Okay, thanks
That sort of trig is way beyond what we study in our maths course, which probably explains my issues with it
I can understand how you got there; but I'd hate to have to reproduce it (thankfully, as I said, we won't have to)
Okay, thanks
That sort of trig is way beyond what we study in our maths course, which probably explains my issues with it
I can understand how you got there; but I'd hate to have to reproduce it (thankfully, as I said, we won't have to)
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I wasn't talking about the distance the wagon had traveled, but it's momentary speed. I figured bob could wait until the wagon had a momentary speed equal to half of his constant speed. 0.5*8=4 which is half of bob's speed.
Anyway, thanks alot. Doesn't look that hard now that you showed how you would do it, but that's always the case isn't it
Last edited by Mathias; October 01, 2007 at 11:26 AM.
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You didn't cover tan² x + 1 = sec² x? That's one of the most basic trigonometric identities. You'd normally have it drilled into you along with sin² x + cos² x = 1 and cot² x + 1 = csc² x. All three equations are equivalent, of course (simply multiple or divide by the square of sine or cosine).
Oh, I see, I misunderstood you. That's correct, yes. It's quite easy to see graphically, at least with some basic calculus concepts (area under curve = distance traveled), and applies to any situation of constant acceleration. I could do a more detailed derivation if you like, but you're right, certainly.
Nope, for some inexplicable reason our trig course doesn't cover sec/csc/cotYou didn't cover tan² x + 1 = sec² x? That's one of the most basic trigonometric identities. You'd normally have it drilled into you along with sin² x + cos² x = 1 and cot² x + 1 = csc² x. All three equations are equivalent, of course (simply multiple or divide by the square of sine or cosine).
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Yeah, the first time I covered those identities and functions in detail was Cal 1... though I never had a trig-specific course, it pretty much went from algebra and geometry (and the basic trig functions) to Cal 1&2, where, of course, you would have a hard time without those identities.
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I'm quite surprised. When trigonometry was introduced to me for the first time, I was told there were six trigonometric functions, which were duly enumerated.
well, we were told of them, but I would hardly say that they, especially the identities, were "drilled" until Cal 1.
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Keep in mind that this is just high school math/phys level
We started out with right angled trig (ie really basic stuff) a couple of years ago, and the only real stuff we touch on this year, when looking at triangles in general, is relatively simple stuff like the sine/cosine/area of a triangle rules; basically it's there to supplement the other courses that we do that require trig, not as a study in itself.
At this level, the only really theoretical stuff we're doing is generally pure algebra, in relation to logarithmic functions/general calculus, and everything else is done from a practical point of view.
I've no idea why there's such a contrast between what we do for theoretical purposes and for practical, but it's there
We weren't even drilled on the basic trigonometric identities, really (of which we only did a few, even fewer of which I can remember offhand )
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Anyone know anything about Statistics?
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I would bet someone does... but you should probably be more specific...
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Hi, folks (mainly Simetrical and drak ), I hope you have all had a wonderful day. I have a problem with my mathematics homework. Again. So, without furtherbutt kissingdelay, here is the question.
The diagram (on paper, but I can draw it if needed) shows a sketch of the graph of y=x3-9x+4 and two parallel tangents drawn at P and Q. (P is situated ever so slightly to the left of the maximum turning point, and Q is situated slightly to the right of the minimum turning point, the graph makes an N shape, but has no numbers on any of the axes.)
- Find the equations of the tangents to the curve y=x3-9x+4 which have gradient 3
- Show that the shortest distance between the tangents is (16√10)/5
However, I managed to get a start to a.
I differentiated the original equation.
y'=3x²-9 is what I got.
Therefore, I found the gradient:
3=3x²-9
12=3x²
x=2
Which gave me these coordinates (2,-6)
And then this equation for one of the tangents
y-3x=-8
But that's as far as I can go.
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You missed a solution here:
x=+-2. -2 works as well. This will give you your other tangent line.Originally Posted by Shaun
Once you have both tangent lines you need to construct a 3rd line, perpendicular to both. Since it is perpendicular, finding it's intersections with the 2 tangent lines and finding the distance between them will give you the shortest distance.
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I know the math being discussed here is farrrrrrrr beyond me, but I hope my insignificant and meager question could be answered by one of the math masters here.
I have a small question concerning second order determinants (Cramer's Law) to solve systems of equations. Could somebody type a simple explanation of it and perhaps show an example of it being utilized? I believe I may have taken bad notes from the teacher because I never seem to get the correct answer for any problems by using this method, and the book doesn't help at all (gets me more confused). Thanks in advance!
P.S. I am inept at math (the only subject I suck at), so I am sorry if it is an extremely simple question.
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yesAnyone know anything about Statistics?
Sure I've been called a xenophobe, but the truth is Im not. I honestly feel that America is the best country and all other countries aren't as good. That used to be called patriotism.
EDIT: Never mind the question now... not important
Last edited by LSJ; November 01, 2007 at 02:54 PM.
Last edited by chriscase; April 17, 2008 at 05:24 PM.
Quick question, is this math statement true or false?
The equation:
y=(x^2-9) / (x-3) has a vertical asymptote.
My friends say it's FALSE, because once you simplify the equation you get y=x+3, which is a straight line with no asymptote.
I maintain that even if it is a straight line, it will have an open circle at x=3 because it's undefined at that point, therefore making the statement TRUE.
Who's right?
It's not just on the unit circle, it's the Pythagorean theorem anywhere. Pick either non-right angle on a right triangle, call it x. The Pythagorean theorem tells you
adjacent² + opposite² = hypotenuse²
so
(adjacent/hypotenuse)² + (opposite/hypotenuse)² = 1
sin² x + cos² x = 1
Of course, this "proof" only works with angles of 0 to π/2, exclusive, and lacks rigor generally, but it's a decent mnemonic.
As always in mathematics, refer to the definition of a vertical asymptote. The usual definition for real-valued functions of a single real variable goes something like: the function f has a vertical asymptote at a if and only if limx → a |f(x)| = ∞. By that definition, clearly your example does not have a vertical asymptote at 3 (the limit is 6, not ∞). It's merely not defined there; it has a point discontinuity.
If you're using some other definition of asymptote, like if you haven't learned what a limit is, results may vary. I think any rigorous definition of a limit would agree with what I said. The way to think about it is that things like asymptotes are properties of the function itself, not of the way you write it down. Just write your function like this: "f(3) is undefined, and f(x) = x + 3 for all other x." Then clearly you wouldn't say it has a vertical asymptote, but it's the exact same function.