Question

Find all points on the curve\({{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy = 2}}\) where theslope of the tangent line is \({\rm{ - 1}}\)

Short answer

The points at which the slope of the tangent line is \({\rm{ - 1}}\) are \(\left( {{\rm{ - 1, - 1}}} \right)\) and \(\left( {{\rm{1,1}}} \right)\)

Step-by-step solution

Step\({\rm{1}}\): Given information

The given equation of the curve is \({{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy = 2}}\)with the slope of the tangent line is \({\rm{ - 1}}\)

Step\({\rm{2}}\): Definition of implicit differentiation

Implicit differentiation:This method consists of differentiating both sides of the equation with respect to\({\rm{x}}\)and then solving the resulting equation for\({\rm{y'}}\)

Step\({\rm{3}}\): Differentiate the given equation with respect to \({\rm{x}}\)

The given equation of the curve is \({{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy = 2}}\)

Differentiate both sides with respect to \({\rm{x}}\)

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy}}} \right){\rm{ = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{2}} \right)\)

The Product rule for differentiation

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right){\rm{ \times g}}\left( {\rm{x}} \right)} \right){\rm{ = f'}}\left( {\rm{x}} \right){\rm{ \times g}}\left( {\rm{x}} \right){\rm{ + f}}\left( {\rm{x}} \right){\rm{ \times g'}}\left( {\rm{x}} \right)\)

Chain rule: The chain rule states that the derivative of\({\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)\)is equal to\({\rm{f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{ \times g'}}\left( {\rm{x}} \right)\)

\(\begin{array}{l}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{xy}}} \right){\rm{ = 0}}\\\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{2}}}} \right){\rm{ \times }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{y}}^{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right){\rm{ \times y + }}\left( {\rm{x}} \right){\rm{ \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{y}} \right){\rm{ = 0}}\\{\rm{2x}}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{ \times 2y \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{y + 1 \times y + xy' = 0}}\\{\rm{2x}}{{\rm{y}}^{\rm{2}}}{\rm{ + 2}}{{\rm{x}}^{\rm{2}}}{\rm{yy' + y + xy' = 0}}\end{array}\)

Step\({\rm{4}}\): Find the value of \({\rm{x}}\)

As slope of tangent line is \({\rm{ - 1}}\) that means \({\rm{y' = - 1}}\)

\(\begin{array}{l}{\rm{2x}}{{\rm{y}}^{\rm{2}}}{\rm{ + 2}}{{\rm{x}}^{\rm{2}}}{\rm{y}}\left( {{\rm{ - 1}}} \right){\rm{ + y + x}}\left( {{\rm{ - 1}}} \right){\rm{ = 0}}\\{\rm{2x}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2}}{{\rm{x}}^{\rm{2}}}{\rm{y + y - x = 0}}\\{\rm{2xy}}\left( {{\rm{y - x}}} \right){\rm{ + y - x = 0}}\\\left( {{\rm{y - x}}} \right)\left( {{\rm{2xy + 1}}} \right){\rm{ = 0}}\end{array}\)

So it has to be satisfied \({\rm{y - x = 0}}\) or \({\rm{2xy + 1 = 0}}\)

Step\({\rm{5}}\): Find the points for the condition \({\rm{y - x = 0}}\)

Substitute \({\rm{y = x}}\) in the given curve \({{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy = 2}}\)

\(\begin{array}{l}\left( {{{\rm{x}}^{\rm{2}}}} \right){\rm{ \times }}\left( {{{\rm{x}}^{\rm{2}}}} \right){\rm{ + }}\left( {\rm{x}} \right){\rm{ \times }}\left( {\rm{x}} \right){\rm{ = 2}}\\{{\rm{x}}^{\rm{4}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{ - 2 = 0}}\end{array}\)

Now, let \({\rm{t = }}{{\rm{x}}^{\rm{2}}}\)

The equation becomes \({{\rm{t}}^{\rm{2}}}{\rm{ + t - 2 = 0}}\)

Quadratic formula: If\({\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c = 0}}\)be a quadratic equation, where\(a,b\)the coefficients,\(x\)is the variable and\(c\)is the constant term then quadratic formula is equal to\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

By quadratic formula

\(\begin{array}{l}{\rm{t = }}\frac{{{\rm{ - 1 \pm }}\sqrt {{{\rm{1}}^{\rm{2}}}{\rm{ - 4 \times 1 \times }}\left( {{\rm{ - 2}}} \right)} }}{{{\rm{2 \times 1}}}}\\{\rm{ = }}\frac{{{\rm{ - 1 \pm }}\sqrt {{\rm{1 + 8}}} }}{{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{ - 1 \pm 3}}}}{{\rm{2}}}\end{array}\)

Thus \({\rm{t = 1}}\) and \({\rm{,t = - 2}}\)

As \({\rm{t = }}{{\rm{x}}^{\rm{2}}}\) this implies that,

\(\begin{array}{l}{\rm{1 = }}{{\rm{x}}^{\rm{2}}}\\{\rm{x = \pm 1}}\end{array}\)

But \({\rm{ - 2 = }}{{\rm{x}}^{\rm{2}}}\) is never satisfied because \({{\rm{x}}^{\rm{2}}}{\rm{ > 0}}\)

So, substitute \({\rm{x = \pm 1}}\) in \({\rm{y = x}}\)

\({\rm{y = \pm 1}}\)

Therefore, the points at which the slope of the tangent line is \({\rm{ - 1}}\) are \(\left( {{\rm{ - 1, - 1}}} \right)\) and \(\left( {{\rm{1,1}}} \right)\)

Step\({\rm{6}}\): Find the points for the condition \({\rm{2xy + 1 = 0}}\)

Substitute \({\rm{x = - }}\frac{{\rm{1}}}{{{\rm{2y}}}}\) in the given curve \({{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + xy = 2}}\)

\(\begin{array}{l}{\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2y}}}}} \right)^{\rm{2}}}{\rm{ \times }}\left( {{{\rm{y}}^{\rm{2}}}} \right){\rm{ + }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2y}}}}} \right){\rm{ \times }}\left( {\rm{y}} \right){\rm{ = 2}}\\\frac{{\rm{1}}}{{{\rm{4}}{{\rm{y}}^{\rm{2}}}}}{\rm{ \times }}{{\rm{y}}^{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 2}}\\\frac{{\rm{1}}}{{\rm{4}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 2}}\\{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ = 2}}\end{array}\)

Which is never satisfy

Therefore, the points at which the slope of the tangent line is \({\rm{ - 1}}\) are \(\left( {{\rm{ - 1, - 1}}} \right)\) and \(\left( {{\rm{1,1}}} \right)\)