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I ran into a problem of finding the anti-derivative of a function which involves the absolute value function. Here is the problem:

Question: If $v(t) = \langle|t^2-1|,4t-3\rangle$ is the velocity vector, find $s(t)$ the position vector. Given that $s(0) = \langle 1,1\rangle$.

My answer: $s(t) = \displaystyle \int_{0}^t v(x)dx$. Do we split this into $2$ cases ? $0 \le t \le 1$ and $t \ge 1$ ?

To be quite honest here my elementary differential geometry is quite rusty. I would appreciate to see a full analysis on this. Thanks. WY.

Wang YeFei
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1 Answers1

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Upgrading my earlier comment to an answer:

Your thought process is correct.

  • When $0\le t<1$, we simplify $|x^2-1|=1-x^2$, and

$$\begin{align*} s(t) &= s(0) + \int_0^t v(x) \, dx \\ &= \langle1,1\rangle + \int_0^t \left\langle 1-x^2, 4x-3\right\rangle \, dx \\ &= \left\langle-\frac{t^3}3+t+1, 2t^2 - 3t+1\right\rangle \end{align*}$$

  • When $t\ge1$, we have $|x^2-1|=x^2-1$ in the integral with $x\ge1$.

$$\begin{align*} s(t) &= s(0) + \left\{\int_0^1 + \int_1^t\right\} v(x) \, dx \\ &= \left\langle\frac53,0\right\rangle + \int_1^t \left\langle x^2-1,4x-3\right\rangle \, dx \\ &= \left\langle \frac{t^3}3 - t + \frac73, 2t^2-3t+1\right\rangle \end{align*}$$

user170231
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