Question

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

Answer 1

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.