You shouldn't have to solve for t here if you only want the zeros. 0 is special because if f(x)y(x) = 0 then you can be sure that f(x)=0 or g(x) = 0. so you need to find the zeros of f(x) = 12 - 25.2t and g(x) = exp(-2.1t) |
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An electron moving along the x axis has a position given by x = 12te^(-2.1 t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops? |
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Last edited by Invader; 09-07-2009 at 03:12 AM.
You shouldn't have to solve for t here if you only want the zeros. 0 is special because if f(x)y(x) = 0 then you can be sure that f(x)=0 or g(x) = 0. so you need to find the zeros of f(x) = 12 - 25.2t and g(x) = exp(-2.1t) |
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Previously PhilosopherStoned
Heh, that's right, I could have done it that way. Though I think you misread the |
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Last edited by Invader; 09-07-2009 at 08:02 AM.
That's a really good way to look at it. As the functions get more difficult though, one needs to learn tricks too But keep looking at it geometrically as long as you can. Going between the two without having to rely on either is key to being good at math as it gets more advanced. |
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Previously PhilosopherStoned
See, that was never explained to me, even through calc 1. I always assumed |
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No problem. I love this stuff. I could use a review on all this stuff anyway. I quit math for years so I'm rusty on all levels. I've been working on it again for the past month or so though so I've got a lot of the kinks worked back out. |
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Previously PhilosopherStoned
Yeah just treat exp(x) as a constant and get rid of it, it's never 0. |
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it's more general than treating one of them as a constant though. it works any way that you can factor a function. take sin^5(x)cos^2(x)x^2/(x^2 + 1). |
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Previously PhilosopherStoned
tbh I don't know of any other ways to solve an equation other than setting each factor to zero. What's inversion in this context? |
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if f is a function, g is its inverse if f(g(x)) = x. so the inverse of y = 2x is x = 1/2y. It's essentially expressing x as a function of y. We can only do it globally when dy/dx is never zero. If it's zero somewhere, we have to break it into pieces on either side of where it's zero. for example, if y = x^2, then dy/dx = 2x which is 0 when x=0. So we have one inverse of y on either side of 0, these are x = sqrt(y) for x > 0 and x = -sqrt(y) for x > 0. |
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Last edited by PhilosopherStoned; 09-08-2009 at 12:23 AM.
Previously PhilosopherStoned
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There are two angles that will hit the goal. These angles bounds of the range that you are looking for. So you can solve them one at a time. Do you know how to solve those problems? |
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Last edited by PhilosopherStoned; 09-21-2009 at 10:47 PM.
Previously PhilosopherStoned
I think I need the two angles that are going to hit the goal 3.44m above the ground. And I don't know how to solve for them individually. I was given example problems in place of this question, both of which I solved quickly, but they weren't exactly the same as what I'm being presented with here. |
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Last edited by Invader; 09-21-2009 at 10:53 PM.
Resolve the initial velocity vector (u) and hence get equations for the horizontal and vertical displacement: |
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yeah, that's it xei (except you forgot the part where you integrate to get the equations you wrote down ). Fuck I hate trig equations.... |
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Previously PhilosopherStoned
Some of them are funky. IIRC cos@ + cos2@ + cos 3@ = 0 is quite tricky. |
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Got @ = 17.59, 76.69 deg btw, but it's a shit question. Probs slipped up somewhere. |
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0 = -1.01tan^2(θ) + 46tanθ -4.05 |
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I'd recommend you use x, y, u, and @, until you have obtained the general projectile equation. Post it up here and I'll check. Then it's just a matter of setting x = 46, y = 3.44, u = 32 and solving for tan@. |
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Apologies for the late response. I'll show you how I've been trying to work this out, and you can point out where I'm making my blunder. |
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There's way too many numbers in there. They tend to confuse the situation. Do what Xei said and just use letters. That way, it's much easier to spot mistakes. |
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Last edited by PhilosopherStoned; 09-23-2009 at 10:53 PM.
Previously PhilosopherStoned
I think that last term should look more like: |
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Last edited by Invader; 09-23-2009 at 11:03 PM.
-(gx2/2u2)tan2θ + xtanθ - [y + (gx2/2u2)] |
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Last edited by PhilosopherStoned; 09-23-2009 at 11:21 PM.
Previously PhilosopherStoned
"because -(gx2/2u2) distributes over (1 + tan2θ)" |
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I believe you should be aiming for |
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