Question

27-36Useimplicit differentiationto find an equation of the tangent line to the curve at the given point.

29.\({x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = 4,\,\,\left( { - 3\sqrt 3 ,1} \right)\)

Short answer

The required equation of the tangent line is \(y = \frac{1}{{\sqrt 3 }}x + 4\).

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, \({x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = 4\).

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

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\\frac{d}{{dx}}\left( {{x^{\frac{2}{3}}}} \right) + \frac{d}{{dx}}\left( {{y^{\frac{2}{3}}}} \right) &= \frac{d}{{dx}}\left( 4 \right)\\\frac{2}{3}{x^{ - \frac{1}{3}}} + \frac{2}{3}{y^{ - \frac{1}{3}}} \cdot \frac{{dy}}{{dx}} &= 0\\\frac{1}{{\sqrt(3){x}}} + \frac{{\frac{{dy}}{{dx}}}}{{\sqrt(3){y}}} &= 0\\\frac{{dy}}{{dx}} &= - \frac{{\sqrt(3){y}}}{{\sqrt(3){x}}}\end{aligned}\)

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

\(\left( { - 3\sqrt 3 ,1} \right)\)

Put \(x = - 3\sqrt 3 \) and \(y = 1\)into\(\frac{{dy}}{{dx}} = - \frac{{\sqrt(3){y}}}{{\sqrt(3){x}}}\).

\(\begin{aligned}{r}\frac{{dy}}{{dx}} &= - \frac{{\sqrt(3){1}}}{{\sqrt(3){{ - 3\sqrt 3 }}}}\\ &= - \frac{1}{{{{\left( { - 3\sqrt 3 } \right)}^{\frac{1}{3}}}}}\\ &= \frac{1}{{{3^{\frac{1}{2}}}}}\\ &= \frac{1}{{\sqrt 3 }}\end{aligned}\)

Now put \(x = - 3\sqrt 3 \), \(y = 1\) and \(\frac{{dy}}{{dx}} = 1\) into \(y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)\).

\(\begin{aligned}{c}y - 1 &= \frac{1}{{\sqrt 3 }}\left( {x + 3\sqrt 3 } \right)\\y - 1 &= \frac{1}{{\sqrt 3 }}x + 3\\y &= \frac{1}{{\sqrt 3 }}x + 4\end{aligned}\)

Thus, anequation of the tangent line to the curve at the given point, \(\left( { - 3\sqrt 3 ,1} \right)\) is \(y = \frac{1}{{\sqrt 3 }}x + 4\).