Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ x=\frac{1}{3} \sqrt{y}(y-3), \quad 1 \leq y \leq 4 $$

Short answer

The arc length of the graph of the function over the indicated interval [1, 4] can be found by evaluating the integral \(\int_{1}^{4}\sqrt{1+\left(\frac{1}{3}[1/2\sqrt{y}(y-3)+\sqrt{y}]\right)^2}dy\)

Step-by-step solution

Compute the derivative

We first compute the derivative \( dx/dy \) of the given function. Let's differentiate \( x=\frac{1}{3}\sqrt{y}(y-3) \) with respect to y. Using the product rule gives \( dx/dy=\frac{1}{3}[(1/2\sqrt{y}(y-3)+\sqrt{y}(1)] \).

Square the derivative

Now square the derivative computed in the previous step to get \((dx/dy)^2 = [\frac{1}{3}(1/2\sqrt{y}(y-3)+\sqrt{y})]^2\)

Insert into the arc length formula

The formula for the arc length includes the square root of 1 plus the derivative squared. Inserting, it leads to the integral \(\int_{1}^{4}\sqrt{1+\left(\frac{1}{3}[1/2\sqrt{y}(y-3)+\sqrt{y}]\right)^2}dy\)

Evaluate the integral

Now that we've formulated the integral, evaluate it from the limits 1 to 4 to get the arc length of the graph of the function over the indicated interval.