Calculus

Applications of the Derivative

Applications of Derivative Logo

Envelope of a Family of Curves

Solved Problems

Example 3.

Find the envelope of the family of straight lines, which together with the line segments intercepted by the coordinate axes form triangles of equal area.

Solution.

Write the equation of the straight line in intercept form:

\[\frac{x}{a} + \frac{y}{b} = 1,\]

where \(a, b\) are \(x\)-intercept and \(y\)-intercept, respectively. By symmetry, it suffices to consider the lines in the first quadrant, i.e. we assume that \(a \gt 0,\) \(b \gt 0.\) The area of the right triangle (Figure \(5\)) is

\[S = \frac{{ab}}{2}.\]
Envelope of a family of straight lines
Figure 5.

Hence, \(b = \frac{{2S}}{a},\) and the equation of the family of straight lines is written as

\[\frac{x}{a} + \frac{{ay}}{{2S}} = 1,\]

where the line segment \(a\) is a parameter.

Differentiate this equation in \(a:\)

\[ - \frac{x}{{{a^2}}} + \frac{y}{{2S}} = 0\]

and express \(a\) in terms of the other variables:

\[- \frac{x}{{{a^2}}} + \frac{y}{{2S}} = 0,\;\;\Rightarrow {a^2}y = 2Sx,\;\;\Rightarrow {a^2} = \frac{{2Sx}}{y},\;\;\Rightarrow a = \sqrt {\frac{{2Sx}}{y}} .\]

If we substitute the expression for \(a\) in the first equation, we get the equation of the envelope (here we assume that the family of straight lines has no singular points):

\[\frac{x}{a} + \frac{{ay}}{{2S}} = 1, \;\;\Rightarrow \frac{x}{{\sqrt {\frac{{2Sx}}{y}} }} + \frac{{\sqrt {\frac{{2Sx}}{y}} y}}{{2S}} = 1, \;\;\Rightarrow \frac{{\sqrt {xy} }}{{\sqrt {2S} }} + \frac{{\sqrt {xy} }}{{\sqrt {2S} }} = 1, \;\;\Rightarrow 2\sqrt {xy} = \sqrt {2S} , \;\;\Rightarrow \sqrt {xy} = \sqrt {\frac{S}{2}} .\]

Given that \(x \gt 0,\) \(y \gt 0\) in the first quadrant, we can write

\[xy = \frac{S}{2}.\]

As seen, the equation of the envelope is a hyperbola.

Example 4.

Find the envelope of the family of ellipses \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\] having the same area.

Solution.

The area of an ellipse is

\[S = \pi ab.\]

Then the equation of the family of curves can be written as

\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{{\left( {\frac{S}{{\pi a}}} \right)}^2}}} = 1\;\;\text{or}\;\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{\pi ^2}{y^2}{a^2}}}{{{S^2}}} = 1,\]

where the semi-axis \(a\) is a parameter. By differentiating with respect to \(a,\) we get another equation:

\[- \frac{{2{x^2}}}{{{a^3}}} + \frac{{2{\pi ^2}{y^2}a}}{{{S^2}}} = 0,\;\; \Rightarrow \frac{{{x^2}}}{{{a^3}}} = \frac{{{\pi ^2}{y^2}a}}{{{S^2}}},\;\; \Rightarrow {S^2}{x^2} = {\pi ^2}{y^2}{a^4},\;\; \Rightarrow S\left| x \right| = \pi \left| y \right|{a^2}.\]

We express here \({a^2}\) and substitute into the first equation:

\[{a^2} = \frac{{S\left| x \right|}}{{\pi \left| y \right|}},\;\; \Rightarrow \frac{{{x^2}}}{{\frac{{S\left| x \right|}}{{\pi \left| y \right|}}}} + \frac{{\frac{{{\pi ^2}{y^2}S\left| x \right|}}{{\pi \left| y \right|}}}}{{{S^2}}} = 1,\;\; \Rightarrow \frac{{\pi \left| x \right|\left| y \right|}}{S} + \frac{{\pi \left| x \right|\left| y \right|}}{S} = 1,\;\; \Rightarrow \frac{{2\pi \left| x \right|\left| y \right|}}{S} = 1,\;\; \Rightarrow 2\pi \left| x \right|\left| y \right| = S,\;\; \Rightarrow \left| y \right| = \frac{S}{{2\pi \left| x \right|}}.\]

This equation consists of \(4\) hyperbolic branches and describes the envelope of the family of ellipses (Figure \(6\)).

Envelope of a family of ellipses with the same area
Figure 6.

Example 5.

Find the envelope of the family of straight lines given by the normal equation: \[x\cos C + y\sin C - p = 0.\]

Solution.

In this equation, \(\cos C\) and \(\sin C\) are the directional cosines of the normal vector to the straight line (Figure \(7\),) and \(p\) is the distance from the line to the origin.

Envelope of a family of straight lines given by the normal equation.
Figure 7.

We differentiate the equation of the family of straight lines with respect to the parameter \(C:\)

\[{\left( {x\cos C + y\sin C - p} \right)^\prime } = 0,\;\; \Rightarrow - x\sin C + y\cos C = 0,\;\; \Rightarrow x\sin C = y\cos C,\;\; \Rightarrow \tan C = \frac{y}{x}.\]

Express \(\cos C\) and \(\sin C\) in terms of \(\tan C:\)

\[\cos C = \pm \frac{1}{{\sqrt {1 + {{\tan }^2}C} }},\;\;\;\sin C = \pm \frac{{\tan C}}{{\sqrt {1 + {{\tan }^2}C} }}.\]

Then

\[\cos C = \pm \frac{1}{{\sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} }} = \pm \frac{1}{{\sqrt {1 + \frac{{{y^2}}}{{{x^2}}}} }} = \pm \frac{1}{{\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} }} = \pm \frac{x}{{\sqrt {{x^2} + {y^2}} }},\]
\[\sin C = \pm \frac{{\frac{y}{x}}}{{\sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} }} = \pm \frac{{\frac{y}{x}}}{{\sqrt {1 + \frac{{{y^2}}}{{{x^2}}}} }} = \pm \frac{{\frac{y}{x}}}{{\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} }} = \pm \frac{y}{{\sqrt {{x^2} + {y^2}} }}.\]

Substituting this into the equation of the family of straight lines, we find:

\[x\cos C + y\sin C = p,\;\; \Rightarrow x\left( { \pm \frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right) + y\left( { \pm \frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right) = p,\;\; \Rightarrow \pm \frac{{{x^2} + {y^2}}}{{\sqrt {{x^2} + {y^2}} }} = p,\;\; \Rightarrow \pm \sqrt {{x^2} + {y^2}} = p,\;\; \Rightarrow {x^2} + {y^2} = {p^2}.\]

Thus, the envelope of the family of straight lines is the circle with radius \(p\) and centered at the origin.

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