what is the Horizontal Translation of Tangent?
106150
I am not sure what you mean. I assume you mean what is tan(x+a)=sin(x+a)/cos(x+a)\n=[sin(a)cos(x)+sin(x)cos(a)] / [cos(x)cos(a)-sin(x)sin(a)]\n=[tan(a)+tan(x)] / [1-tan(x)tan(a)]\n\nSo tan(x), if the argument x is translated by a, becomes this expression.
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...
95376
SOHCAHTOA....geometry term Sine Opposite Over Hypotenuse,Cosine adjacent over hypotenuse, Tangent opposite Over Adjacent
1035212
I know of two:\n\nTranslation and rotation: \n\n1) translation :\n\n Take a rectangle and cut a diagonal and extended the lines  from all three side of the triangle at right angle to it  from the side that is made from cutting the rectangle until those lines equal the lenght of the side of the triangle drawn from  and it  form three boxes  around all three sides of the triangle. Now let take the big side an calulate the square  of it ,then calulate the squares of the other two sides and add themup,  it should equal the square of the big side and that proves the pythagorean theorem. \n\nRotation: \n\nSee the show : Project Mathematics by Cal tech : Entitled "The Pythagorean Theorem" it explain better  than I can, It worth watching. Very good show on Math & Geometry
1186712
arctan x should be interpreted as "the angle whose tangent is x". So draw a right triangle and include an angle whose tangent is x. Then figure out the sine of that angle in terms of x and you're done.\n\nEasiest way to do this is to let the side opposite the angle be x and the side adjacent to the angle be 1. Then the hypotenuse will have length sqrt(1+x^2).
52711
shift or change\nfixed or immobile\nvector
787829
SOHCAHTOA\nsinθ = O/H = a/c\ncosθ = A/H = b/c\ntanθ = O/A = a/b
1114168
Ok, here's how i think you do it.\n\nto get the horizontal asymptote, you have to get the limit of the functon when x tends to be infinite, so you would get something like this\n\nlim     (3x-x^2)/(4-x)\nx→∞\n\nthen you substitute ∞ on the x's and you get\n\n∞/∞, that's not defined, so what you use is l'hopital law where you derive both the nominator and the denominator of a fraction, so you will get\n\nlim    3-2x/-1\nx→∞\n\nnow you can substitute ∞ and you get\n\n(3-2*∞)/-1 =∞\n\nand you do the same with -∞ and you get as an answer -∞\n\nwhich means there is no horizontal asymptote.\n\nI hope this helps.
973710
"Main Idea: If f(x) is a function that has a derivative at a point x = a (i.e., f'(a) exists), then the line tangent to the graph of f at x = a, given by:\n\ny - f(a) = f'(a)(x - a);\nor, equivalently y = f(a)+f'(a)(x - a) can be an exceptionally good approximation of f(x) as long as x is close to a.\n\nAs a definition, we call the tangent line\nL(x) = f(a)+f'(a)(x - a)\nthe linearization (or linear approximation) of f at a. The\npoint x = a is the center of the approximation." (Ref 1)\n\nSo for f(x) = lnx at x = 8\n\nL(x) = ln8 + 1/8 (x - 8) (as f(8) = ln 8 and f'(x) = 1/x so f'(8) = 1/8)\n= 1/8 x + ln 8 - 1\n\nThis is useful to say estimate ln 8.1 if you know ln 8 (≈ 2.07944)\n\nSo ln (8.1) ≈ (8.1)/8 + 2.07944 - 1\n= 1.0125 + 2.07944 - 1\n= 2.09194\n\nBy calculator ln 8.1 = 2.09186..... so its pretty close (8^O)
715400
Angle of line from horizontal position.
619642
Think about it. tangent is an odd function such that\nf(-x) = -f(x)\ntan(-3pi/4) = -tan(3pi/4)\n\nObviously we have a measure in radians, we want degrees.\nYou should know that:\n1 radian = 180/pi degrees\n1 degree = pi/180 radians\n\nConvert to radians\n3pi/4 radians\n3pi/4 (180/pi degrees)\n45 x 3\n135\n\nThus 3pi/4 radians = 135 degrees\n\nPlug back in \n-tan (135)\n\nWe have a reference angle of 45 degrees with the x axis. Tangent in the second quadrant is negative. What is the tangent of 45? It's 1, but because we're in the second quadrant, tan (135) = -1.\n\nHowever, don't forget we have that -tan. So, -1 x -1 = 1.\n\nAs a result, tan(-3pi/4) = 1.
889456
I'm not sure I understand the question, but if you want to translate the vector from its current position to the origin, then just add (4,-5) to its beginning and end points.\n The vector will now be pointing from the origin to (6,1).\nHope this helps.
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).
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
984494
sine opposite/hypotenuse\ncosine adjacent/hypotenuse\ntangent opposite/adjacent
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
355882
The relationship is defined in one dimension as f'(x) being the slope of the tangent line to the graph of f(x) at the point x. In symbols, f'(x) is the limit as h approaches 0 of the slope of a secant to the graph passing through x: [f(x + h) - f(x)]/h.\nf'(x) being the slope of the tangent line to the graph of f at x has its advantages. When f'(x) is 0, we know the tangent line is horizontal there, meaning the graph is either at a minimum, a maximum, or a temporary flat area. Higher derivatives (treating the derivative as a function itself) can tell us which of the above it is.\nDerivatives are most concrete in physical examples. If x(t) is the position of an object at time t, then the velocity of the object at time t can be idealized as the derivative of x(t) at t (instead of the average velocity, which is the slope of a secant line to the graph of x(t)). Similarly, the acceleration at time t is given by the second derivative.\nIn higher dimensions, where x = (x1, x2, ..., xn) is a point and f(x) = (f1(x), f2(x), ..., fn(x)) is also a point, this is generalized to f'(x) being the unique linear function that vanishes with the change in f at x. In symbols, the limit as ||h|| vanishes of [f(x + h) - f(x) - f'(h)]/||h|| vanishes. This coincides with the definition in 1 dimension above. In higher dimensions, f'(x) corresponds to the basis of a tangent space at x, as it did in 1 dimension.\nIf f is a map from R^2 into R, f'(x) is a linear map from R^2 into R, which is a plane. This plane is a tangent space to the graph of f at x. Higher-dimensional tangent spaces are not as visualizable.