A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?a)3/8*(rate of change of area of any face of the cube)b)3/4*(rate of change of area of any face of the cube)c)3/10*(rate of change of area of any face of the cube)d)3/2*(rate of change of area of any face of the cube)Correct answer is option 'D'. Can you explain this answer?

Question

Answer

The volume of a cube is given by the formula V = s^3, where s is the length of one side of the cube. We are given that the cube changes its volume while maintaining its shape, so the side length also changes. Let's denote the rate of change of the side length as ds/dt, and the rate of change of the volume as dV/dt.

To find the rate of change of the volume, we can differentiate the volume formula with respect to time:

dV/dt = d/dt (s^3)

To simplify the differentiation, we can use the chain rule:

dV/dt = 3s^2 * ds/dt

Now we need to relate the rate of change of the volume to the rate of change of the area of any face of the cube. The area of a face of the cube is given by the formula A = s^2. We can differentiate this formula with respect to time to find the rate of change of the area:

dA/dt = d/dt (s^2)

Using the chain rule again:

dA/dt = 2s * ds/dt

We can rewrite this equation as:

ds/dt = (1/2s) * dA/dt

Substituting this expression for ds/dt into the equation for dV/dt, we get:

dV/dt = 3s^2 * [(1/2s) * dA/dt]

Simplifying:

dV/dt = 3/2 * s * dA/dt

Since the volume of the cube is 1 (given in the question as 'unit volume'), we have s = 1. Substituting this value into the equation, we get:

dV/dt = 3/2 * dA/dt

Therefore, the rate of change of the volume is equal to 3/2 times the rate of change of the area of any face of the cube. Hence, the correct answer is option 'D' - 3/2 times the rate of change of the area of any face of the cube.

Answer

Let x be the length of a side of the cube.
If v be the volume and s the area of any face of the cube, then
v = x³ and s = x²
Thus, dv/dt = dx³/dt = 3x²(dx/dt)
And ds/dt = dx²/dt = 2x(dx/dt)
Now, (dv/dt)/(ds/dt) = 3x/2
Or, dv/dt = (3x/2)(ds/dt)
Now, for a cube of unit volume we have,
v = 1
⇒ x = 1 [as, x is real]
Therefore, for a cube of unit volume [i.e. for x = 1], we get,
dv/dt = (3/2)(ds/dt)
Thus the rate of change of volume = 3/2*(rate of change of area of any face of the cube)

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