Why do we use calculus to determine the exact expressions for derivatives? For example, to determine the derivative of y = ax^2 + bx + c we typically consider two points on this curve, (x,y) and (x+dx,y+dy), where dx and dy are initially regarded as small finite quantities. So we have the two equations y = ax^2 + bx + c (y+dy) = a(x+dx)^2 + b(x+dx) + c Subtracting the first from the second gives dy = 2ax dx + a(dx)^2 + b dx So far, so good. Then, since dx is considered to be non-zero, it is legitimate to divide by dx, giving the result dy/dx = 2ax + a dx + b Now for the interesting part: "Let dx go to zero". Of course if dx actually EQUALS zero then we weren't justified in dividing the previous equation by zero. To avoid this little inconsistency the epsilon-delta limit device is used to establish that the LIMIT of the ratio dy/dx is 2ax + b as dx APPROACHES zero. It could be argued that the "limit" concept has been introduced here because we had reached the "limit" of our ability to algebraically solve equations. To illustrate, consider again the case y = ax^2 + bx + c We want the slope q of the line y = qx + p tangent to this curve. Equating y values, we have ax^2 + (b-q)x + (c-p) = 0 Solving this quadratic for x gives the point(s) of intersection (q-b) +- sqrt[ (b-q)^2 - 4a(c-p) ] x = ----------------------------------- 2a There is a single point of intersection (i.e., the line is tangent to the curve) iff the sqrt term vanishes, which leaves q = 2ax + b. There are no divisions by "something approaching zero", no "limits", "differentials", or "infintesimals" involved in this derivation. Of course, this purely algebraic approach to finding derivatives becomes unwieldy when dealing with more complicated functions, and the "x+dx" trick is a far more efficient way of determining the derivative of a function F(x) without having to solve explicitly for the roots of F(x)-L(x). On the other hand, if our objective was an explicit algebraic solution of F(x)-L(x), then calculus doesn't help. Hence it could be said that we resort to calculus when algebra becomes too difficult (somewhat as we resort to algebraic numbers when integer arithmetic becomes too difficult). What about integration? Suppose we want to know the average height of all the point in the triangle shown below: ______ | /\ | / \ | / \ / \ H / \ / \ | / \ ___|__ /_____________________\ | ------- B -----------| The straight-forward calculus approach (without much thinking) would probably be to treat the width of the triangle w(h) at height h above the base as the "weight" at h, and evaluate the mean as the weighted average given by the ratio of integrals H INT h w(h) dh h=0 h_mean = ---------------- H INT w(h) dh h=0 The width varies linearly from a value of B at h=0 to a value of 0 at h=H. Therefore, we have w(h) = B(1-h/H). Inserting this expression into the above formula and evaluating the integrals gives / 1 1 B \ |H ( - B h^2 - - - h^3 ) | \ 2 3 H / |0 h_mean = --------------------------- / 1 B \ |H ( B h - - - h^2 ) | \ 2 H / |0 B H^2 (1/2 - 1/3) = ----------------- = H/3 B H (1 - 1/2) Of course, with a bit of foresight we could have made the substutition of variables h = H - h' (so h' is the vertical distance from the TOP of the triangle), which would give the simpler expression w(h') = (B/H)h', leading to the formula H INT (B/H) (h')^2 dh' h'=0 (H^3)/3 h'_mean = --------------------- = ------- = 2H/3 H (H^2)/2 INT (B/H) (h') dh' h'=0 Thus we again have h_mean = H - h'_mean = H/3. This isn't too bad, but do we really need calculus to solve this simple problem? The ancient Greeks would probably have noted that any triangle can be split into four precisely similar triangles as shown below ______ | /\ | / \ | / \ /_________\ H / \ /\ / \ / \ | / \ / \ ___|__ /__________\/_________\ Suppose the average height of a triangle of this shape is k times its height. It follows that the average height of the overall triangle is kH, and the average heights of the four sub-triangles (above the original base) are kH/2, H/2-kH/2, kH/2, and H/2+kH/2, so the average height of the combined set is (H+kH)/4, and this must equal kH for the original triangle. Thus we have H + kH = 4kH and so k = 1/3. This illustrates how it's often possible, by exploiting some symmetry in a problem, to solve a "calculus problem" without resorting to calculus. The most impressive example of this is Isaac Newton's Principia, in which hundreds of complex problems of kinematics and dynamics are solved by Newton's own "synthetic" methods, modelled on the geometrical presentations of the ancient Greeks. Regardless of whether this was the original form in which he conceived of his results, or (as he later suggested in the midst of the calculus priority dispute) simply reformulations of results he had derived by means of his fluxional calculus, it's tremendously impressive to behold how he was always able to see a "synthetic" form of demonstration. Of course, the very brilliance and vaiety of the insights needed to produce the demonstrations in the Principia reveals the reason that calculus prevailed over synthetic methods. As has been said, Newton was a giant, but the instruments he used are too heavy for anyone else to lift. We can't all dream up an inspired synthetic approach each time we're faced with a new problem. So we use calculus. It may not always be elegant, but it's easy and can be applied to a wide variety of problems without much thought.

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