Question

Find the arc length of the curve on the given interval. $$ x=t, \quad y=\frac{t^{5}}{10}+\frac{1}{6 t^{3}} \quad 1 \leq t \leq 2 $$

Short answer

The arc length of the curve on the given interval is 1.

Step-by-step solution

Recognize the formula

The formula for the arc length \( s \) of a curve defined by parametric equations \( x = f(t) \) and \( y = g(t) \) on the interval \( a \leq t \leq b \) is: \[ s = \int_a^b \sqrt { [f'(t)]^2 + [g'(t)]^2 } dt \] Here, \( x = t \) and \( y = \frac{t^{5}}{10} + \frac{1}{6 t^{3}}, \) and the interval of \( t \) is from 1 to 2.

Compute derivatives%

Find the derivatives of \( x \) and \( y \) with respect to \( t \). For \( x \), \( x' = 1 \). For \( y \), by the power rule, \( y' = \frac{1}{2}t^{4} - \frac{1}{2}t^{4} \).

Plug into the formula

Plug the derivatives into the formula for the length: \[ s= \int_1^2 \sqrt{(1)^2 + (\frac{1}{2}t^{4}- \frac{1}{2}t^{4})^2} dt \]. Simplifying that inside the square root yields: \[ s = \int_1^2 \sqrt{1 + t^8 - t^8} dt \]. Simplified further, this becomes: \[ s= \int_1^2 \sqrt{1} dt \].

Evaluate the integral

The integral of \( s = \int_1^2 dt \) is simply \( t \), evaluated from 1 to 2. Subtracting the lower limit from the upper limit gives the arc length. \( s = 2 - 1 = 1 \)