Wikipedia:Reference desk/Archives/Mathematics/2010 September 21

= September 21 =

Projectile Motion problem
Hey mathies--

I'm having trouble on a projectile motion problem, perhaps you can see where I'm going wrong:

A diver leaves a 3-m board on a trajectory that takes her 2.5 m above the board and then into the water a horizontal distance of 2.8 m from the end of the board. Find the horizontal distance of the diver from the end of the board, when she is at her maximum height.

I know that the magnitude of her velocity should be 0 at her max height, so I plan to find the displacement when her velocity is zero. I figured I'd need to solve first for her initial velocity components:  -V(y)^2=-2g(yfinal-yinitial) but that's where I got stuck. Is her yfinal=0 (level of water) and her yinitial=3? because then I end up taking the square root of a negative number...help!24.63.107.0 (talk) 02:56, 21 September 2010 (UTC)


 * I don't follow how you derived your formula (but that may just be me). If you post your steps leading up to that, perhaps we can better help.  I would expect that you would need a formula that defines the height over time in terms of (1) the initial height, (2) the vertical component of the initial velocity, and (3) acceleration due to gravity.  Be sure you have your signs correct − if up is positive, acceleration due to gravity will be negative.  It's worth noting that sometimes it is easier (although less elegant) to break the problems such as this down into simpler components.  Consider calculating the time it takes to travel vertically to the peak separately from the time it takes to drop to the water and then apply those numbers to the horizontal component.  --  Tom N (tcncv) talk/contrib 05:02, 21 September 2010 (UTC)


 * The formula you're using is derived from the formula for the vertex of a parabola. yinitial is the height at time 0 (which is 3), and yfinal is the height at the peak of the jump (5.5).--203.97.79.114 (talk) 06:28, 21 September 2010 (UTC)


 * (ec)The trajectory y = f(x) is a parabola f(x) = ax2+bx+c satisfying the 4 equations f(0)&minus;3 = f(2.8) = f(d)&minus;5.5 = f '(d) = 0 in the 4 unknowns a,b,c,d. (The time does not enter into the problem). Eliminating c=3, b=&minus;2ad, and a=&minus;2.5d&minus;2 leaves the quadratic equation 15d2+70d&minus;98 = 0. Bo Jacoby (talk) 06:43, 21 September 2010 (UTC).

Integration by Parts
How do you tell your students (or how do you do it) to choose u and dv when doing integration by parts? I realize there is a rule-of-thumb, LIATE. It seems to work often but not always. The rule I remember was simple: Choose for dv the most complicated thing you can integrate without integration by parts and then u is everything else. What does that mean, though, most complicated? Well, I guess the LIATE sort of describes one sort of order of most complicated things and you just need to realize that sometimes LIATE doesn't describe enough to decide what the most complicated thing is. Do you know of examples where choosing the "most complicated" thing you can integrate doesn't work? Thanks for any thoughts, no real right answers here. StatisticsMan (talk) 20:47, 21 September 2010 (UTC)


 * It seems to me that it's just intuition that you gain by practicing it over and over. Rules of thumb are helpful, but they only go so far, as you note; the point of doing homework problems is to gain experience in recognizing when these rules of thumb don't work and what to do in those situations.
 * A story about explaining rules of thumb for integration by parts: A year ago I had just moved from Nebraska to Pittsburgh and was teaching a calculus recitation. One of the examples I did at the board was $$\int e^x\sin x\,dx$$ or something like that—an integral where you do integration by parts twice and then solve for the answer. It doesn't matter which of $$e^x$$ and $$\sin x$$ you choose to be u and which you choose to be dv, but you must make the same choice for both integrations by parts, or else you will end up in the second integration canceling out the work you did in the first. I was explaining this to my students, and I summarized with, "What I'm saying here is, don't change horses in the middle of the stream." Apparently this saying isn't as widely known here in Pittsburgh as it is in Nebraska, because my entire class just looked at me with bemused looks on their faces. Several times afterward I had students ask me for more horse stories. —Bkell (talk) 03:26, 22 September 2010 (UTC)

Whatever you call u should be something that you can differentiate, and dv should be something that you can antidifferentiate, and sometimes that's enough to decide the issue. Michael Hardy (talk) 21:32, 22 September 2010 (UTC)


 * You might also be interested in Integration_by_parts discussed recently on this desk.   D b f i r s   22:30, 22 September 2010 (UTC)

Asymptote and limit
Please what the difference is between the asymptote and the limit. Thanks you. 24.92.78.167 (talk) 23:41, 21 September 2010 (UTC)


 * Briefly, limit is the broader term, referring to the general concept, applicable to many kinds of functions, infinite series, and so on, while an asymptote arises from a specific kind of limit behaviour in which a curve approaches a straight line. Odysseus1479 (talk) 09:02, 22 September 2010 (UTC)
 * Asymptotes always describe some sort of limiting behavior, but there are functions like $$f(x)=x+\frac1{x^2+1}$$ that have oblique asymptotes; they don't cleanly correspond to any particular limit ($$\lim_{x\rightarrow\pm\infty}x^3=\pm\infty$$, but it has no asymptotes). --Tardis (talk) 14:02, 22 September 2010 (UTC)

Well..... We have
 * $$ \lim_{x\to 3} \frac{x^2 + 2x - 15}{x - 3} = 8 $$

and that has nothing directly to do with an asymptote. And also see above. When you observe that
 * $$ \lim_{x\to\pm\infty} \frac{2x + 9}{x} = 2 $$

then the limit is just a number: 2; whereas the asymptote is a horizontal line, not just a number. And then there are vertical asymptotes:
 * $$ \lim_{x\to 4} \frac{1}{(x-4)^2} = \infty $$

and here the limit is &infin;, and the asymptote is the line x = 4. Now if we observe that
 * $$ \lim_{x\to 6} \frac{1}{(x-6)^2} = \infty $$

then again the limit is &infin; but the asymptote is not the same as in the previous example even though the limit is the same. Michael Hardy (talk) 21:30, 22 September 2010 (UTC)


 * Completely agree with all of Michael's exampels. More generally, an asymptote may be a straight line i.e. a linear function, but does not have to be - it can be any function at all. Usually we want to describe the asymptotic behaviour of a function in terms of a family of simpler functions e.g. polynomials, exponentials etc.
 * To see how it is possible to have an asymptote without a limit, consider the function
 * $$f(x) = x + \sin \left( \frac {1}{x} \right)$$
 * For large positive values of x, f(x) asymptotically approaches the line y=x, and we could describe this as an oblique asymptote. Working in the extended real line, we could say that
 * $$\lim_{x \rightarrow \infty} f(x) = +\infty$$
 * However, for small positive values of x, f(x) asymptotically approaches the curve y=sin(1/x). In this direction, the limit as x approaches 0 does not exist, so we have an asymptote without a limit. Gandalf61 (talk) 10:14, 23 September 2010 (UTC)