UPSC NDA Math Model Paper 9

At any instant t, let the length of each side of the cube be x, V be its volume and S be its surface area.
Given: dV/dt = 7 cm‌3/sec
As we know that, volume of a cube is given by: a , where a is the length of each side of the cube.
i.e V=x3
Now by differentiating V with respect to t, we get
⇒

dV

dt

=

dV

dx

.

dx

dt

⇒

dV

dx

=

d(x3)

dx

=3x2
By substituting the value of dV/dx in dV/dt we get
⇒

dV

dt

=3x2.

dx

dt

Now by substituting the dV/dt = 7 cm‌3/sec in the above equation we get,
⇒7=3x2.

dx

dt

⇒

dx

dt

=

7

3x2

As we know that, surface area of a cube is given by: 6a2 , where a is the length of each side of the cube.
i.e S = 6a2
Now by differentiating S with respect to t we get,
⇒

dS

dt

=

dS

dx

.

dx

dt

⇒

dS

dx

=

d(6x2)

dx

=‌12x
Now by substituting the value of dS/dx in dS/dt, we get
⇒

dS

dt

=12x.

dx

dt

By substituting the value of dx∕dt=7∕3x2
in the abve equation we get
⇒

dS

dt

=12x.‌

7

3x2

=

28

x

Now by substituting x = 12 cm in the above equation we get,
⇒[

dS

dt

]x=12=

28

12

=‌2

1

3

‌cm2∕sec