Question

27-36Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

33. \({x^2} + {y^2} = {\left( {2{x^2} + 2{y^2} - x} \right)^2},\,\,\,\left( {0,\frac{1}{2}} \right)\)

Short answer

The required equation of the tangent line is \(y = x + \frac{1}{2}\).

Step-by-step solution

Write the definition of implicit differentiation, the formula of the derivatives of trigonometric functions and the product rule

Implicit differentiation:Differentiate the both sides of the equation with respect to \(x\) and then solve the obtained result for \(\frac{{dy}}{{dx}}\) this process is known as Implicit differentiation.

The product rule:If \(f\) and \(g\) are differentiable functions, then \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

Theequation of the tangent lineto the curve at the given point \(\left( {{x_1},{y_1}} \right)\) is \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

Find \(\frac{{dy}}{{dx}}\)by implicit differentiation

Consider the equation, \({x^2} + {y^2} = {\left( {2{x^2} + 2{y^2} - x} \right)^2}\).

Differentiate both the sides of the equation w.r.t. \(x\) and simplify by using the formula of the product rule.

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) &= \frac{d}{{dx}}{\left( {2{x^2} + 2{y^2} - x} \right)^2}\\\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}{\left( {2{x^2} + 2{y^2} - x} \right)^2}\\2x + 2y\frac{{dy}}{{dx}} &= 2\left( {2{x^2} + 2{y^2} - x} \right)\left( {4x + 4y\frac{{dy}}{{dx}} - 1} \right)\end{aligned}\)

Step 3: Find an equation of the tangent line to the curve at the given point, \(\left( {0,\frac{1}{2}} \right)\)

Put \(x = 0\) and \(y = \frac{1}{2}\)into \(2x + 2y\frac{{dy}}{{dx}} = 2\left( {2{x^2} + 2{y^2} - x} \right)\left( {4x + 4y\frac{{dy}}{{dx}} - 1} \right)\).

\(\begin{aligned}2\left( 0 \right) + 2\left( {\frac{1}{2}} \right)\frac{{dy}}{{dx}} &= 2\left( {2{{\left( 0 \right)}^2} + 2{{\left( {\frac{1}{2}} \right)}^2} - \left( 0 \right)} \right)\left( {4\left( 0 \right) + 4\left( {\frac{1}{2}} \right)\frac{{dy}}{{dx}} - 1} \right)\\\frac{{dy}}{{dx}} &= 2\frac{{dy}}{{dx}} - 1\\\frac{{dy}}{{dx}} &= 1\end{aligned}\)

Now put \(x = 0\), \(y = \frac{1}{2}\) and \(\frac{{dy}}{{dx}} = 1\) into \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

\(\begin{aligned}y - \frac{1}{2} &= 1\left( {x - 0} \right)\\y &= x + \frac{1}{2}\end{aligned}\)

Thus, an equation of the tangent line to the curve at the given point, \(\left( {0,\frac{1}{2}} \right)\) is \(y = x + \frac{1}{2}\).