Question

The edge of a cube is increasing at the rate of 0.3 cm/s, the rate of change of its surface area
when edge is 3 cm is
(a) 10.8cm s
(b) 11.2cm's
c) 12.4cm s
(d) none of these​

Answer 1

Given:-

Edge of cube is increasing at the rate of 0.3 cm/s.

Total Surface Area of Cube is 6 × Side².

Edge is at 3 cm.

To Find:-

Rate of change of its surface area when its edge is 3 cm.

Solution:-

Here, Let the edge of cube be x

and, Total Surface Area be a.

Given that,

Edge of cube is increasing at the rate of 0.3 cm/s.

Thus, $\dfrac{dx}{dt}$ = 0.3 cm/s                     -------- (i)

We have to calculate the rate of change of its Total Surface Area.

i.e. We need to calculate $\dfrac{dv}{dt}$ when x = 3 cm.

We know that,

Total Surface Area of Cube = ( 6 × Edge² )

i.e.

                 T.S.A = 6x²

Differentiate with respect to time,

⇒  $\dfrac{da}{dt} = \dfrac{d(6x^{2})}{dt}$

⇒  $\dfrac{da}{dt} = \dfrac{d(6x^{2})}{dt}$ × $\dfrac{dx}{dx}$

⇒  $\dfrac{da}{dt} = \dfrac{d(6x^{2})}{dx}$ × $\dfrac{dx}{dt}$

⇒  $\dfrac{da}{dt}$ $= 6$ × $3x^{2-1}$ × $\dfrac{dx}{dt}$           [ Note:- $\dfrac{d(kx^{n})}{dx}$ = $knx^{n-1}$ ]

⇒  $\dfrac{da}{dt} = 12x$ × $0.3$         [ From Equation (i) ]

⇒  $\dfrac{da}{dt} = 3.6x$

When x = 3 cm

⇒  $\dfrac{da}{dt}$ = 3.6 × 3

⇒  $\dfrac{da}{dt}$ = 10.8

Since, Total Surface area will be in cm² and time will be in sec.

⇒  $\dfrac{da}{dt}$ = 10.8 $\dfrac{cm^{2}}{sec}$

⇒  $\dfrac{da}{dt}$ = 10.8 cm²/sec

Hence, The rate of change of its Total Surface Area is 10.8 cm²/sec.

∴ Option ( a ) is Correct.

Some Important Terms:-

• $\dfrac{d(kx^{n})}{dx}$ = $knx^{n-1}$

• $\dfrac{d(x^{n})}{dx} = nx^{n-1}$

• $\dfrac{d(k)}{dx} = 0$

• $\dfrac{d(x)}{dx} = 1$