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Problem 54 Hard Difficulty

Work done by a radial force with constant magnitude A particle moves along the smooth curve $y=f(x)$ from $(a, f(a))$ to $(b$ $f(b)$ ). The force moving the particle has constant magnitude $k$ and always points away from the origin. Show that the work done by the force is
$$\int_{C} \mathbf{F} \cdot \mathbf{T} d s=k\left[\left(b^{2}+(f(b))^{2}\right)^{1 / 2}-\left(a^{2}+(f(a))^{2}\right)^{1 / 2}\right]$$

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Video Transcript

okay for this problem, we can calculate what that is, and we can't get it. What are is are said the he after where teas from a toupee and, uh, has not need to kay and is pointy point half points away from Florida, and so does after the after became Times Axe, your I d advise were this right? Okay. Why your idea about this? So this is F so I far t should be, uh, t Square plus where? Okay, T and chaos he squared plus after square. Okay. And are crimes of the one of prime tea. Okay, so it's just stupid under one to calculate from a to B. Ah, Katie Over Squared T square, plus half square. And plus okay. Uh uh, Brian have a He's square close ups. Where this Yuki. So we want a corner s. So what we want is going to prove is that this is just equals toe t square plus half square to the want, huh? Okay. Has proved destructive is just this. Okay, I'll take a prime so we can have he squared plus half square. And how to t thus too. Times of prime. Right? So here's the two up there, so they're gone. So this is just equals. Okay, times this. So by the Newton lateness formula, this is always true.

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