Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.\(x = si{n^3}\theta ,y = co{s^3}\theta ;\theta = \frac{\pi }{6}\)

Short answer

The equation of the tangent line for given system of equations is: \(y = - \sqrt 3 x + \frac{{\sqrt 3 }}{2}\)

Step-by-step solution

Find \(\frac{{dy}}{{dx}}\)

For this question, we can compute from the function of x:

\(\begin{array}{l}x = {\sin ^3}\theta \\\frac{{dx}}{{d\theta }} = 3{\sin ^2}\theta \cos \theta \end{array}\)

For this question, we can compute from the function of y:

\(\begin{array}{l}y = {\cos ^3}\theta \\\frac{{dy}}{{d\theta }} = - 3{\cos ^2}\theta \sin \theta \end{array}\)

By using the chain rule we can write,

\(\begin{array}{l}\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\\\frac{{dy}}{{dx}} = \frac{{ - 3{{\cos }^2}\theta \sin \theta }}{{3{{\sin }^2}\theta \cos \theta }}\\\frac{{dy}}{{dx}} = \frac{{ - \cos \theta }}{{\sin \theta }}\\\frac{{dy}}{{dx}} = - \cot \theta \end{array}\)

Substitute \(\theta  = \frac{\pi }{6}\) and find \(\frac{{dy}}{{dx}},x,y\)

\(\begin{array}{l}\frac{{dy}}{{dx}} = - \cot \theta \\\frac{{dy}}{{dx}} = - \cot \frac{\pi }{6}\\\frac{{dy}}{{dx}} = - \sqrt 3 \end{array}\)

\(\begin{array}{l}x = {\sin ^3}\theta \\x = {\sin ^3}\frac{\pi }{6}\\x = {\left( {\frac{1}{2}} \right)^3}\\x = \frac{1}{8}\end{array}\)

\(\begin{array}{l}y = {\cos ^3}\theta \\y = {\cos ^3}\frac{\pi }{6}\\y = {\left( {\frac{{\sqrt 3 }}{2}} \right)^3}\\y = \frac{{3\sqrt 3 }}{8}\end{array}\)

Write the equation of tangent

The equation of line can be given as: \(y - {y_1} = m(x - {x_1})\)

We can find the equation of tangent when it pass the point\(({x_1},{y_1}) = (\frac{1}{8},\frac{{3\sqrt 3 }}{8})\)and\(\frac{{dy}}{{dx}} = - \sqrt 3 \)will get:

\(\begin{array}{l}y - 0 = \pi (x + \pi )\\y = - \sqrt 3 x + \frac{{\sqrt 3 }}{2}\end{array}\)

Hence,the equation of tangent when it pass the point\(({x_1},{y_1}) = (\frac{1}{8},\frac{{3\sqrt 3 }}{8})\)will get:

\(y = - \sqrt 3 x + \frac{{\sqrt 3 }}{2}\).