Question

Compare the values of \(d y\) and \(\Delta y\). \(y=0.5 x^{3} \quad x=2 \quad \Delta x=d x=0.1\)

Short answer

The exact change in the function \(dy\) is 0.6 and the approximated change in the function \(\Delta y\) is 0.6305.

Step-by-step solution

Differentiation

The first step involves differentiation of the given function \(y\) with respect to \(x\). Thus, this gives: \(y' = dy/dx = 0.5 \times 3x^2 = 1.5x^2\).

Calculate the value of \(dy\)

Using the value \(x=2\) and \(dx=0.1\), the exact change in \(y\) i.e. \(dy\) can be calculated using the formula: \(dy = y'(2) \times dx = 1.5 \times 2^2 \times 0.1 = 0.6\). Thus, \(dy\) equals 0.6.

Calculate the value of \(\Delta y\)

Now we will calculate \(\Delta y\), the approximate change in \(y\), which is given by: \(\Delta y = (y(x + dx) - y(x))\). Substituting the given values: \(\Delta y = y((2+0.1))-y(2) = 0.5 \times (2.1)^3 - 0.5 \times 2^3 = 0.5 \times 9.261-0.5 \times 8 = 0.6305 - 4 = 0.6305\). Thus, \(\Delta y\) equals 0.6305.