Something to do with tangent?
956162
It depends on which leg you know the length of.  If it's the "adjacent"leg--the one that, along with the hypotenuse, forms the angle--then remember that the cosine of an angle is defined as the length of the adjacent angle divided by the length of the hypotenuse.\n\nIf it's the "opposite" leg--the one that isn't part of the angle--then you use the fact that the sine is defined as the length of the opposite leg divided by the length of the hypotenuse.\n\nEither way, you use the inverse trig functions on your calculator to find the angle.
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...
120626
There's a derivatives tutorial on the following page:\n\nhttp://www.karlscalculus.org/calculus.html#s4
447477
In a triangle the sum of the angles is 180°, but the sum of the tangents is not anything in particular.  The tangent function is 0 for 0°, but it increases without bound as the angle approaches 90°.  So if the triangle contains one angle very very close to 90° and the other 2 are the rest of the 180°, the sum of tangents can be as large as you like.  On the other hand, if two of the angles are very very close to 0 and the third very close to 180, the sum of the tangents can be as close to 0 as you like.
57176
http://www.calculus.org/
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.
826563
multiply the tangent of the given angle by the known side as long as it's not the sloping side (i.e. the hypotenuse). You will need a scientific calculator to do this. I am assuming of course that you are dealing with a right triangle. This will give you the other side of the triangle. Example:\n\nlets say you have side A of a right triangle and an angle of 30 degrees and A = 10. Take your scientific calculator press 30 and then the tan button depending on the type of caculator you have. Now multiply that result by 10 and you will get side B.\n\nSide C is found by A^2 + B^2 = C^2. take the square root for your answer.\n\nNow if the given angle is opposite the given side. Just calculate the angle of the given side by adding the given angle to 90 degrees and subtract the result from 180, and then multiply the tangent of the new new angle by the given side.\n\nTheoretically you can check the answer by multiplying the new side by the tangent of the new angle. You should in theory get the given side. But my trig is so rusty I may have forgotten more than  you will ever know. good luck.
201825
Well um...\n\na = r''\n\na = (tangential acceleration * unit tangent) + (normal acceleration * unit normal) = (at * T) + (an * N)\n\nT and N are your unit tangent and unit normal of the function that defines your position.\n\nat = (r' dotted with r'') / (|r'|)\n\nThose are the tangential acceleration components but for this problem i think that you are going to have to reparamiterize your curve.
324765
What does this have to do with math?
250283
A little more elaboration?
1116918
In Calculus of Varations deals with minimization problems the most elementary of which is the Brachistochrone.  This is the problem of the shape of a wire which provides the fastest path between two points for a marble sliding down it.  I don't know what applications of differentiation is.
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.
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.
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).
761613
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
258618
This is what I wrote for a very close question to this one. Just ignore the graphing part. I didn't feel like doing the question right now, but this is the way to do it:\n\n"If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent."
605893
U mean Plato's CUrsed Triangle?,,\nWell, check out the gradient of the two triangle, ITS different!.\nThe difference of the gradient makes a whole in the next triangle