Letter to Friedrich Engels, end of 1865-beginning of 1866

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TO ENGELS IN MANCHESTER

[London, end of 1865-beginning of 1866]

Appendix

While I was last staying in Manchester you once asked me to explain differential calculus. The thing will be quite clear to you from the following example. The whole of differential calculus arose originally from the task of drawing tangents through any point on any curve. So, that is the example I am going to use for you.

Assume that the line nAo is any curve whose nature (whether it's a parabola, ellipse, etc.) is unknown to us and on which a tangent is to be drawn at point m.

Ax is the axis. We drop the perpendicular mP (the ordinate) onto the abscissa Ax. Now assume that point n is the infinitely closest point on the curve beside to. If I drop a perpendicular np onto the axis, p must be the infinitely closest point to P and np the infinitely closest parallel line to mP. Now drop an infinitely small perpendicular mR onto np. If you now take the abscissa AP...X and the ordinate mP...y, then np = mP (or Rp), increased by an infinitely small increment [nR], or [nR] = dy (differential of y) and mR (=Pp)=dx. Since the part of the tangent mn is infinitely small, it coincides with the corresponding part of the curve itself. I can therefore regard mnR as a Δ(triangle), and the ΔmnR and mTP are similar triangles. Therefore: dy (=nR):dx(=mR) = y(mP):PT (which is the subtangent of the tangent Tn). Thus, the subtangent PT=y [math]\displaystyle{ \frac{dy}{dx} }[/math]. We now have a general differential equation for drawing the tangent at any points on any curves. If I am now to operate with this equation and to determine through it the length of the subtangent PT (once I have found this, I merely need to connect the points T and m by a straight line in order to have the tangent), so I need to know the specific character of the curve. According to its character (parabola, ellipse, cissoid, etc.), it will have a distinct form of general equation for its ordinate and abscissa at any point, which one can find in algebraic geometry. Thus, if the curve mAo, e.g., is a parabola, I know that y² (y=the ordinate at any given point) = ax, where a is the parameter of the parabola and x is the abscissa corresponding to the ordinate y.

If I put this value for y into the equation PT=[math]\displaystyle{ \frac{y \cdot dx}{dy} }[/math], I must then work out dy, i.e. to find the differential of y (which represents an infinitely small increment of y). If y²=ax, I know from the differential calculus that from d(y²)=d(ax) (I have to differentiate both sides of the equation, of course), it follows that 2y dy=adx (d always means differential). Thus dx = [math]\displaystyle{ \frac{2y \, dy}{a} }[/math]. If I put this value of dx into the formula PT=[math]\displaystyle{ \frac{y \cdot dx}{dy} }[/math], I get PT= [math]\displaystyle{ \frac{2y^2 \, dy}{a \, dy} }[/math] = [math]\displaystyle{ \frac{2y^2}{a} }[/math] = [math]\displaystyle{ \frac{2ax}{a} }[/math] (since y²=ax) = 2x. Or the subtangent of any point m on the parabola=twice the abscissa of the same point. The differential values are cancelled in the operation.