Singular points of plane curves

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Find the branch points and the single-valued continuous branches y=y(x)y=y(x) (−1⩽x⩽1)(-1 \leqslant x \leqslant 1) of the multi-valued function yy defined by the equation

(x2+y2)2=x2−y2 (x^{2}+y^{2})^{2}=x^{2}-y^{2}
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Prove that the equation

(x2+y2)2=a2(x2−y2)(a≠0) (x^{2}+y^{2})^{2}=a^{2}(x^{2}-y^{2}) \quad (a \neq 0)

defines two differentiable functions, y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x), in a neighborhood of the point x=0x=0, y=0y=0. Find y1′(0)y_{1}'(0) and y2′(0)y_{2}'(0).

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Let:

(x2+y2)2=3x2y−y3, (x^{2}+y^{2})^{2}=3x^{2}y-y^{3} ,

find the value of y′y' when x=0x=0 and y=0y=0.

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Investigate the type of singular point of the following curve and roughly sketch the curve: y2=ax2+x3y^{2}=ax^{2}+x^{3}.

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Investigate the type of singular point of the following curve and roughly sketch the curve: x3+y3−3xy=0x^{3}+y^{3}-3xy=0.

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Investigate the type of singular point of the following curve and roughly sketch the curve: x2+y2=x4+y4x^{2}+y^{2}=x^{4}+y^{4}.

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Investigate the types of singular points of the following curves and roughly sketch these curves: x2+y4=x6x^{2}+y^{4}=x^{6}.

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Investigate the types of singular points of the following curves and roughly sketch these curves: (x2+y2)2=a2(x2−y2)(x^{2}+y^{2})^{2}=a^{2}(x^{2}-y^{2}).

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Investigate the types of singular points of the following curves and roughly sketch these curves: (y−x2)2=x5(y-x^{2})^{2}=x^{5}.

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Investigate the types of singular points of the following curves and roughly sketch these curves: (a+x)y2=(a−x)x2(a+x)y^{2}=(a-x)x^{2}.

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Investigate the relationship between the values of the parameters a,b,ca,b,c (a⩽b⩽c)(a \leqslant b \leqslant c) and the shape of the curve y2=(x−a)(x−b)(x−c)y^{2}=(x-a)(x-b)(x-c).

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Investigate the singular points of the transcendental curve: y2=1−e−x2y^{2}=1-e^{-x^{2}}.

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Investigate the singular points of the transcendental curve: y2=1−e−x3y^{2}=1-e^{-x^{3}}.

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Investigate the singular points of the transcendental curve: y=xln⁡xy=x\ln x.

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Investigate the singular points of the transcendental curve: y=x1+e1xy=\dfrac {x}{1+e^{\frac{1}{x}}}.

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Investigate the singular points of the transcendental curve: y=arctg⁡(1sin⁡x)y=\operatorname {arctg}\left(\dfrac {1}{\sin x}\right).

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Investigate the singular points of the transcendental curve: y2=sin⁡πxy^{2}=\sin \dfrac {\pi }{x}.

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Investigate the singular points of the transcendental curve: y2=sin⁡x2y^{2}=\sin x^{2}.

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Investigate the singular points of the transcendental curve: y2=sin⁡3xy^{2}=\sin^{3}x.

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For the equation x(x2+y2)−a(x2−y2)=0x(x^{2}+y^{2})-a(x^{2}-y^{2}) = 0 at x=0x=0, y=0y=0, the formula ∂f∂x+∂f∂yy′=0\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y' = 0 does not make it possible to find y′y'; however, y′y' can be found here too, from the formula

∂2f∂x2+2∂2f∂x∂yy′+∂2f∂y2y′2+∂f∂yy′′=0. \frac{\partial ^{2} f}{\partial x^{2}}+2\frac{\partial ^{2} f}{\partial x \partial y}y'+\frac{\partial ^{2} f}{\partial y^{2}}y'^{2}+\frac{\partial f}{\partial y}y'' = 0.

Find y′y'.

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(x2+y2−bx)2=a2(x2+y2)(x^{2}+y^{2}-bx)^{2} = a^{2}(x^{2}+y^{2}). Find y′y' at x=0x=0, y=0y=0.

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The same question for the equation x3+y3−3xy=0x^{3}+y^{3}-3xy = 0.

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In investigating the shape of a curve near a given point (a,b)(a,b) of it, an expansion in a Taylor series is useful. If f(x,y)=0f(x,y)=0 is the equation of the curve, then, expanding the left-hand side in powers of the differences x−a=ξx-a=\xi, y−b=ηy-b=\eta, we obtain, by virtue of the equality f(a,b)=0f(a,b)=0:

pξ+qη+12(rξ2+2sξη+tη2)+…=0, p\xi +q\eta +\frac{1}{2}(r\xi ^{2}+2s\xi \eta +t\eta ^{2})+\ldots = 0 ,

where pp, qq, rr, ss, and tt denote, respectively, the values of the derivatives ∂f∂x\dfrac {\partial f}{\partial x}, ∂f∂y\dfrac {\partial f}{\partial y}, ∂2f∂x2\dfrac {\partial^{2}f}{\partial x^{2}}, ∂2f∂x ∂y\dfrac {\partial^{2}f}{\partial x\, \partial y}, and ∂2f∂y2\dfrac {\partial^{2}f}{\partial y^{2}} at x=ax=a, y=by=b.

If p2+q2>0p^{2}+q^{2}>0, then for small values of ∣ξ∣\left|\xi \right| and ∣η∣\left|\eta \right| the curve is close to the line pξ+qη=0p\xi +q\eta =0, i.e.to the line p(x−a)+q(y−b)=0p(x-a)+q(y-b)=0. This line is the tangent to the curve at the point (a,b)(a,b). If, however, p=0p=0 and q=0q=0, i.e.if at the given point both derivatives ∂f∂x\dfrac {\partial f}{\partial x} and ∂f∂y\dfrac {\partial f}{\partial y} vanish, then the equation above becomes:

12[rξ2+2sξη+tη2]+…=0. \frac{1}{2}\left[r\xi ^{2}+2s\xi \eta +t\eta ^{2}\right]+\ldots = 0 .

The point (a,b)(a,b) is a singular point of the curve.

If, moreover, s2−rt>0s^{2}-rt>0, then the point (a,b)(a,b) is a double point. Two branches of the curve intersect there. The slopes of the tangents to these branches equal the roots of the quadratic equation tσ2+2sσ+r=0t\sigma^{2}+2s\sigma +r=0.

If s2−rt<0s^{2}-rt<0, then the point (a,b)(a,b) is an isolated point: sufficiently near it there are no points of the curve. If s2−rt=0s^{2}-rt=0, then to fully investigate the nature of the singular point one must take into account further terms of the Taylor expansion. Usually in this case (a,b)(a,b) is a cusp of the curve.

Investigate the singular points of the algebraic curve y2=x3y^{2}=x^{3}.

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Investigate the singular points of the algebraic curve y2=bx3+ax2y^{2}=bx^{3}+ax^{2}.

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Investigate the singular points of the algebraic curve y2=ax2+bx5y^{2}=ax^{2}+bx^{5}.

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Investigate the singular points of the algebraic curve x3+y3−xy=0x^{3}+y^{3}-xy=0.

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Investigate the singular points of the algebraic curve x4−2ax2y+2ay3+a2y2=0x^{4}-2ax^{2}y+2ay^{3}+a^{2}y^{2}=0; a≠0a\neq 0.

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Investigate the singular points of the algebraic curve x6−2a2x3y−b3y3=0x^{6}-2a^{2}x^{3}y-b^{3}y^{3}=0.

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Investigate the singular points of the algebraic curve 2y3x−y4=x(y−x)22y^{3}x-y^{4}=x(y-x)^{2}.

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Investigate the singular points of the algebraic curve x4+y4=x2+y2x^{4}+y^{4}=x^{2}+y^{2}.

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For what relation between aa and bb does the curve

y2=x3+ax+b y^{2}=x^{3}+ax+b

have a double point?

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Prove that for 4a3+27b2<04a^{3}+27b^{2}<0 the curve y2=x3+ax+by^{2}=x^{3}+ax+b consists of two separate parts, while for 4a3+27b2>04a^{3}+27b^{2}>0 it is a single connected curve.

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In studying transcendental curves, Taylor's formula is not always applicable, so transcendental curves have singular points of kinds not found on algebraic curves (such are, for example, points of cessation and corner points). Investigate the singular points of the transcendental curve y=xxy=x^{x}.

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Investigate the singular points of the transcendental curve y=xln⁡xy=x\ln x.

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Investigate the singular points of the transcendental curve yln⁡x=1y\ln x=1.

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Investigate the singular points of the transcendental curve y=e1/xy=e^{1/x}.

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Investigate the singular points of the transcendental curve y=11+e1/xy=\dfrac {1}{1+e^{1/x}}.

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Investigate the singular points of the transcendental curve y=x1+e1/xy=\dfrac {x}{1+e^{1/x}}.

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Investigate the singular points of the transcendental curve y=arctg⁡1xy=\arctg \dfrac {1}{x}.

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Investigate the singular points of the transcendental curve y=xarctg⁡1xy=x\arctg \dfrac {1}{x}.

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Investigate the singular points of the transcendental curve y=sin⁡1xy=\sin \dfrac {1}{x}.

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Investigate the singular points of the transcendental curve y=xsin⁡1xy=x\sin \dfrac {1}{x}.

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Investigate the singular points of the transcendental curve y=x2sin⁡1xy=x^{2}\sin \dfrac {1}{x}.

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Investigate the singular points of the transcendental curve xy=yxx^{y}=y^{x}.

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Investigate the singular points of the curves given by the equations. Find the tangents at the singular points.

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(1)

x3+y3−3xy=0x^{3}+y^{3}-3 x y = 0;

(2)

x3−2x2y−y2=0x^{3}-2 x^{2} y-y^{2} = 0;

(3)

x3+y3−3x2=0x^{3}+y^{3}-3 x^{2} = 0;

(4)

x3−2x2+x−y2=0x^{3}-2 x^{2}+x-y^{2} = 0;

(5)

x3−2x2y+2xy2+2x2−2xy+x=0x^{3}-2 x^{2} y+2 x y^{2}+2 x^{2}-2 x y+x = 0;

(6)

x4+xy−y4=0x^{4}+x y-y^{4} = 0;

(7)

y4+x2y2−4x2=0y^{4}+x^{2} y^{2}-4 x^{2} = 0;

(8)

x4+y4−x2−y2=0x^{4}+y^{4}-x^{2}-y^{2} = 0;

(9)

x2y2−(y+1)2(4−y2)=0x^{2} y^{2}-(y+1)^{2}\left(4-y^{2}\right) = 0;

(10)

x4+y4−8x2−10y2+16=0x^{4}+y^{4}-8 x^{2}-10 y^{2}+16 = 0;

(11)

x5−(y−2x2)2=0x^{5}-\left(y-2 x^{2}\right)^{2} = 0;

(12)

y5+5y4−4x2=0y^{5}+5 y^{4}-4 x^{2} = 0;

(13)

x6−(y−2)4−x2=0x^{6}-(y-2)^{4}-x^{2} = 0;

(14)

y6−y4+x2=0y^{6}-y^{4}+x^{2} = 0.

Investigate the singular points of the curves given by the equations. Find the tangents at the singular points.

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(1)

y2=ax2+x5y^{2} = a x^{2}+x^{5};

(2)

x(x2+y2)+a(y2−x2)=0,a≠0x\left(x^{2}+y^{2}\right)+a\left(y^{2}-x^{2}\right) = 0, a \neq 0;

(3)

(2a−x)y2=x(x−a)2,a≠0(2 a-x) y^{2} = x(x-a)^{2}, a \neq 0

(4)

(x2+y2)(y−a)2−b2y2=0,a>0,b>0\left(x^{2}+y^{2}\right)(y-a)^{2}-b^{2} y^{2} = 0, a > 0, b > 0;

(5)

y2=x3+ax+by^{2} = x^{3}+a x+b;

(6)

y2=(x−a)(x−b)(x−c),a⩽b⩽cy^{2} = (x-a)(x-b)(x-c), a \leqslant b \leqslant c.

Investigate the singular points of the curves given by the equations. Find the tangents at the singular points.

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(1)

x2/3+y2/3=22/3x^{2 / 3}+y^{2 / 3} = 2^{2 / 3};

(2)

y2=sin⁡x2y^{2} = \sin x^{2};

(3)

y2=sin⁡3xy^{2} = \sin^{3} x;

(4)

y2=1−e−x2y^{2} = 1-e^{-x^{2}};

(5)

y2=1−e−x3y^{2} = 1-e^{-x^{3}};

(6)

y={x/(1+e1/x), if x≠00, if x=0y = \begin{cases} x /\left(1+e^{1 / x}\right), & \text{ if } x \neq 0 \\ 0, & \text{ if } x = 0 \end{cases}

(7)

y={xln⁡x, if x≠00, if x=0y = \begin{cases} x \ln x, & \text{ if } x \neq 0 \\ 0, & \text{ if } x = 0 \end{cases}

(8)

yx=xyy^{x} = x^{y}.

Determine the order nn of the singular point (0,0)(0,0) of the curve and sketch the curve in a neighborhood of this point:

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(1)

x4+2x2y−xy2+y2=0x^{4}+2 x^{2} y-x y^{2}+y^{2} = 0;

(2)

x4−6x2y+25y2−16x2=0x^{4}-6 x^{2} y+25 y^{2}-16 x^{2} = 0;

(3)

(x2+y2−6x)2−(x2+y2)=0\left(x^{2}+y^{2}-6 x\right)^{2}-\left(x^{2}+y^{2}\right) = 0;

(4)

x4+y4−6y3+8x2y=0x^{4}+y^{4}-6 y^{3}+8 x^{2} y = 0;

(5)

x4+2y3−4x2y=0x^{4}+2 y^{3}-4 x^{2} y = 0;

(6)

x5+y5−xy2=0x^{5}+y^{5}-x y^{2} = 0;

(7)

y3−x2y+x5=0y^{3}-x^{2} y+x^{5} = 0;

(8)

(x2+y2)3−27x2y2=0\left(x^{2}+y^{2}\right)^{3}-27 x^{2} y^{2} = 0.

Determine the order nn of the singular point (0;0)(0 ; 0) of the equation

(y/3)3+(x/5)5=0. (y / 3)^{3}+(x / 5)^{5} = 0.

Does the curve defined by this equation have singular points?

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Find and investigate the discriminant curve of the family of curves:

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(1)

C2y=4(C+1)xC^{2} y = 4(C+1) x;

(2)

y=(x−C)3y = (x-C)^{3};

(3)

y2=(x−C)3y^{2} = (x-C)^{3};

(4)

y3=(x−C)2y^{3} = (x-C)^{2};

(5)

y−C=((x−C)/3)3y-C = ((x-C) / 3)^{3};

(6)

((y+C)/5)5=((x+C)/7)7((y+C) / 5)^{5} = ((x+C) / 7)^{7};

(7)

y=3Cx4/3−C3y = 3 C x^{4 / 3}-C^{3};

(8)

(2−x)(y−C)2=x2(2+x)(2-x)(y-C)^{2} = x^{2}(2+x);

(9)

x3+(y−C)3=3x(y−C)x^{3}+(y-C)^{3} = 3 x(y-C);

(10)

(y−(x−C)2)2=(x−C)5\left(y-(x-C)^{2}\right)^{2} = (x-C)^{5}.

References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.