Which of the following functions define the volume of a cube? x 0 2 4 6 y 6 5 4 3 17 a. V = 3s, where s is the length of the edge b. V = s 3 , where s is the length of the edge c. V = 2s 3 , where s is the length of the edge d. V = s 3 , where s is the length of the edge
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Answer:
Let x be the length of a side, V be the volume, and S be the surface area of the cube.
Then, V=x
3
and S=6x
2
It is given that
dt
dV
=8cm
3
/s.
Then, by using the chain rule, we have:
∴8=
dt
dV
=
dt
d
(x
3
)⋅
dx
d
=3x
2
⋅
dt
dx
⇒
dt
dx
=
3x
2
8
.........(1)
Now,
dt
dS
=
dt
d
(6x
2
)⋅
dx
d
=(12x)⋅
dt
dx
[By chain rule]
=12x⋅
dt
dx
=12x⋅(
3x
2
8
)=
x
32
Thus, when x=12 cm,
dt
dS
=
12
32
cm
2
/s=
3
8
cm
2
/s.
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of
3
8
cm
2
/s.
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