Question

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

30. \({y^2}\left( {6 - x} \right) = {x^3},\,\,\left( {2,\sqrt 2 } \right)\)

Short answer

The required equation of the tangent line is \(y = \frac{7}{{4\sqrt 2 }}x - \frac{3}{{2\sqrt 2 }}\).

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 \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

The equation of the tangent line to 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, \({y^2}\left( {6 - x} \right) = {x^3}\).

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

\(\begin{aligned}\frac{d}{{dx}}\left( {{y^2}\left( {6 - x} \right)} \right) &= \frac{d}{{dx}}\left( {{x^3}} \right)\\{y^2}\frac{d}{{dx}}\left( {6 - x} \right) + \left( {6 - x} \right)\frac{d}{{dx}}\left( {{y^2}} \right) &= \frac{d}{{dx}}\left( {{x^3}} \right)\\{y^2}\left( { - 1} \right) + \left( {6 - x} \right) \cdot 2y \cdot \frac{{dy}}{{dx}} &= 3{x^2}\\\frac{{dy}}{{dx}}2y\left( {6 - x} \right) &= 3{x^2} + {y^2}\\\frac{{dy}}{{dx}} &= \frac{{3{x^2} + {y^2}}}{{2y\left( {6 - x} \right)}}\end{aligned}\)

Find an equation of the tangent line to the curve at the given point,

\(\left( {2,\sqrt 2 } \right)\)

Put \(x = 2\) and \(y = \sqrt 2 \)into\(\frac{{dy}}{{dx}} = \frac{{3{x^2} + {y^2}}}{{2y\left( {6 - x} \right)}}\).

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{3{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}{{2\left( {\sqrt 2 } \right)\left( {6 - 2} \right)}}\\ &= \frac{{12 + 2}}{{2\sqrt 2 \cdot 4}}\\ &= \frac{7}{{4\sqrt 2 }}\end{aligned}\)

Now put \(x = 2\), \(y = \sqrt 2 \) and \(\frac{{dy}}{{dx}} = \frac{7}{{4\sqrt 2 }}\) into \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

\(\begin{aligned}y - \sqrt 2 &= \frac{7}{{4\sqrt 2 }}\left( {x - 2} \right)\\y &= \frac{7}{{4\sqrt 2 }}x - \frac{7}{{2\sqrt 2 }} + \frac{4}{{2\sqrt 2 }}\\y &= \frac{7}{{4\sqrt 2 }}x - \frac{3}{{2\sqrt 2 }}\end{aligned}\)

Thus, an equation of the tangent line to the curve at the given point, \(\left( {2,\sqrt 2 } \right)\) is \(y = \frac{7}{{4\sqrt 2 }}x - \frac{3}{{2\sqrt 2 }}\).