Question

Suppose a particle is moving in space with a position described by the position vector r(t) = cos(t^2)I +sin(t^2)j+t^2k Find the arc length of the function s(t) from time 0 to time t(assume t is positive)

Answer 1

s(t) = √(2) · t²

Step-by-step explanation:

To solve the problem of finding the arc length s(t) of a particle moving in space with the given position vector from time = 0 to time = t (assuming 't' is positive), we'll follow these steps:

1. Compute the Velocity Vector: Differentiate the position vector 'r(t)' with respect to 't' to find the velocity vector 'v(t)'.
2. Calculate the Speed: Find the magnitude of the velocity vector to get the speed of the particle.
3. Integrate to Find Arc Length: Use the formula for arc length to integrate the speed over the interval from 0 to t.

Given position vector:

`\vec r(t) = \cos(t^2)\hat i +\sin(t^2)\hat j+t^2\hat k`

`\hrulefill`

Step 1: Compute the Velocity Vector
`\hrulefill`

To find the velocity vector 'v(t)', differentiate each component of r(t) with respect to 't':

`\Longrightarrow (d)/(dt)[\vec r(t) = \cos(t^2)\hat i +\sin(t^2)\hat j+t^2\hat k]\\\\\\\\\Longrightarrow (d)/(dt)[\vec r(t)] = (d)/(dt)[\cos(t^2)]\hat i +(d)/(dt)[\sin(t^2)]\hat j+(d)/(dt)[t^2]\hat k\\\\\\\\\therefore \vec v(t) = -2t\sin(t^2)\hat i +2t\cos(t^2)\hat j+2t\hat k`

Thus, we have our velocity vector.

`\hrulefill`

Step 2: Calculate the Speed
`\hrulefill`

The speed of the particle is the magnitude of the velocity vector 'v(t)'.

If we have a velocity vector,

`\vec v(t)=\vec v_x \hat i + \vec v_y \hat j + \vec v_z \hat k`

Then the magnitude of this vector is,

`\big|\big|\vec v(t)\big|\big|=√((\vec v_x)^2+(\vec v_y)^2+(\vec v_z)^2)`

So computing the magnitude of our velocity vector we get:

`\Longrightarrow \big|\big|\vec v(t)\big|\big|=√((-2t\sin(t^2))^2+(2t\cos(t^2))^2+(2t)^2)\\\\\\\\\Longrightarrow √(4t^2\sin^2(t^2)+4t^2\cos(t^2)+4t^2)\\\\\\\\\Longrightarrow √(4t^2(\sin^2(t^2)+\cos(t^2))+4t^2)\\\\\\\\\Longrightarrow √(4t^2(1)+4t^2) \ \Big[\because \sin^2A+\cos^2A=1\Big]\\\\\\\\\Longrightarrow √(8t^2)\\\\\\\\`

`\therefore \big|\big|\vec v(t)\big|\big|=2√(2)t`

Thus, we have taken the magnitude of our velocity vector.

`\hrulefill`

Step 3: Integrate to Find Arc Length
`\hrulefill`

The arc length 's(t)' from time = 0 to time = t is given by the integral of the speed:

`\displaystyle s(t)=\int\limits^t_0 \big \, dt`

Now, let's perform the calculations.

`\displaystyle\Longrightarrow s(t)=\int\limits^t_0 {2√(2)t} \, dt\\\\\\\\\Longrightarrow 2√(2)\int\limits^t_0 {t} \, dt\\\\\\\\\Longrightarrow 2√(2)\left[(1)/(2)t^2 \right]\limits^t_0\\\\\\\\\Longrightarrow 2√(2)\left[(1)/(2)t^2-(1)/(2)(0)^2 \right]\\\\\\\\\Longrightarrow 2√(2)\left[(1)/(2)t^2 \right]\\\\\\\\\therefore s(t) = \boxed{√(2)t^2}`

We find that s(t) = √(2) · t². Assuming you are familiar with basic derivative and integration rules, I did not include the rules in this answer.