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Posted 2 months ago

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.)
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Answer from Sia
Posted 2 months ago
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=ab(dxdt)2+(dydt)2dt L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt where x=t22 x = \frac{t^2}{2} and y=(2t+1)3/23 y = \frac{(2t+1)^{3/2}}{3} , and 0t10 0 \leq t \leq 10
step 2
First, we find dxdt\frac{dx}{dt}: dxdt=ddt(t22)=t \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^2}{2} \right) = t
step 3
Next, we find dydt\frac{dy}{dt}: dydt=ddt((2t+1)3/23)=32(2t+1)1/223=(2t+1)1/2 \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 dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} into the arc length formula: L=010t2+(2t+1)dt L = \int_{0}^{10} \sqrt{t^2 + (2t+1)} \, dt L=010t2+2t+1dt L = \int_{0}^{10} \sqrt{t^2 + 2t + 1} \, dt L=010(t+1)2dt L = \int_{0}^{10} \sqrt{(t+1)^2} \, dt L=010t+1dt L = \int_{0}^{10} |t+1| \, dt
step 5
Since t t ranges from 0 to 10, t+1 t+1 is always positive, so t+1=t+1|t+1| = t+1: L=010(t+1)dt L = \int_{0}^{10} (t+1) \, dt L=[t22+t]010 L = \left[ \frac{t^2}{2} + t \right]_{0}^{10} L=(1022+10)(022+0) L = \left( \frac{10^2}{2} + 10 \right) - \left( \frac{0^2}{2} + 0 \right) L=(50+10)0 L = \left( 50 + 10 \right) - 0 L=60 L = 60
The length of the curve is 60.
Key Concept
Arc Length of Parametric Curve
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.

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