91Ó°ÊÓ

In calculus, differentiating functions is a fundamental process that allows us to find the rate at which a function is changing at any given point. Differentiation involves calculating the derivative of a function, which represents the slope of the tangent line to the function's graph at a particular point. For the function in the exercise, given by \( y = \frac{3}{2} x^{2 / 3}+4 \), the derivative is found by applying the power rule: \( f'(x) = \frac{d}{dx}\left(\frac{3}{2} x^{2 / 3}\right) \).

Through a series of simplifications, the derivative is obtained as \( f'(x) = x^{-1/3} \). The derivative tells us how the function's rate of change in \(y\) with respect to a change in \(x\) and is an essential component when calculating the arc length of the curve.

The arc length of a curve is a measure of the distance along the curve from one point to another. The arc length formula for a function \( y = f(x) \) from \(x=a\) to \(x=b\) is given by: \( L = \int_a^b \sqrt{1 + (f'(x))^2} dx \).

This formula originates from the Pythagorean theorem, considering an infinitesimally small segment of the curve as a right triangle with sides \( dx \) and \( dy \) and the hypotenuse representing an infinitesimally small part of the arc. By integrating the length of these segments over the interval, we find the total length of the arc. In the exercise provided, after differentiating the function, we plug in the derivative squared into the formula to set up the integral that will yield the arc length.

Evaluating integrals is a core operation in calculus, particularly when finding areas, volumes, and, as seen in our example, arc lengths. It involves determining the total accumulation of quantities, often requiring techniques like substitution to simplify the integral before solving it.

In the case of the arc length exercise, we encountered an integral that is not straightforward to evaluate. By using a substitution \( u = x^{2/3} \), we simplify the integral considerably, changing the variable and the bounds accordingly. The new bounds \( u=1 \) and \( u=9 \) correspond to the original \( x \) bounds of \( x=1 \) and \( x=27 \). The resulting integral, \( \frac{2}{3}\int_{1}^{9}\frac{du}{\sqrt{u}} \), is much simpler to evaluate. After solving, we get the final arc length, showcasing the crucial role of integral evaluation in solving such problems.