Find the length of the curve.
$$
x=\frac{t^{2}}{2}, y=\frac{(2 t+1)^{3 / 2}}{3}, 0 \leq t \leq 10
$$

The length of the curve is $\square$ (Simplify your answer.)

Question

Find the length of the curve.
$$
x=\frac{t^{2}}{2}, y=\frac{(2 t+1)^{3 / 2}}{3}, 0 \leq t \leq 10
$$

The length of the curve is $\square$ (Simplify your answer.)

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the length of the curve, we use the formula for the arc length of a parametric curve: $$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$ where $ x = \frac{t^2}{2} $ and $ y = \frac{(2t+1)^{3/2}}{3} $, and $ 0 \leq t \leq 10 $. ‖ ‖ step 2 ⋮ First, we find $\frac{dx}{dt}$: $$ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^2}{2} \right) = t $$ ‖ ‖ step 3 ⋮ Next, we find $\frac{dy}{dt}$: $$ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{(2t+1)^{3/2}}{3} \right) = \frac{3}{2} \cdot \frac{(2t+1)^{1/2} \cdot 2}{3} = (2t+1)^{1/2} $$ ‖ ‖ step 4 ⋮ Substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the arc length formula: $$ L = \int_{0}^{10} \sqrt{t^2 + (2t+1)} \, dt $$ $$ L = \int_{0}^{10} \sqrt{t^2 + 2t + 1} \, dt $$ $$ L = \int_{0}^{10} \sqrt{(t+1)^2} \, dt $$ $$ L = \int_{0}^{10} |t+1| \, dt $$ ‖ ‖ step 5 ⋮ Since $ t $ ranges from 0 to 10, $ t+1 $ is always positive, so $|t+1| = t+1$: $$ L = \int_{0}^{10} (t+1) \, dt $$ $$ L = \left[ \frac{t^2}{2} + t \right]_{0}^{10} $$ $$ L = \left( \frac{10^2}{2} + 10 \right) - \left( \frac{0^2}{2} + 0 \right) $$ $$ L = \left( 50 + 10 \right) - 0 $$ $$ L = 60 $$ ‖ ∻Answer∻ ⚹ The length of the curve is 60. ⚹ ∻Key Concept∻ ⚹Arc Length of Parametric Curve⚹ ∻Explanation∻ ⚹To find the length of a curve given by parametric equations, we use the arc length formula, which involves integrating the square root of the sum of the squares of the derivatives of the parametric equations.⚹◊What is the formula for finding the length of a curve given parametric equations $x = f(t)$ and $y = g(t)$?⍭ Generate me a similar question◊

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