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We shall try to state what it is exactly that makes one uneasy. However,we shall first point out a more geometric construction of the tangent to a curve atone of its points P0 (see Fig. 5.3).Take an arbitrary point P of the curve different from P0 . The line determinedby the pair of points P0 and P , as already noted, is called a secant in relation tothe curve.
We now force the point P to approach P0 along the curve. If the secanttends to some limiting position as we do so, that limiting position of the secant isthe tangent to the curve at P0 .Despite its intuitive nature, such a definition of the tangent is not available to usat the moment, since we do not know what a curve is, what it means to say that6 Thisis a slight deviation from the more common notation Tx0 R or Tx0 (R).1825Differential Calculus“a point tends to another point along a curve”, and finally, in what sense we are tointerpret the phrase “limiting position of the secant”.Rather than make all these concepts precise, we point out a fundamental difference between the two definitions of tangent that we have introduced.
The secondwas purely geometric, unconnected (at least until it is made more precise) withany coordinate system. In the first case, however, we have defined the tangent toa curve that is the graph of a differentiable function in some coordinate system.The question naturally arises whether, if the curve is written in a different coordinate system, it might not cease to be differentiable, or might be differentiable butyield a different line as tangent when the computations are carried out in the newcoordinates.This question of invariance, that is, independence of the coordinate system, always arises when a concept is introduced using a coordinate system. The question applies in equal measure to the concept of velocity, which we discussed inSect. 5.1.1 and which, as we have mentioned already, includes the concept of atangent.Points, vectors, lines, and so forth have different numerical characteristics in different coordinate systems (coordinates of a point, coordinates of a vector, equationof a line).
However, knowing the formulas that connect two coordinate systems, onecan always determine from two numerical representations of the same type whetheror not they are expressions for the same geometric object in different coordinatesystems. Intuition suggests that the procedure for defining velocity described inSect. 5.1.1 leads to the same vector independently of the coordinate system in whichthe computations are carried out. At the appropriate time in the study of functionsof several variables we shall give a detailed discussion of questions of this sort. Theinvariance of the definition of velocity with respect to different coordinate systemswill be verified in the next section.Before passing to the study of specific examples, we now summarize some of theresults.We have encountered the problem of the describing mathematically the instantaneous velocity of a moving body.This problem led us to the problem of approximating a given function in theneighborhood of a given point by a linear function, which on the geometric level ledto the concept of the tangent.
Functions describing the motion of a real mechanicalsystem are assumed to admit such a linear approximation.In this way we have distinguished the class of differentiable functions in the classof all functions.The concept of the differential of a function at a point has been introduced. Thedifferential is a linear mapping defined on displacements from the point under consideration that describes the behavior of the increment of a differentiable functionin a neighborhood of the point, up to a quantity that is infinitesimal in comparisonwith the displacement.The differential df (x0 )h = f (x0 )h is completely determined by the numberf (x0 ), the derivative of the function f at x0 , which can be found by taking thelimit5.1 Differentiable Functionsf (x0 ) =183limEx→x0f (x) − f (x0 ).x − x0The physical meaning of the derivative is the rate of change of the quantity f (x)at time x0 ; its geometrical meaning is the slope of the tangent to the graph of thefunction y = f (x) at the point (x0 , f (x0 )).5.1.5 Some ExamplesExample 1 Let f (x) = sin x.
We shall show that f (x) = cos x.Proof2 sin( h2 ) cos(x + h2 )sin(x + h) − sin x= lim=h→0h→0hhsin( h )h= lim cos x +· lim h 2 = cos x.h→02 h→0 ( 2 )limHere we have used the theorem on the limit of a product, the continuity of thefunction cos x, the equivalence sin t ∼ t as t → 0, and the theorem on the limit of acomposite function.Example 2 We shall show that cos x = − sin x.Proof−2 sin( h2 ) sin(x + h2 )cos(x + h) − cos x= lim=h→0h→0hhsin( h )h= − lim sin x +· lim h 2 = − sin x.h→02 h→0 ( 2 )limExample 3 We shall show that if f (t) = r cos ωt, then f (t) = −rω sin ωt.Proofh−2 sin( ωhr cos ω(t + h) − r cos ωt2 ) sin ω(t + 2 )= r lim=h→0h→0hhlimsin( ωh )h= −rω lim sin ω(t + ) · lim ωh2 =h→02 h→0 ( 2 )= −rω sin ωt.Example 4 If f (t) = r sin ωt, then f (t) = rω cos ωt.1845Differential CalculusProof The proof is analogous to that of Examples 1 and 3.Example 5 (The instantaneous velocity and instantaneous acceleration of a pointmass) Suppose a point mass is moving in a plane and that in some given coordinatesystem its motion is described by differentiable functions of timex = x(t),y = y(t)or, what is the same, by a vectorr(t) = x(t), y(t) .As we have explained in Sect.
5.1.1, the velocity of the point at time t is the vectorv(t) = ṙ(t) = ẋ(t), ẏ(t) ,where ẋ(t) and ẏ(t) are the derivatives of x(t) and y(t) with respect to time t.The acceleration a(t) is the rate of change of the vector v(t), so thata(t) = v̇(t) = r̈(t) = ẍ(t), ÿ(t) ,where ẍ(t) and ÿ(t) are the derivatives of the functions ẋ(t) and ẏ(t) with respectto time, the so-called second derivatives of x(t) and y(t).Thus, in the sense of the physical problem, functions x(t) and y(t) that describethe motion of a point mass must have both first and second derivatives.In particular, let us consider the uniform motion of a point along a circle ofradius r. Let ω be the angular velocity of the point, that is, the magnitude of thecentral angle over which the point moves in unit time.In Cartesian coordinates (by the definitions of the functions cos x and sin x) thismotion is written in the formr(t) = r cos(ωt + α), r sin(ωt + α) ,and if r(0) = (r, 0), it assumes the formr(t) = (r cos ωt, r sin ωt).Without loss of generality in our subsequent deductions, for the sake of brevity,we shall assume that r(0) = (r, 0).Then by the results of Examples 3 and 4 we havev(t) = ṙ(t) = (−rω sin ωt, rω cos ωt).From the computation of the inner product+,v(t), r(t) = −r 2 ω sin ωt cos ωt + r 2 ω cos ωt sin ωt = 0,as one should expect in this case, we find that the velocity vector v(t) is orthogonalto the radius-vector r(t) and is therefore directed along the tangent to the circle.5.1 Differentiable Functions185Fig.
5.4Next, for the acceleration, we havea(t) = v̇(t) = r̈(t) = −rω2 cos ωt, −rω2 sin ωt ,that is, a(t) = −ω2 r(t), and the acceleration is thus indeed centripetal, since it hasthe direction opposite to that of the radius-vector r(t).Moreover,22a(t) = ω2 r(t) = ω2 r = |v(t)| = v ,rrwhere v = |v(t)|.Starting from these formulas, let us compute, for example, the speed of a lowaltitude satellite of the Earth. In this case r equals the radius of the earth, that is,r = 6400 km, while |a(t)| = g, where g ≈ 10 m/s2 is the acceleration of free fall atthe surface of the earth.Thus, v 2 = |a(t)|r ≈ 10 m/s2 × 64 × 105 m = 64 × 106 (m/s)2 , and so v ≈8 × 103 m/s.Example 6 (The optic property of a parabolic mirror) Let us consider the parabola1 2y = 2px (p > 0, see Fig.
5.4), and construct the tangent to it at the point (x0 , y0 ) =1 2(x0 , 2px0 ).Since f (x) =1 22p x ,we havef (x0 ) = lim1 22p xx→x0−1 22p x0x − x0=11lim (x + x0 ) = x0 .2p x→x0pHence the required tangent has the equationy−1 2 1x = x0 (x − x0 )2p 0 por1x0 (x − x0 ) − (y − y0 ) = 0,pwhere y0 =1 22p x0 .(5.25)1865Differential CalculusThe vector n = (− p1 x0 , 1), as can be seen from this last equation, is orthogonalto the line whose equation is (5.25). We shall show that the vectors ey = (0, 1)and ef = (−x0 , p2 − y0 ) form equal angles with n. The vector ey is a unit vectordirected along the y-axis, while ef is directed from the point of tangency (x0 , y0 ) =1 2(x0 , 2px0 ) to the point (0, p2 ), which is the focus of the parabola.
Thuscos eyn =&ey , n'1=,|ef ||n| |n|pp1 21 21 2&ef , n'1p x0 + 2 − 2p x02 + 2p x0===.cos efn =pp1 2 21 2 2|ey ||n||n|2|n| x0 + ( 2 − 2p x0 )|n| ( 2 + 2p x0 )Thus we have shown that a wave source located at the point (0, p2 ), the focus ofthe parabola, will emit a ray parallel to the axis of the mirror (the y-axis), and thata wave arriving parallel to the axis of the mirror will pass through the focus (seeFig. 5.4).Example 7 With this example we shall show that the tangent is merely the bestlinear approximation to the graph of a function in a neighborhood of the point oftangency and does not necessarily have only one point in common with the curve,as was the case with a circle, or in general, with convex curves.
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