tag:blogger.com,1999:blog-8972120889629675714.post3301572254114951470..comments2021-09-28T02:36:57.070-04:00Comments on Beyond Easy: Permaculture, proselytizing, power problemsPatrick Rhttp://www.blogger.com/profile/02410016566636603639noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-8972120889629675714.post-17403021660224675652012-05-29T15:09:26.736-04:002012-05-29T15:09:26.736-04:00Onward and upwards!
I cede to John my share of an...Onward and upwards!<br /><br />I cede to John my share of any chickens.Anonymoushttps://www.blogger.com/profile/11366190775659089627noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-18416079860826737622012-05-28T21:39:02.638-04:002012-05-28T21:39:02.638-04:00Good call. Flipped 15 pages ahead to "hey, he...Good call. Flipped 15 pages ahead to "hey, here's a list of shortcuts that will make everything fifty times easier!" The problem crumbled in 90 seconds and I nearly threw the book out the window.<br /><br />Thanks for your help, gentlemen! Who wants drawings of chickens?Patrick Rhttps://www.blogger.com/profile/02410016566636603639noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-40314095866330517302012-05-27T21:57:38.607-04:002012-05-27T21:57:38.607-04:00Jeez, this nonsense again? Asking you to find comp...Jeez, this nonsense again? Asking you to find complex derivatives with the limit approach is ridiculous. I agree with Matt, it's better to move on to the next chapter and learn how you're 'actually' going to be taking derivatives for the rest of your calculus career. I commend you for your thoroughness though.John Thyerhttps://www.blogger.com/profile/17629076787737164071noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-69336861225733603322012-05-27T20:17:28.451-04:002012-05-27T20:17:28.451-04:00Gah. Is your textbook available online, by any ch...Gah. Is your textbook available online, by any chance? I'd like to see what they expect you to know at this point. You can get the derivative that way ... kind of.<br /><br />Substitute your f(a) into the definition of the derivative and you get the expression 2/h*[(3-a-h)^(-1/2) - (3-a)^(-1/2)].<br /><br />Factor out (3-a)^(-1/2) to get (skipping some algebra ...) 2/h*(3-a)^(-1/2)*[(1-h/(3-a))^(-1/2) - 1].<br /><br />Now, for small x, (1-x)^(-1/2) can be expanded in an infinite series as 1 + x/2 + (3/8)*x^2 + ... .<br /><br />Keep just the first two terms and substitute them into our expression (where x = h/(3-a)): 2/h*(3-a)^(-1/2)*[1 + (1/2)*h/(3-a) - 1] .<br /><br />In the expression in square brackets the +1 and -1 terms cancel. In the remaining term the 2/h and the h/2 cancel, and you're left with (3-a)^(-3/2). (Tadaa!)<br /><br />It's way too late so I might be missing something obvious, but I think that's how you'd have to approach it. The thing is, that series expansion comes from a Taylor series -- which is based on derivatives! Which is why I'm curious to see the textbook; there might be another approach to infinite series that I'm forgetting.<br /><br />But there's no reason to find complicated derivatives with this approach, unless you like pain. If you've got the idea that you're making ever-closer approximations of the slope of the curve (as h -> 0), then just move on.Anonymoushttps://www.blogger.com/profile/11366190775659089627noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-19362839444700741552012-05-27T18:49:33.127-04:002012-05-27T18:49:33.127-04:00Oops. Wikipedia image transparencies. Trying again...Oops. Wikipedia image transparencies. Trying again:<br /><br />http://i.imgur.com/aQzzu.png><br /><br />Is it *possible* to solve that problem with only this formula? I'm driving myself crazy.Patrick Rhttps://www.blogger.com/profile/02410016566636603639noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-7303589664137814372012-05-27T18:45:37.336-04:002012-05-27T18:45:37.336-04:00Yeah...here's the thing. The textbook hasn'...Yeah...here's the thing. The textbook hasn't actually gotten into *any* of that yet. The only derivative formula it's given me so far is this one:<br /><br />http://upload.wikimedia.org/wikipedia/en/math/4/2/c/42cf4f4861ae1266b13104c4115e7b5d.png<br /><br />I don't feel like I can move on in the text until I can solve this sort of problem with that formula, and I just. can't. do it.Patrick Rhttps://www.blogger.com/profile/02410016566636603639noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-67006160626546593552012-05-27T14:12:23.515-04:002012-05-27T14:12:23.515-04:00Well, I was talking about the chain rule too, not ...Well, I was talking about the chain rule too, not u-substitution. My line, df/dx = df/du * du/dx, <b>is</b> the chain rule expressed in Leibniz notation. I think that notation makes it clearer <b>why</b> the mechanics of the chain rule work the way they do.<br /><br />But u-substitution and the chain rule are very similar processes. You can even think of u-substitution as <a href="http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/USubstitution.html" rel="nofollow">the chain rule in reverse</a>.Anonymoushttps://www.blogger.com/profile/11366190775659089627noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-51103368868929712542012-05-27T11:22:57.397-04:002012-05-27T11:22:57.397-04:00I've never used u-substitution on a derivative...I've never used u-substitution on a derivative problem before, but that's definitely a way to do it!<br /><br />I'd simplify the problem with exponents. Turn it into: <br /><br />2*(3-x)^(-1/2)<br /><br />The 2 is a constant that isn't affected by taking the derivative, so just focus on the (3-x) part. Use the CHAIN RULE to take the derivative. d/dx[2*(3-x)^(-1/2)] = 2*(-1/2)*(3-x)^(-3/2)*(-1) = <br /><br />(3-x)^(-3/2)<br /><br />Did you catch it? We took the derivative of (3-x)^(-1/2) just like we were taking the derivative of x (multiply it by the value of the exponent, subtract 1 to actual exponent), except right at the end we multiplied it by the derivative of the INSIDE of (3-x), that is to say, -1.<br /><br />You see the same thing if you take the derivative of (5-x^2)^4:<br /><br />4 * (5-x^2)^3 * (-2x)<br /><br />Remember, the derivative of a constant is just zero. The problems can be as convoluted as you want! 17(x^3+3x^2+2x+5)^(10000) becomes:<br /><br />17*10000*(x^3+3x^2+2x+5)^(9999)*(3x^2+6x+2)<br /><br />That's it really -- remove the confusing fractional element by turning it into an exponent (in situations where you can't just turn it into an exponent, that's where you'll have to understand the quotient rule), then use the chain rule to find the derivative. Unless you're confused about the f'(a) part in which case... hell if I know.John Thyerhttps://www.blogger.com/profile/17629076787737164071noreply@blogger.comtag:blogger.com,1999:blog-8972120889629675714.post-53981639717441475022012-05-27T05:03:48.409-04:002012-05-27T05:03:48.409-04:00I don't know enough about permaculture to say ...I don't know enough about permaculture to say anything intelligent, so I'll skip to the calculus! Sorry for the plain-ASCII notation.<br /><br />I assume 'a' is some known value, and the difficulty is just with differentiating f(x)? If so, the key is the chain rule.<br /><br />Define a temporary variable u = 3-x.<br /><br />You can write your function in terms of u, f(u) = 2 * u^(-1/2).<br /><br />Now you have f in terms of u, but you want df/dx. Multiply and divide by du (which doesn't change anything, since du/du = 1) and then rearrange: df/dx = df/dx * du/du = df/du * du/dx.<br /><br />Getting df/du is easy using the power rule: 2 * (-1/2) * u^(-3/2) = -u^(-3/2).<br /><br />du/dx is also easy: -1<br /><br />So, df/dx = (-u^(-3/2) * (-1) = u^(-3/2).<br /><br />Now, just substitute the definition of u back in to get the answer in terms of x: df/dx = (3-x)^(-3/2).<br /><br />If you don't have math software, you can check yourself using Wolfram Alpha: http://www.wolframalpha.com/input/?i=differentiate+2%2Fsqrt%283-x%29 . In fact, if you click on the 'show steps' button to the right of the derivative, you'll see essentially the same thing as what I've written.Anonymoushttps://www.blogger.com/profile/11366190775659089627noreply@blogger.com