Find the distance traveled by a particle with position $(x, y)$ as $t$ varies in the given time interval. $x = 4\sin^2(t), \quad y = 4\cos^2(t), \quad 0 \leq t \leq 4\pi$

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Compare with the length $L$ of the curve.

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Neetesh Kumar | January 3, 2025

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This is the solution to Math 1c
Assignment: 10.2 Question Number 17
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Step-by-step solution:

Step 1: Parametrize the motion

The particle's position is given as:

$x = 4\sin^2(t), \quad y = 4\cos^2(t)$

From the trigonometric identity $\sin^2(t) + \cos^2(t) = 1$, we observe that:

$x + y = 4$

This means the particle moves along the straight line $x + y = 4$ in the coordinate plane.

Step 2: Distance traveled by the particle

To find the total distance traveled by the particle, we need the particle's speed.

First, compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$\frac{dx}{dt} = \frac{d}{dt}\left(4\sin^2(t)\right) = 8\sin(t)\cos(t) = 4\sin(2t)$

$\frac{dy}{dt} = \frac{d}{dt}\left(4\cos^2(t)\right) = -8\sin(t)\cos(t) = -4\sin(2t)$

The speed $v$ is given by:

$v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

Substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$v = \sqrt{\left(4\sin(2t)\right)^2 + \left(-4\sin(2t)\right)^2}$

$v = \sqrt{16\sin^2(2t) + 16\sin^2(2t)} = \sqrt{32\sin^2(2t)}$

$v = 4\sqrt{2} |\sin(2t)|$

The total distance traveled is the integral of the speed over $t \in [0, 4\pi]$:

$\text{Distance} = \int_0^{4\pi} v \, dt = \int_0^{4\pi} 4\sqrt{2} |\sin(2t)| \, dt$

Since $|\sin(2t)|$ is periodic with a period of $\pi$, there are $4$ full periods in $[0, 4\pi]$.

Compute the integral over one period $[0, \pi]$:

$\int_0^\pi |\sin(2t)| \, dt = 2$

Thus, the total distance traveled is:

$\text{Distance} = 4 \cdot 4\sqrt{2} \cdot 2 = 32\sqrt{2}$

Step 3: Length of the curve

The particle moves along the straight line $x + y = 4$, and the curve is the diagonal segment between the endpoints of motion.

Find the endpoints of the segment by evaluating $x$ and $y$ at $t = 0$ and $t = \pi/2$:

1. At $t = 0$:

   $x = 4\sin^2(0) = 0, \quad y = 4\cos^2(0) = 4$

   Point: $(0, 4)$

2. At $t = \pi/2$:

   $x = 4\sin^2(\pi/2) = 4, \quad y = 4\cos^2(\pi/2) = 0$

   Point: $(4, 0)$

The curve is the straight line segment between $(0, 4)$ and $(4, 0)$.

Using the distance formula:

$L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute:

$L = \sqrt{(4 - 0)^2 + (0 - 4)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Final Answer:

Distance traveled by the particle:

$\boxed{32\sqrt{2}}$

Length of the curve $L$:

$\boxed{4\sqrt{2}}$

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