Question

Math

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:
$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
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.

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