The edge of a cube is increaing at the rate of 0.05 centimeters per second. In terms of the edge of the cube, s, what is the rate of change of the volume of the cube, in cubic centimeters per second?
a. 0.05^3
b. 0.05s^2
c. 0.05s^3
d. 0.15s^2
e. 3s^2
2006-12-31 03:07:55 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
V=s^3
dV/dt=dV/ds*ds/dt
=3s^2*0.05 cc per sec
=0.15s^2
choice d
2006-12-31 03:12:57 · answer #1 · answered by raj 7 · 1⤊ 0⤋
the edge of cube is s, thus the volume is s^3
the edge after increasing s + 0..05, thus the volume is (s+0.05)^3
the rate of change of volume
(s+0.05)^3 - s^3 = 0.05(s^2 + 0.05s + 0.0025)
2006-12-31 11:16:48 · answer #2 · answered by James Chan 4 · 0⤊ 1⤋
ds/dt=.05
v=s^3
so dv/dt=3s^2*ds/dt
dv/dt=.15s^2
choice D
2006-12-31 13:15:34 · answer #3 · answered by The Q Meister 2 · 0⤊ 0⤋
ds/dt= 0.05 cm/s
dV/dt= dV/ds*ds/dt
V= s^3
thus,
dV/ds= 3s^2
thus dV/dt = 3s^2 * 0.05 = 0.15s^2
so ur answer is (d)
2006-12-31 11:37:32 · answer #4 · answered by Riddhi 2 · 1⤊ 0⤋
None of those is right and it has nothing to do with calculus.
2006-12-31 11:23:48 · answer #5 · answered by Nomadd 7 · 0⤊ 1⤋