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The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product = For example, if a force of 10 newtons ( F = 10 N ) acts along a point that travels 2 metres ( s = 2 m ), then W = Fs = (10 N) (2 m) = 20 J .
The cross product on the right-hand side of this expression is a (not necessarily unital) surface normal determined by the parametrisation. This formula defines the integral on the left (note the dot and the vector notation for the surface element).
Hence the constant "k" is the product of x and y. The graph of two variables varying inversely on the Cartesian coordinate plane is a rectangular hyperbola. The product of the x and y values of each point on the curve equals the constant of proportionality (k). Since neither x nor y can equal zero (because k is non-zero), the graph never ...
The curve is a cycloid, and the time is equal to π times the square root of the radius (of the circle which generates the cycloid) over the acceleration of gravity. The tautochrone curve is related to the brachistochrone curve, which is also a cycloid.
Since the angular velocity ω = v/r is constant, the area swept out in a time Δt equals ω r 2 Δt; hence, equal areas are swept out in equal times Δt. In uniform linear motion (i.e., motion in the absence of a force, by Newton's first law of motion), the particle moves with constant velocity, that is, with constant speed v along a line.
Considering that the object is a person sitting inside a plane moving in a circle, the two forces (weight and normal force) will point down only when the plane reaches the top of the circle. The reason for this is that the normal force is the sum of the tangential force and centripetal force.
Geometrically, when the scalar field f is defined over a plane (n = 2), its graph is a surface z = f(x, y) in space, and the line integral gives the (signed) cross-sectional area bounded by the curve and the graph of f. See the animation to the right.
From this derivative equation, in the one-dimensional case it can be seen that the area under a velocity vs. time (v vs. t graph) is the displacement, s. In calculus terms, the integral of the velocity function v(t) is the displacement function s(t). In the figure, this corresponds to the yellow area under the curve.