Integral Calculus - Granville - Revised.djvu/284]] 260 DIFFEEENTIAL CALCULUS which shows that the curve extends to the right only of OF, for negative values of x make y imaginary. The origin is therefore a cusp, and since the branches lie on oppo- site sides of the common tangent, it is a cusp of the first kind. Placing the terms of lowest (second) degree equal to zero, we get y 2 = 0, showing that the two common tangents coincide with OX. ILLUSTRATIVE EXAMPLE 3. Examine (y x 2 ) 2 = x 5 for singular points. Solution. Proceeding as in the last example, we find a cusp at (0, 0), the common tangents to the two branches coinciding with OX. Solving for y, If we let x take on any value between and 1, y takes on two different positive values, showing that in the vicinity of the origin both branches lie above the common tangent. Hence the singular point (0, 0) is a cusp of the second kind. . Conjugate or isolated points. ( *- ) --L-L<Q. dxdy / dx z dy 2 In this case the values of the slope found are imaginary. Hence there are no real tangents ; the singular point is the real intersection of imaginary branches of the curve, and the coordinates of no other real point in the immediate vicinity satisfy the equation of the curve. Such an isolated point is called a conjugate point. ILLUSTRATIVE EXAMPLE 1. Examine the curve y 2 =x s x 2 for singular points. Solution. Here (0, 0) is found to be a singular point of the curve at which = V 1. Hence the origin is a conjugate point. Solving the dx equation for ?/, y=xVx 1. This shows clearly that the origin is an isolated point of the curve, for no values of x between and 1 give real values of y. . Transcendental singularities. A curve whose equation involves transcendental functions is called a transcendental curve. Such a curve may have an end point at which it terminates abruptly, caused by a discontinuity in the function ; or a salient point at which two branches of the curve terminate without having a common tan- gent, caused by a discontinuity in the derivative. ILLUSTRATIVE EXAMPLE 1. Show that y = x logx has an end point at the origin. Solution, x cannot be negative, since negative numbers have no logarithms; hence the curve extends only to the right of OF. When x = 0, y = 0. Thera being only one value of y for each positive value of x, the curve consists of a single branch terminating at the origin, which is therefore an end point.
