Question

Write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ x=2 t-t^{2}, \quad y=2 t^{3 / 2} \quad 1 \leq t \leq 2 $$

Short answer

The integral that represents the arc length of the curve on the interval from 1 to 2 is \(L = \int_{1}^{2} \sqrt{4t^{2} - 8t + 13} dt\).

Step-by-step solution

Formulate Curve's Derivatives

Firstly, differentiate \(x\) and \(y\) with respect to \(t\). \(x'\) is the derivative of \(x\) with respect to \(t\), and \(y'\) is the derivative of \(y\) with respect to \(t\). For \(x = 2t - t^2\), \(x'\) is equal to \(2 - 2t\). For \(y = 2t^{3/2}\), \(y'\) is equal to \(3t^{1/2}\).

Implement Arc Length Formula

The arc length for the parametric curve is given by, \[L = \int_{a}^{b} \sqrt{(dx/dt)^2 + (dy/dt)^2} dt\]where \(a\) and \(b\) are the given interval values. Substitute the values of \(x'\) and \(y'\) obtained in the above step into this formula. So,\[L= \int_{1}^{2} \sqrt{(2 - 2t)^2 + (3t^{1/2})^2} dt\]

Simplify Integral

Lastly, simplify the integral's square root term. Simplify \((2 - 2t)^2\) to \(4t^2 - 8t + 4\) and \((3t^{1/2})^2\) to \(9t\). Thus, the integral becomes\[L = \int_{1}^{2} \sqrt{4t^{2} - 8t + 4 + 9t} dt \]or\[L = \int_{1}^{2} \sqrt{4t^{2} - 8t + 13} dt\]