tangent line to the curve at each point and measuring the slope (see figure). We can find that at $t=1$, $C(1)\approx-1.25$ gal/day; at $t=4$, $C(4)\approx-3.33$ gal/day; and at $t=8$, $C(8)\approx-0.75$ gal/day. c. The maximum rate of gasoline consumption happened at about $t=4$ days, and its value is about $-3.33$ gal/day. In terms of consumption which already includes the meaning of negative change, we can write the absolute values as our answers.