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Find the points on the lemniscate in Exercise 23 where the tangent is horizontal.

Short Answer

Expert verified

The points on the lemniscate curve in Exerice 23 where the tangent is horizontal are \(\left( { \pm \frac{{{\rm{5}}\sqrt {\rm{3}} }}{{\rm{4}}}, \pm \frac{{\rm{5}}}{{\rm{4}}}} \right)\).

Step by step solution

01

Given Information

The given equation of lemniscate curveis \({\rm{2}}{\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)^{\rm{2}}} = {\rm{25}}\left( {{x^{\rm{2}}} - {y^{\rm{2}}}} \right)\).

02

Definition of Derivative

The derivative of a real-valued function measures the sensitivity of the function's value (output value) to changes in its argument in mathematics (input value).

03

Horizontal tangent of a curve

Differentiateand find the horizontal tangent of the curve at \(\frac{{dy}}{{dx}} = 0\).

Differentiate the lemniscate curve with respect to \(x\).

\(\begin{array}{c}{\rm{2}}{\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)^{\rm{2}}} = {\rm{25}}\left( {{x^{\rm{2}}} - {y^{\rm{2}}}} \right)\\{\rm{4}}\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)\left( {{\rm{2}}x + {\rm{2}}y\frac{{dy}}{{dx}}} \right) = {\rm{25}}\left( {{\rm{2}}x - {\rm{2}}y\frac{{dy}}{{dx}}} \right)\\{\rm{4}}\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)x = {\rm{25x}}\\{x^{\rm{2}}} + {y^{\rm{2}}} = \frac{{{\rm{25}}}}{{\rm{4}}}\end{array}\)

04

Points of horizontal tangent

Substitute \({x^{\rm{2}}} + {y^{\rm{2}}} = \frac{{{\rm{25}}}}{{\rm{4}}}\) in equation of curve \({\rm{2}}{\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)^{\rm{2}}} = {\rm{25}}\left( {{x^{\rm{2}}} - {y^{\rm{2}}}} \right)\).

\(\begin{array}{c}{\rm{2}}{\left( {\frac{{{\rm{25}}}}{{\rm{4}}}} \right)^{\rm{2}}} = {\rm{25}}\left( {\frac{{{\rm{25}}}}{{\rm{4}}} - {\rm{2}}{y^{\rm{2}}}} \right)\\\frac{{{\rm{25}}}}{{\rm{8}}} = \frac{{{\rm{25}}}}{{\rm{4}}} - {\rm{2}}{y^{\rm{2}}}\\{y^{\rm{2}}} = \frac{{{\rm{25}}}}{{{\rm{16}}}}\\y = \pm \frac{{\rm{5}}}{{\rm{4}}}\end{array}\)

Substitute \(y = \pm \frac{{\rm{5}}}{{\rm{4}}}\) in \({x^{\rm{2}}} + {y^{\rm{2}}} = {\left( {\frac{{\rm{5}}}{{\rm{2}}}} \right)^{\rm{2}}}\) and get the values \({\rm{x}} = \pm \frac{{{\rm{5}}\sqrt {\rm{3}} }}{{\rm{4}}}\).

Therefore, the points on the lemniscate curve \({\rm{2}}{\left( {{x^{\rm{2}}} + {y^{\rm{2}}}} \right)^{\rm{2}}} = {\rm{25}}\left( {{x^{\rm{2}}} - {y^{\rm{2}}}} \right)\), where the tangent is horizontal, are \(\left( { \pm \frac{{{\rm{5}}\sqrt {\rm{3}} }}{{\rm{4}}}, \pm \frac{{\rm{5}}}{{\rm{4}}}} \right)\).

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