Determine the instantaneous change in water volume at t = 4 ( 2004 ) t = 4(2004) and t = 9 ( 2009 ) t = 9(2009) , if the volume of the water in a reservoir is modeled by the function V ( t ) = 500 e 0.05 t V(t)=500e^{0.05t} , where t t represents the number of years after 2000 2000 . At which of these times was the instantaneous change in water volume greater?

V ′ ( 4 ) = 30.54 V^{\prime}(4)=30.54 ; V ′ ( 9 ) = 39.21 V^{\prime}(9)=39.21 ; 2009 2009

V ′ ( 4 ) = 78.42 V^{\prime}(4)=78.42 ; V ′ ( 9 ) = 30.54 V^{\prime}(9)=30.54 ; 2009 2009

V ′ ( 4 ) = 30.54 V^{\prime}(4)=30.54 ; V ′ ( 9 ) = 78.42 V^{\prime}(9)=78.42 ; 2004 2004

V ′ ( 4 ) = 39.21 V^{\prime}(4)=39.21 ; V ′ ( 9 ) = 30.54 V^{\prime}(9)=30.54 ; 2004 2004

Table of contents

4. Applications of Derivatives

Related Rates

4. Applications of Derivatives

Related Rates: Study with Video Lessons, Practice Problems & Examples

3PRACTICE PROBLEM

ANSWERS OPTIONS

$V prime of 4 equals 78.42$ ; $V prime of 9 equals 30.54$ ; $2009$

$V prime of 4 equals 30.54$ ; $V prime of 9 equals 78.42$ ; $2004$

$V prime of 4 equals 39.21$ ; $V prime of 9 equals 30.54$ ; $2004$

$V prime of 4 equals 30.54$ ; $V prime of 9 equals 39.21$ ; $2009$