Question

(a) find the slope of the tangent line to the astroid \(x = aco{s^3}\theta ,y = asi{n^3}\theta \) , in terms of \(\theta \)

(b) At what points is the tangent horizontal or vertical ?

(c)At what point does the tangent have slope 1 or -1?

Short answer

The slope of the tangent line to the trochoid of the function can be given as follows:

\(\frac{{dy}}{{dx}} = - \tan \theta \)

Step-by-step solution

Find \(\frac{{dx}}{{d\theta }},\frac{{dy}}{{d\theta }}\)

We are given that

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

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

Apply the chain rule and find  \(\frac{{dy}}{{dx}}\)

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

Therefore, the slope of the tangent line is:

\(\frac{{dy}}{{dx}} = - \tan \theta \).

Answer(b):

For \((0,a),(0, - a)\) the tangent is vertical and for \((a,0)\)the tangent is horizontal.

(b) Step 1: Find the vertical tangent 

Put \(\frac{{dy}}{{dx}} = \infty \)For vertical tangent

\(\begin{array}{l}\frac{{dy}}{{dx}} = \infty \\\frac{{ - \sin \theta }}{{\cos \theta }} = \infty \\\cos \theta = 0\\\theta = \frac{\pi }{2},\pi + \frac{\pi }{2},2\pi + \frac{\pi }{2},3\pi \frac{\pi }{2}.......\end{array}\)

Substitute \(\theta = \frac{\pi }{2}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\frac{\pi }{2}\\x = 0\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\frac{\pi }{2}\\y = a\end{array}\)

Substitute \(\theta = \pi + \frac{\pi }{2}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\left( {\pi + \frac{\pi }{2}} \right)\\x = 0\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\left( {\pi + \frac{\pi }{2}} \right)\\y = - a\end{array}\)

If\(\theta = \frac{\pi }{2},\pi + \frac{\pi }{2},2\pi + \frac{\pi }{2},3\pi \frac{\pi }{2}.......\) we are getting the same values of x and y.

The points in which the tangent is vertical: \((0,a),(0, - a)\)

Answer(c):

Whenthe slope of tangent is\( \pm 1\),the points are given below:

\((\frac{a}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }})\) and \((\frac{a}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }})\).

(c) Step 1: Find the points when slope is +1

\(\begin{array}{l}\frac{{dy}}{{dx}} = 1\\ - \tan \theta = 1\\\theta = - \frac{\pi }{4},\pi - \frac{\pi }{4}\end{array}\)

Substitute \(\theta = - \frac{\pi }{4}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\left( { - \frac{\pi }{4}} \right)\\x = a\left( {\frac{1}{{2\sqrt 2 }}} \right)\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\left( { - \frac{\pi }{4}} \right)\\y = a\left( {\frac{{ - 1}}{{2\sqrt 2 }}} \right)\end{array}\)

Substitute \(\theta = \pi - \frac{\pi }{4}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\left( {\pi - \frac{\pi }{4}} \right)\\x = - a\left( {\frac{1}{{2\sqrt 2 }}} \right)\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\left( {\pi - \frac{\pi }{4}} \right)\\y = a\left( {\frac{1}{{\sqrt 2 }}} \right)\end{array}\)

The points in which the slope is +1: \((\frac{a}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }})\)

(c) Step 2: Find the points when slope is -1

\(\begin{array}{l}\frac{{dy}}{{dx}} = - 1\\ - \tan \theta = - 1\\\theta = \frac{\pi }{4},\pi + \frac{\pi }{4}\end{array}\)

Substitute \(\theta = \frac{\pi }{4}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\left( {\frac{\pi }{4}} \right)\\x = a\left( {\frac{1}{{2\sqrt 2 }}} \right)\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\left( {\frac{\pi }{4}} \right)\\y = a\left( {\frac{1}{{2\sqrt 2 }}} \right)\end{array}\)

Substitute \(\theta = \pi + \frac{\pi }{4}\)

\(\begin{array}{l}x = a{\cos ^3}\theta \\x = a{\cos ^3}\left( {\pi + \frac{\pi }{4}} \right)\\x = - a\left( {\frac{1}{{2\sqrt 2 }}} \right)\\y = a{\sin ^3}\theta \\y = a{\sin ^3}\left( {\pi + \frac{\pi }{4}} \right)\\y = - a\left( {\frac{1}{{\sqrt 2 }}} \right)\end{array}\)

The points in which the slope is -1: \((\frac{a}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }})\)

Hence, the points when the slope of tangent is +1: \((\frac{a}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }})\)and the points when the slope of the tangent is-1: \((\frac{a}{{2\sqrt 2 }},\frac{a}{{2\sqrt 2 }}),(\frac{{ - a}}{{2\sqrt 2 }},\frac{{ - a}}{{2\sqrt 2 }})\).