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# Find the Exact Length of the Curve. x = 1/3 √y (y − 3), 9 ≤ y ≤ 25

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

## Answer: The Exact Length of the Curve x = 1/3 √y (y − 3), 9 ≤ y ≤ 25 is 92/3.

Let's solve this step by step.

**Explanation:**

Given, x = 1/3 √y (y − 3), 9 ≤ y ≤ 25

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

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 = \(\int_9^{25} \sqrt{1 + [f′(x)]^2} dx \)

=\(\int_9^{25} \sqrt{1 + [\frac{y-1}{2\sqrt{y}}]^2} dx \)

=\(\int_9^{25} \sqrt{1 + \dfrac{(y-1)^2}{4y}} dx \)

=\(\int_9^{25}\sqrt{ \dfrac{4y + (y-1)^2}{4y}} dx \)

=\(\int_9^{25}\sqrt{ \dfrac{4y + y^2 + 1-2y}{4y}} dx \)

=\(\int_9^{25}\sqrt {\dfrac{2y + y^2+1}{4y}} dx \)

=\(\int_9^{25} \sqrt{\dfrac{ (y+1)^2}{4y}} dx \)

= \(\int_9^{25} \dfrac{ (y+1)}{2\sqrt{y}}dx \)

=\(\int_9^{25} \dfrac{\sqrt{y}}{2}+\dfrac{1}{2\sqrt{y}}dx \)

= [(y)^{3/2 }/3 + √y]_{9}^{25}

= [(25)^{3/2 }/3 + √25] - [(9)^{3/2 }/3 + √9]

= [125/3 + 5] - [27/3 + 3]

= 98/3 - 2

= 92/3

### Hence, the exact length of the curve x = 1/3 √y (y − 3), 9 ≤ y ≤ 25 is 92/3.

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