application of differential calculus in astronomy?
93183
Differential (and integral) calculus is used EVERYWHERE in astronomy. Without it, one would be able to do very little. \nPlaces where calculus is used in astronomy include:\n\n1. Design of the telescopes \n2. Design of the mirrors and lenses\n3. Celestial mechanics laws objects obey\n4. Techniques for processing collected data to remove noise, fit models, etc\n5. Computing models of limb-darkening, stellar atmospheres, star spots, Roche lobes, general relativistic lensing, stellar magnetic fields, planetary atmospheres; just about anything used to model the data\n\nOne could come up with a HUGE long list. Any single project in astronomy uses calculus all the way from the design of the equipment, to the data processing, modelling and analysis. \n\nTo go into explicit detail on all or any of these here is probably beyond the scope of what you are looking for and not really possible here.
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
308389
this question is miscategorized.  this is not a question about astronomy.\n\nhttp://en.wikipedia.org/wiki/Astronomy
805613
Since this is a slice of a sphere, it makes sense to do things all in sperical coordinates.  From your limits, that is what you are doing. All of your limits look fine.  All that remains is to convert the funciot to spherical coordinates using:\n\nx = p*sin(phi)*cos(theta)\ny = p*sin(phi)*sin(theta)\n\nAnd use- a differential volume of dv = p^2 * sin(phi) * dp dphi dtheta\n\nAfter integration I get:\n\n(R^4)/4 * (atan(4/3) - 4/25) ( 3*sin(b) + cos(b) - 3)\n\nWhere R = 25, b = atan(sqrt(277/123))
816697
I havent a clue what your are asking... are we talking about astronomy terms?  Elevation and declination?  Or are you asting about Lattitude and Longitude?
598106
you can use the following proportion:\n\ndegree/180 = radians/pi
1105564
Don't know try it and tell us.  \n\nWould you sniff this?\n\nhttp://plaza.ufl.edu/kbeaz/severe_calculus.jpg
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
711470
Try here:/\n\nhttp://archives.math.utk.edu/visual.calculus/
1178962
To solve this it's a pretty straightforward application of Laplace transforms. You can do this question using convolutions as follows:\n\nFirst take the laplace transfom of the ODE:\n\nlet L{y(t)}(s) = Y(s) be the laplace transform of y(t).  Then we have:\ns^2(Y(s))  - 2s + 3Y(s) = 1/s^2 - (e^-s)/s\n\nrearranging we have Y(s) = (1/s^2 - e^-s/s + 2s) (1/(s^2+3)) = G(s) H(s) where G(s) = (1/s^2 - (e^-s)/s)  and H(s) =  (1/(s^2+3)) \n\nnow using the inverse laplace formula for the convolution of two functions:\n\ny(t) = L(^-1){Y(s)} = integral from 0 to t of (h(t-x)g(x)dx) + 2cos(sqrt(3)t)\n\n where h(t) = 1/sqrt(3) sin(sqrt(3)t) and g(t) = (t - u1(t))\n\nand the integral can be evaluated. \n\nSorry for the readability but I don't think yahoo answers can render math text properly. Also, as a note, this question would be far easier to solve using a partial fraction method rather than unecesarily using convolutions.
392380
All kinds of simple, easy calculus examples in real life (especially engineering!).  These depend on the level of math knowledge, but for University Calc. II:\n\nOptimization: (i.e. give a box with a volume, find minimum surface area) or (here's a certain amount of fencing, make the largest enclosure you can).  (Lagrangian Multipliers? did I even spell that right?  We looked at those when there are a lot of constraints on the problem (i.e. relationship between box dimensions must follow .....).\n\nForces: for example, average fluid pressure on a dam, given a function of distributed force, or fluid pressure on a submerged object. (Can be done in single or multiple integrals).\n\nRelated Rates:  i.e. two cars approach at right angles, different speeds, when will they meet.  Or, dropping a ball near a light source, find speed of shadow as function of angle made with light source, distance from ground, etc.  Differetial calculus example.\n\nArea and Volume: Double and triple integrals to find the volumes and areas of shapes. (Or single integrals, volumes of revolution, areas from curve to axis, if multi-variable calculus not covered).\n\nCurve Fitting: really good real-life use, take a set of data points from an experiment, select a curve type (i.e. quadratic, cubic, straight line, logarithmic, etc) and use minimization to get the best curve fit (set up an equation for the residuals, sum of (function) - (data point), and take derivatives with respect to each unknown coefficient in the function, set all derivatives equal to zero, solve simulatneously for curve coefficients. Now we can predict the outcome of the experiment at intermediate values, (as long as the curve fit is good).\n\nCentre of Mass, Centroids, etc: Find the balancing point on certain objects or groups of objects using integration. Can be single or multiple integrals, depending on complexity of shape.\n\nFields:  i.e. graviational fields, electrical fields, etc.  Or make up some fields.  Integration to find work required by a particle to traverse a path through some field.  Conservative and non-conservative fields.  Can use as an example of energy needed to move satellites along paths (equations) in space near earth's gravity, an electron through an electrical field, etc.  I guess some of these examples the field is conservative and it becomes kinda trivial if you work with end points, but don't have to.\n\nThese are some of the applications we used in our Calculus II-type course.  The hardest part is always setting the problem up, not solving, (which makes for good application problems I guess).  If multi-variable calculus isn't covered yet, then some of these examples would be kinda rough (i.e. fields).  Anyway, hope these are useful. \n\nCheers,
236377
Let the two foci be at (a, b) and (c, d) which we assume are distinct -- if they are identical then you have a circle and a much easier problem. Afterwards we can check the result to see if relaxing that restriction would introduce a division by zero.\n\nLet the sum of the distances from the two foci (in the case of a circle, the diameter) be s.\n\nLet theta be the angle from (a, b) following the usual convention that an angle of zero goes from (a, b) to (a + infinity, b) and theta increases counterclockwise.\n\nThe ray going out from (a, b) in the angle theta direction can be parameterized as the curve\n\n(x(t), y(t)) = (a + t cos theta, b + t sin theta) for t >= 0\n\nWe want to find t such that (x(t), y(t)) satisfies that the sum of its distances from the two foci is s:\n\nThe distance from (a, b) is\n\nsqrt((x(t) - a)^2 + (y(t) - b)^2)\n = sqrt((t cos theta)^2 + (t sin theta)^2)\n = sqrt( t^2 (cos^2 theta + sin^2 theta) )\n = sqrt( t^2 )\n = t\n\nImagine that. :-)  (We don't need to say the absolute value of t because we know t >= 0).\n\nSo we want to find t such that\n\nt + sqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s\n\nsqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s - t\n\n(a + t cos theta - c)^2 + (b + t sin theta - d)^2 = (s - t)^2\n\nt^2 cos^2 theta + 2 (a - c) t cos theta + (a - c)^2 + t^2 sin^2 theta + 2 (b - d) t sin theta + (b - d)^2 = s^2 - 2 s t + t^2\n\nWell look at that, on the left hand side the quadratic term in t is t^2 (cos^2 theta + sin^2 theta) = t^2 whic is also the quadtratic term in t on the right.  So we can drop those and we get\n\nt (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s) = s^2 - (a - c)^2 - (b - d)^2\n\nIf s is at least the distance between the two foci then the right hand side is >= 0 because it is s^2 minus the square of the distance between the two foci.\n\nSo let T = (s^2 - (a - c)^2 - (b - d)^2) / (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s)\n\n(assuming the denonimator is nonzero) and the point you want is\n\n(x(T), y(T)) = (a + T cos theta, b + T sin theta)
665570
I assume delta(x, y, z) = 1 is the mass density ... isn't rho usually used for that?  Anyway, you need to integrate mass times square of perpendicular distance from axis of rotation.\n\nLooks like it will be messy.\n\nMaybe outer x from 0 to 1, middle y from 0 to 1 - x^2, inner z from 0 to (12 - 4x - 3y)/2, of delta(x, y, z) (x^2 + y^2) dz dy dx.
826981
It looks to me like it's .5d which is not too far from your answer.\n\nFor simplicity, use a unit of length in which d = 1 and a unit of time in which the speed of each of the two bugs is also 1. Let r be the distance between them and let h be the angle between the y direction and the vector from S to R. In other words h starts out pi/2 and approaches 0.\n\nThe time derivatives are:\n\ndh/dt = -sin(h)/r , dr/dt = -1 + cos(h)\n--->\ndr/dh = r(1-cos(h))/sin(h)\n\nSeparating the variables r and h to opposite sides of the equation and integrating both sides with the initial condition (r=1 at h=PI/2) gives:\n\nr = tan(h/2)/sin(h)\n\nYou want the distance between the two bugs as t-->oo or equivalently as h --> 0 which is r=1/2 (use L'hopital's rule for example), or d/2 in terms of the initial distance d.\n\nOf course S never actually reaches the y axis but approaches it asymptotically.\n\n(Maybe I should add, I first tried to formulate the problem in terms of the Cartesian coordinates and was able to write a differential equation for the curve but it looked very difficult to solve. Finally tried using the above coordinates (r,h) and it became much easier)
222928
There is no function that exists that has as its derivative e^(x^2).  As someone else pointed out, you can use the Maclaurin series representation and integrate term by term.  However, this series converges very slowly and would require a lot of terms for two place decimal accuracy, especially if x is large.
796855
if you mean an ellipse....\n\narea is equal to pi * minor radius * major radius
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.