Question

# $x(t)=t^{1 / 3}, y(t)=t^{2}$ from t=1 to t=1.1.

Solution

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The parametric equations for are given as

\begin{align*} x(t) & = t^{1/3} \\ y(t) & = t^2 \end{align*}

We need to use the differential $ds$ to approximate the arc length for the interval $1\le t\le 1.1$. Therefore,

\begin{align*} dx & = x(1.1) - x(1) \\ dx & = (1.1)^{1/3} - (1)^{1/3} \\ dx & = 1.0323 - 1 \\ dx & = 0.0323 \end{align*}

Similarly, we have

\begin{align*} dy & = y(1.1)- y(1) \\ dy & = (1.1)^2-(1)^2 \\ dy & = 1.21- 1 \\ dy & = 0.21 \end{align*}

Therefore, the differential $ds$ is used to approximate the arc length as

\begin{align*} s & \approx ds \\ s & \approx \sqrt{(dx)^2 + (dy)^2} \\ s & \approx \sqrt{(0.0323 )^2 + (0.21)^2} \\ s & \approx 0.21247 \\ &\hspace*{-3mm}\boxed{s \approx 0.21247 } \end{align*}

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