# Find the arc length function for the curve with starting point y = 2x^{3/2} with starting point P\(_0\)(1, 2)

A part of a curve is called an arc and arc length is defined as the distance along the curved line that makes up the arc.

## Answer: The arc length function for the curve with starting point y = 2x^{3/2} with starting point P\(_0\)(1, 2) is L(t) = (2/27)[(1+9t)^{3/2} - 10^{3/2}]

Let's go through the steps to find the arc length of a given curve.

**Explanation:**

Given function ⇒ y = 2x^{3/2}

Point = P(1, 2)

Firstly, we have to differentiate the given function (y = 2x^{3/2}) with respect to “x”

dy/dx = d(2x^{3/2})/dx

dy/dx = 2(3/2)x^{(1/2)}

dy/dx = 3x^{(1/2)} ------------- (1)

To find arc length, we use the following formula for the length of the arc(L),

L = \(\int_{x_0}^{t}\sqrt{(1+ (\dfrac{dy}{dx})^2 )} dx\)

Putting the value of dy/dx in length of curve formula from (1)

L= \((\int_{1}^{t}\sqrt{(1+ (3x^{1/2})^2)}dx\)

L = \(\int_{1}^{t}\sqrt{(1+9x)}dx\)

Substitute 1 + 9x = z and 9dx = dz ⇒ dx = dz/9

At x = 1, z = 10 and x = t, z = 1 + 9t

Putting the value of “z” and “dz” in the above equation, we get:

L = \(\int_{10}^{1+9t}\dfrac{z^{1/2}}{9}dz\)

L = \(\left[\dfrac{z^{3/2}}{9.\dfrac{3}{2}}\right]_{10}^{1+9t}\)

L = \(\dfrac{2}{27}[z^{3/2}]_{10}^{1+9t}\)

L = \(\dfrac{2}{27}[(1+9t)^{3/2} -10^{3/2}]\)

L = \(\dfrac{2}{27}[(1+9t)^{3/2} - 10\sqrt{10}]\)