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Example text

Hence the equation to the tangent to a curve at the point x y on the curve is ly 1 V ~ y = the ^ at the point. And the equation to the normal is y~ -9- = — the — at the point. x —x ay x Exercise ^. Find the tangent to the curve x"Hj' = a, at 1 the point x\, y on the curve. Answer, —a? + - y= m + n. yi l 2. Exercise Find the normal to the same curve. Answer, - (x - x\) - - (y - y,) = 0. yi *i Exercise 3. Find the tangent and normal to the parabola y" = 4

I f you ink these curves, let the y curve be black and the ~- curve be red. Notice that the slope or ax at any point, is 2a multiplied by the of the point. We can investigate this algebraically. As before, for any value of x calculate y. Now take a greater value of x which I shall call x + Bx and calculate the new y, calling it y + Sy. We have then y+Sy = a (x + Sx)= a {x- + 2x. Sx + (Sx) }. 2 Subtracting; By = a [2x. 8x + (Bu)-}. Divide by Bx, ^| = 2ax + a . Bx. (relatively to the vessel from which it flows), the jet being 0-1 s

How do I prove this ? By differentiating (2) I obtain (1), therefore I know that (2) is the integral of (1). Only I ought to add a constant in (2), any constant whatever, an arbitrary constant as it is called, because the differential coefficient of a constant is 0. i:-, cur , cx-, aak When one has a list of differential coefficients it is not wise to use them in the reversed way as if it were a list of integrals, for things are seldom given so nicely arranged. l m+1 m 5 For instance jia 3 . v — x*.