how to explain anyone about tangents and normal?
814174
TANGENT                                                                                        Tangent is a line which touches the circle and the radius is always perpendicular to the tangent at the point of intersection between circle and tangent\n\nNORMAL\nNormal is a line which is perpendicular to the tangent at the point of intersection between circle and tangent and it always passes through the center of circle
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
226613
What are the things that you don't understand in Calculus B?
711470
Try here:/\n\nhttp://archives.math.utk.edu/visual.calculus/
488283
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
502657
Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.
319720
explain the something very clearly
893970
x=t^2 - 1\ny=t^4 - 4t\n\ntangent vector: ( 2t, 4t^3 -4)\nthe tangent vector is horizontal if:4t^3 -4 =0, so t^3=1, so t=1,\nin this case\nx= 0, and y = 1-4=-3\n\nthe tangent vector is vertical if: 2t=0, so t=0\nin this case\nx= -1 and y= 0\n`
918737
The equation for the tangent is:\ny = f'(a) * (x-a) + f(a)\n\nYou need to calculate the derivative h'(t) for t=3s and plug that into the equation, adapted to your problem:\n\ny = h'(3) * (t-3) + h(3)\n\nI'll let you do the maths...
527822
What's your sine? Oops. I seem to be going off on a tangent. I can't function without you. I don't know Y, but you remind me of my X. It's plane to see that we'd make a great ordered pair, if you get by point. I'm not line. I really have no axis to grind. I've tried every angle I can think of to get this right. I don't mean to be obtuse, but I think we're complementary.
735957
Well. You can keep drawing tangents until one which you find is the steepest. From your info, i have the idea that the curve is in the shape of a bell. So perhaps you can try drawing tangents at the mid-range of the curve.\n\nHope this helps=)
1010092
Aaaaaargh you need CALCULUS aaaargh....\n\nIt took me 3 classses years ago to get a grip on that, but I've lost what I had.  You must be in high school or you would have been exposed to it by now - try looking up some calculus courses online, or possibly getting (yes I'm serious) one of those "Made Easy" textbooks, specifically "Calculus Made Easy".  Or look for basic calculus sites on the web, I found a few, but not specifically what you're looking for.  If I find something useful, I'll come back & post it.\n\nI think what you need to focus on are Limits, which are about breaking up the space under a curve into rectangles.\n\nI can say no more...\n\nFound Something!  It's Integrals.  Check out the link.\n\nHere's a better site:
127894
I can probably help u with it....I am in grade 10, skipped grades 9...reasonably good at Algebra and Trig, Geomerticaly retarded though T,T"\ndamnedatheist@gmail.com can help u
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
743164
Remember the definitons of tangent, cotangent, etc. as they relate to right triangles.  Draw a triangle for both α and β.  Use the pythagorean theorem to finr the hypotneuses.  You will find the sides of the α triangle to be 3, 4, 5.  so:\n\nsin α = 4/5\ncos α = 3/5\n\nDo the same for β and the triangle is 5, 12, 13 with:\n\nsin β = 12/13\ncos β = 5/13\n\nFinaly, use the identity:\n\n sin(α-β) = sin α  cos β - sin β cos α \n\nSubstituting:\n\nsin(α-β) = (4/5)(5/13) - (12/13)(3/5) = (20 - 36)/65 = - 16/65
629401
If the coordinates of the centre of the circle be (a,b) and its radius be 'r', then the equation representing this circle on a rectangular coordinate system is (x-a)^2 + (y-b)^2 = r^2. If the tangent has got to be vertical it has to touch the circle on either left or right of the circle. Further as any tangent to a circle is perpendicular to the radius of the circle throught the point of tangency so the radius, here, must be horizontal (as tangent is vertical). So the equation of the vertical tangent is either x = a + r (tangent on right side of circle) or x = a - r (tangent on right side of circle).