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Problem 55 Easy Difficulty

In the theory of relativity, the mass of a particle is
$$ m = \dfrac{m_0}{\sqrt{1 - v^2/c^2}} $$
where $ m_0 $ is the rest mass of the particle, $ m $ is the mass when the particle moves with speed $ v $ relative to the observer, and $ c $ is the speed of light. Sketch the graph of $ m $ as a function of $ v $.

Answer

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Video Transcript

were sketching our function m as a function of V. So I just called the f a V. Um, so this is our function. So first, to find her domain, we have that zero to seize our domain because it must be positive. Under the radical intercepts we have from zero to em initial, because if we set our extra zero, we just have a minute Shal over one symmetry none. Assam toots. We have a vertical ass in tow vehicle, See? And the reason for that is if he does equal, see, we have one. So one minus 10 and we cannot have our function over zero. All right, so of prime of the equals. So we're doing the first derivative to see where it's increasing. This is her first derivative C squared, minus B squared, cubed. All right. And we see that this is increasing over our entire domain, which is from zero to see. So there are no local men's or Max is. And if we find a double prime Seacon cavity you see, see on initial times to V squared plus C squared over V squared minus C squared. That's all squared times radical C squared minus B squared. And we see that's also above zero everywhere over our our domain zero to see. So that means it's Kong cave up over our domain. Right? And now we can graft or function. So you know it's gonna be our X axis is gonna be V. And our Y axis is gonna be because here M is equal to F V. All right, so we haven't intercept at zero. I am initial. We haven't asked him to at sea, and our graph will approach but not touch, See?