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Find the velocity, acceleration, and speed of a particle with the given position function.

$ r(t) = \langle t^2 + t, t^2 - t, t^3 \rangle $

$\left\langle 2 t+1,2 t-1,3 t^{2}\right\rangle$$\langle 2,2,6 t\rangle$$\sqrt{\left(\frac{\partial x}{\partial t}\right)^{2}+\left(\frac{\partial y}{\partial t}\right)^{2}+\left(\frac{\partial z}{\partial t}\right)^{2}}$

Calculus 3

Chapter 13

Vector Functions

Section 4

Motion in Space: Velocity and Acceleration

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Okay, so this question gives us the position function of a particle, and it wants us to find expressions for the velocity, acceleration and speed off the particle. So for velocity, we just need to take a derivative of the position. And remember, we do this component by component. So the derivative of the first component is to t plus farm. The derivative of the second component is to t minus one, and the derivative of the third component is three t squared. So that's our expression for velocity and then for acceleration, we just need to take another derivative. So the derivative of the X component of velocity is too. The derivative of the white component of velocity is too, and the derivative for the Z component of velocity is 60. So now we have expressions for both our velocity and acceleration. But we're not done yet because we also need the speed. So speed is equal to the magnitude of velocity. So we're just gonna have to find the magnitude of our velocity function. So the magnitude of e of t is equal to Well, we just use our version of the Pythagorean theorem. So again, that's just t x t t squared plus d Y t t squared plus d C D t squared And we know each of these components from the last part So give us a big square root to work with here. So our X component is to t plus one are y component is to t minus one and r C component is three t squared. So if we do that, what's foil these out and see what we can do Simplifying these so we get for t squared plus 40 plus one plus 40 squared minus 40 plus one men plus nine t to the fourth and now we can see one quick simplification. This cancels with this. So now we get the square root of eight t squared plus two plus nine t to the fourth or registry Raying this in a more standard form the square root of nine T to the fourth plus eight t squared plus two. And this is our final answer

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