This preview shows page 1. Sign up to view the full content.
0.2
2.00 s
3
g
1kg
60 s
4.00(5.00)
3.00 g/s
0.101g/s
0.101
s 1000 g 1 min
6.05 10
kg/min.
t
dm
dt
−
=
−
ªº
=
−
=−
⋅
⋅
¬¼
×
30. To solve the problem, we note that the first derivative of the function with respect to
time gives the rate. Setting the rate to zero gives the time at which an extreme value of
the variable mass occurs; here that extreme value is a maximum.
(a) Differentiating
0.8
( )
5.00
3.00
20.00
mt
t
t
+
with respect to
t
gives
0.2
4.00
3.00.
dm
t
dt
−
The water mass is the greatest when
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Mass

Click to edit the document details