Question

(a) The curve with equation \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\) is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point \(\left( {{\rm{1,}} - {\rm{2}}} \right)\).

(b) At what points does this curve have a horizontal tangent?

(c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.

Short answer

The equation of tangent line to curve \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\)is \( - \frac{{\rm{9}}}{{\rm{4}}}\), its horizontal tangent are at points \(\left( { - {\rm{2}}, - {\rm{2}}} \right)\) and \(\left( { - {\rm{2}},{\rm{2}}} \right)\), and, graph of curve with the tangent lines is,

Step-by-step solution

Given Information

The given equation of Tschirnhausen curve is \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\) and a point on the curve is \(\left( {{\rm{1,}} - {\rm{2}}} \right)\).

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).

Derivative of curve

(a) The equation that is found by differentiating a curve is the equation of tangent.

Differentiate the equation of curve\({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\) with respect to \(x\).

\(\begin{array}{c}{y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\\{\rm{2}}y\frac{{dy}}{{dx}} = {\rm{3}}{{\rm{x}}^{\rm{2}}} + {\rm{6x}}\\\frac{{dy}}{{dx}} = \frac{{{\rm{3x}}\left( {x + {\rm{2}}} \right)}}{{{\rm{2}}y}}\end{array}\)

Find the equation of tangent at \(\left( {{\rm{1,}} - {\rm{2}}} \right)\) point.

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{{{\rm{3x}}\left( {x + {\rm{2}}} \right)}}{{{\rm{2}}y}}\\{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{\rm{1,}} - {\rm{2}}} \right)}} = \frac{{{\rm{3}}\left( {\rm{1}} \right)\left( {{\rm{1}} + {\rm{2}}} \right)}}{{{\rm{2}}\left( { - {\rm{2}}} \right)}}\\{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{\rm{1,}} - {\rm{2}}} \right)}} = - \frac{{\rm{9}}}{{\rm{4}}}\end{array}\)

Point of horizontal tangent

(b) The horizontal tangent will be when its slope is zero, \(\frac{{dy}}{{dx}} = {\rm{0}}\).

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{{{\rm{3x}}\left( {x + {\rm{2}}} \right)}}{{{\rm{2}}y}}\\{\rm{0}} = \frac{{{\rm{3x}}\left( {x + {\rm{2}}} \right)}}{{{\rm{2}}y}}\\{\rm{3x}}\left( {x + {\rm{2}}} \right) = {\rm{0}}\\{\rm{x}} = - {\rm{2}}\end{array}\)

Substitute value \({\rm{x}} = - {\rm{2}}\) in curve \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\) and get \(y = \pm {\rm{2}}\).

The curve has horizontal tangent at points \(\left( { - {\rm{2}}, - {\rm{2}}} \right)\)and \(\left( { - {\rm{2}},{\rm{2}}} \right)\).

Graph of tangent lines

(c) Graph the curve \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\) and its tangent at point \(\left( {{\rm{1,}} - {\rm{2}}} \right)\) and its horizontal tangents.

Therefore, (a) the equation of tangent line to curve \({y^{\rm{2}}} = {x^{\rm{3}}} + {\rm{3}}{{\rm{x}}^{\rm{2}}}\)is \( - \frac{{\rm{9}}}{{\rm{4}}}\), (b) its horizontal tangent are at points \(\left( { - {\rm{2}}, - {\rm{2}}} \right)\) and \(\left( { - {\rm{2}},{\rm{2}}} \right)\), and, (c) graph of curve with the tangent lines is,