0.06 x3 m3
0.6 x3 m3
0.09 x3 m3
0.9 x3 m3
The volume of a cube (V) of side x is given by V =x3
\(\frac{dv}{(\frac{dv}{dx})}\)Δx
=(3x2)Δx
=(3x2)(0.03x) [As 3% of x is 0.03]
=0.09x3m3
Hence, the approximate change in the volume of the cube is 0.09x3 m3.
The correct answer is C.
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
The rate of change of quantities can be expressed in the form of derivatives. Rate of change of one quantity with respect to another is one of the major applications of derivatives. The rate of change of a function with respect to another quantity can also be done using chain rule.
If some other quantity ‘y’ causes some change in a quantity of certain ‘x’, in view of the fact that an equation of the form y = f(x) gets always satisfied, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
This is also called the Average Rate of Change.
If the rate of change of a function is to be defined at a specific point i.e. a specific value of ‘x’, it is known as the Instantaneous Rate of Change of the function at that point. From the definition of the derivative of a function at a point, we have
From this, it is to be concluded that the instantaneous Rate of Change of the function is represented by the derivative of a function. From the rate of change formula, it represents the case when Δx → 0. Thus, the rate of change of ‘y’ with respect to ‘x’ at x = x0 = (dy/dx)x = x0