Question

In Exercises 5 and \(6,\) use the information to evaluate and compare \(\Delta y\) and \(d y\). $$ y=\frac{1}{2} x^{3} \quad x=2 \quad \Delta x=d x=0.1 $$

Short answer

For the given function \(y=\frac{1}{2} x^{3}\), when \(x=2\), \(Δx=0.1\), the differential \(dy\) is \(0.6\) and the increment \(Δy\) is \(0.6415\). Hence, \(Δy\) > \(dy\) for the given values.

Step-by-step solution

Differentiate the Given Function

Given that \(y=\frac{1}{2} x^{3}\). The first derivative of \(y\) with respect to \(x\) is \(y' = \frac{3}{2}x^{2}\).

Compute dy

Now calculate the approximate change in \(y\), denoted by \(dy\). From the formula for differentials \(dy = y' dx\), we find, \(dy = \frac{3}{2}*2^{2}*0.1 = 0.6\).

Compute Δy

The increment in \(y\) (Δy) is computed by the formula \(Δy = f(x+Δx) - f(x)\). Substituting the given values into \(y=\frac{1}{2} x^{3}\) results in \(Δy = \frac{1}{2} *(2.1)^{3} - \frac{1}{2} *2^{3} = 0.6415\).

Compare Δy and dy

Both Δy and dy represent approximations of how \(y\) changes as \(x\) increases by \(0.1\). However, Δy is a more accurate approximation than dy, as it uses the actual function \(y=\frac{1}{2} x^{3}\). In this case, we can see that Δy = 0.6415 is slightly greater than dy = 0.6