Calc help: approximating arctan (K) where K =some constant. please help!!!!!?
346140
Well there are the obvious values that are known of arctan. But if your constant K doesn't fall into these categories I'm not sure how you would do this without taylor expansion.\n\nTry this link, it gives the well known values on tan(x) so reading it backwards gives the arctan values
58408
You use a Taylor series expansion for arctan(x) and then plug in 1 since arctan(1) = pi/4.\n\narctan(x) = x - x^3/3 + x^5/5 - x^7/7 + \n\nso pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...\n\nNote: This converges very slowly. There are other functions whos Taylor series converges much faster.
748887
The distance from lamp to P is r = √[h^2+s^2], where s = dist from center to P (10m).  The illumination is inversely proportional to r^2, and directly proportional to cos(theta), where theta is going to be arctan(s/h).  So the formula is\n\nI(h) = cos(arctan(s/h))/(r^2+s^2)\n\nBy drawing a right triangle with legs l and h, you can find that cos(arctan(l/h)) = h/√[h^2+s^2] so the formula is then\n\nI(h) = h/(h^2+s^2)^(3/2)  this is the function to maximize w.r.t. h, the independent variable. \n\nTo maximize, differentiate and set to zero, you will get h = s/√2, and you can check it by putting into the formula h = (s/√2)+∆ and h = (s/√2)-∆ where ∆ is a small deviation.
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
864388
some formulas that can help :\nhttp://www.eskridgeinc.com/support/formulas.html
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
164104
maybe :\n\nKa = 1.3*10^-2
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
885147
You have to note that if f(x) = arctan(x), then f ' = 1/(1+x^2).\n\nFurthermore integral (2 f '  f dx) = f^2.\n\nTherefore integral (arctan (x) /(1+x^2)) = (1/2) (arctan(x))^2
380391
Let alpha = arcsin 6x\nThen sin(alpha) = sin(arcsin 6x) = 6x\n\nNow note d(sin(alpha))/dx = d(sin(alpha))/d(alpha) * d(alpha)/dx (Chain Rule)\n\nd(6x)/dx = cos(alpha) * d(alpha)/dx\n6 = cos(alpha) * d(alpha)/dx\nTherefore, d(alpha)/dx = 6/cos(alpha)\n\nNow consider a right triangle with angle alpha < pi/2.  From before, we know sin(alpha) = 6x so we know the following:\n\nlength of leg opposite angle (a)/length of hypotenuse (c) = 6x\n\nso that...\n\na = 6cx\n\nUsing the Pythagorean Theorem, we get the length of the other leg b:\n\nb = sqrt(c^2 - 36c^2x^2) = c*sqrt(1 - 36x^2)\ncos alpha = b/c = sqrt(1 - 36x^2)\n\nTherefore d(alpha)/dx = 6/cos(alpha) = 6/sqrt(1 - 36x^2)\n\nNow you can proceed to solve using L'Hopital OR...\n-----------\nMake this problem more interesting and recall one of the first indeterminate forms you probably solved using the Rule (i.e. lim x-> 0 sin x/x = lim x -> 0 cos x/1 = 1/1 = 1).  This tells us that sin x ~ x for small angles x.  Therefore, arcsin x ~ x for small angles x so that that lim x -> 0 arcsin 6x/5x = lim x->0 6x/5x = lim x->0 6/5 = 6/5.  This can be made more precise with delta-epsilon definitions.
740147
If you have a TI-83 you can graph it and find it easily...\n\ny= button (plug in equation) ---> hit graph\n\nthen hit the second button(yellow)--->calc(same as trace button)\n\nbecause the graph was concave up you want to find a minimum...so click that...and simply enter once on the left side of the graph, once on the right, and then guess in the middle.  It will then give you the exact answer.\n\n(1,6) should be the correct coordinate.
830888
OK, first let's try to figure out the arctan part.  Arctan(-1) means the angle whose tangent is -1, right?  And the tangent is the sine divided by the cosine.  So you're looking for angles whose sines and cosines are equal, but with opposite signs.  They point out to the right and down, or out to the left and up, at 45-degree angles.  That means the angles are 135 degrees and 315 degrees, or 3pi/4 and 7pi/4 radians.\n\nNow, the question asked you to find the sines of these two angles.  The sine of 3pi/4 is sqrt(2)/2, and the sine of 7pi/4 is -sqrt(2)/2.  So your answer is +/- sqrt(2)/2.
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
904086
I would use substitution method on this one...\n\nFirst, break it up into the sum of 2 integrals...\n\nInt(cost/(1+sin^2t))dt + Int(-2sint/(1+cos^2t))dt\n\nFor the first integral let u = sint\n\ndu = cost\n\nThen you have an integral that you can evaluate ...\n\nInt(1/(1 + u^2)du = arctan(u)\n\nSubstitute cos(t) back in for u and you get arctan(cos(t)), evaluate that from 0 to pi/2.\n\nFollow the same reasoning for the 2nd one!
561110
If you're talking about GRADIAN (a measure of angle, 400 to a full circle):\n\n100 Gra = PI/2 Rad = 90 Deg\n\nFrom Gra:\n\n  to Rad: x PI/200\n  to Deg: x 0.9\n\nTo Gra:\n\n  from Rad: x 200/PI\n  from Deg: x 10/9\n\nIf you're talking about GRADIENT expressed as (vertical distance / horizontal distance), you need to use the tangent function (TAN) and inverse tangent (ARCTAN, TAN-1 or INVTAN, depending on your nomenclature)\n\nGradient = TAN(degrees)\n\nGradient = TAN(180.radians/PI)\n\nDegrees = ARCTAN(gradient) or\n          TAN-1(gradient) or\n          INVTAN(gradient)\n\nRadians= (PI/180).ARCTAN(gradient) or\n         (PI/180).TAN-1(gradient) or\n         (PI/180).INVTAN(gradient)
726975
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...