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# Find the arc length function for the curve y = 2x^{3/2} with starting point P(16, 128)?

**Solution:**

Given, y = \(2x^{\frac{3}{2}}\)

Starting point P(16, 128)

Differentiating with respect to x,

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

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

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

The arc length can be found by using the formula,

\(L=\int_{x_{0}}^{t}\sqrt{[1+(\frac{dy}{dx})^2]dx}\)

Putting the value of dy/dx in the formula,

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

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

Let 1 + 9x = z

9dx = dz

dx = dz/9

At x = 16,

z = 1 + 9(16)

z = 1 + 144

z = 145

At x = t,

z = 1 + 9t

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

\(L=\int_{145}^{1+9t}\frac{z^{\frac{1}{2}}}{9}\, dz\)

\(L=[\frac{z^{\frac{3}{2}}}{9.\frac{3}{2}}]_{145}^{1+9t}\)

\(L=\frac{2}{27}[z^{\frac{3}{2}}]_{145}^{1+9t}\)

\(L=\frac{2}{27}[(1+9t)^{\frac{3}{2}}-(145)^{\frac{3}{2}}]\)

Therefore, the arc length of the given curve is \(L=\frac{2}{27}[(1+9t)^{\frac{3}{2}}-(145)^{\frac{3}{2}}]\)

## Find the arc length function for the curve y = 2x^{3/2} with starting point P(16, 128)?

**Summary:**

The arc length function for the curve y = 2x^{3/2} with starting point P(16, 128) is \(L=\frac{2}{27}[(1+ 9t)^{\frac{3}{2}}-(145)^{\frac{3}{2}}]\)

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