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# Find the Exact Length of the Curve. x = 1/3 √y (y − 3), 1 ≤ y ≤ 9

We will be using the formula of the exact length of the curve to solve this.

## Answer: The Exact Length of the Curve x = 1/3 √y (y − 3), 1 ≤ y ≤ 9 is 32/3 units.

Let's solve this step by step.

**Explanation:**

Given, x = (1/3) √y (y − 3), 1 ≤ y ≤ 9

Length of the curve x = f(y) from y = a to y = b is given by: \( \int_a^b √1 + [f′(y)]^2 dy\)

Let's find the first derivative of x.

x = (1/3) √y (y − 3)

dx/dy = (1/3) y^{1/2} + (1/3) (1/2√y) (y − 3)

dx/dy = (1/3) [2y + y - 3] / 2√y

dx/dy = (1/3) [3y - 3] / 2√y

dx/dy = (y - 1)/2√y

Length of curve = \(\int_1^9 \sqrt{1 + [f′(y)]^2} dy\)

= \(\int_1^9 \sqrt{1 + (\dfrac{y-1}{2\sqrt{y}}})^2 dy\)

= \(\int_1^9 \sqrt{1 + \dfrac{(y-1)^2}{4y}} dy\)

= \(\int_1^9 \sqrt{\dfrac{4y + (y-1)^2}{4y}} dy\)

= \(\int_1^9 \sqrt{\dfrac{4y + y^2 - 2y + 1}{4y}} dy\)

= \(\int_1^9 \sqrt{\dfrac{2y + y^2 + 1}{4y}} dy\)

= \(\int_1^9 \sqrt{\dfrac{(y+1)^2}{4y}} dy\)

= \(\int_1^9 \dfrac{y+1}{2\sqrt{y}}dy\)

= \(\int_1^9 \dfrac{\sqrt{y}}{2}+\dfrac{1}{2\sqrt{y}}dy\)

= [(y)^{3/2 }/3 + √y]_{1}^{9}

= [(9)^{3/2 }/3 + √9] - [(1)^{3/2 }/3 + √1]

= [27/3 + 3] - [1/3 + 1]

= 12 - 4/3

= 32/3

### Hence, the exact length of the curve x = 1/3 √y (y − 3), 1 ≤ y ≤ 9 is 32/3 units.

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