Question

To find the exact value of \(y\)-coordinates of the highest point on curve \(r = \sin 2\theta \) using a graph, also use calculus to find the exact value.

Short answer

The exact value of \(y\)-coordinates of the highest point on curve \(r = \sin 2\theta \) is \(y = 0.77\).

Step-by-step solution

Given data

The exact value of \(y\)-coordinates of the highest point on curve \(r = \sin 2\theta \)is given.

Definition of Polar equation

The equation which defines an algebraic curve expressed in polar coordinatesis known as a polar equation.

Sketch the graph of \(r = \sin 2\theta \)

Use online graphing calculator and sketch the graph of \(r = \sin 2\theta \) as shown in Figure below.

From figure, the exact value of \(y\)-coordinates of the highest point on curve \(r = \sin 2\theta \) is \(y = 0.77\).

Solve for the variables \(x\)–axis and \(y\)–axis

The equation for the variable \(x\) is as, \(x = r\cos \theta \). …… (1)

The equation for the variable \(y\) is as, \(y = r\sin \theta \). …… (2)

Substitute \((\sin 2\theta )\) for \(r\) in equation (1) as:

\(x = r\cos \theta \)

\(x = (\sin 2\theta )\cos \theta \) …… (3)

Substitute \((\sin 2\theta )\) for \(r\) in equation (2) as:

\(y = r\sin \theta \)

\(y = (\sin 2\theta )\sin \theta \) …… (4)

Differentiate the variable \(y\)–axis

Use calculus method to obtain the exact value of maximum height.

Differentiate the variable \(y\) with respect to \(\theta \).

\(\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}(\sin \theta \sin 2\theta )\)

Differentiate the above equation.

\(\begin{aligned}{l}\frac{{dy}}{{d\theta }} &= 2\sin \theta \cos 2\theta + \cos \theta \sin 2\theta \\\frac{{dy}}{{d\theta }} &= 2\sin \theta \left( {2{{\cos }^2}\theta - 1} \right) + \cos \theta (2\sin \theta \cos \theta )\\\frac{{dy}}{{d\theta }} &= 2\sin \theta \left( {2{{\cos }^2}\theta - 1 + {{\cos }^2}\theta } \right)\\\frac{{dy}}{{d\theta }} &= 2\sin \theta \left( {3{{\cos }^2}\theta - 1} \right)\end{aligned}\)

To obtain maximum height, the derivative has to be \(0\).

That is, \(\frac{{dy}}{{d\theta }} = 0\).

Observe that since the curve is symmetric about the origin, the case of first can be applied here.

In the first quadrant \(\frac{{dy}}{{d\theta }} = 0\) when \(\left( {3{{\cos }^2}\theta - 1} \right) = 0\).

\(\begin{aligned}{c}\left( {3{{\cos }^2}\theta - 1} \right) &= 0\\3{\cos ^2}\theta &= 1\\{\cos ^2}\theta &= \frac{1}{3}\\\cos \theta &= \pm \frac{1}{{\sqrt 3 }}\end{aligned}\)

Calculation of the value of \(\sin \theta \)

Note that since it is in first quadrant \(\frac{{dy}}{{d\theta }} = 0\) when \(\cos \theta = \frac{1}{{\sqrt 3 }}\).

Therefore, the value of \(\sin \theta \) is given below.

\(\begin{aligned}{c}{\sin ^2}\theta &= 1 - {\cos ^2}\theta \\{\sin ^2}\theta &= 1 - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}\\{\sin ^2}\theta &= \frac{2}{3}\\\sin \theta &= \sqrt {\frac{2}{3}} \end{aligned}\)

Substitute \(\cos \theta = \frac{1}{{\sqrt 3 }}\) and \(\sin \theta = \sqrt {\frac{2}{3}} \) in equation (4):

\(\begin{aligned}{l}y &= (\sin 2\theta )\sin \theta \\y &= 2\sin \theta \cos \theta \sin \theta \\y &= 2{\left( {\sqrt {\frac{2}{3}} } \right)^2}\left( {\frac{1}{{\sqrt 3 }}} \right)\\y &= \frac{4}{{3\sqrt 3 }}\\y &= 0.7698\end{aligned}\)

Therefore, the exact value of \(y\)-coordinates of the highest point on curve \(r = \sin 2\theta \) is \(y = 0.77\).