Copyright

The parametric curve x = \sqrt 3 \sin t + \frac{1}{2} \cos t, \quad y = - \sin t + \frac{\sqrt...

Question:

The parametric curve {eq}x = \sqrt 3 \sin t + \frac{1}{2} \cos t, \quad y = - \sin t + \frac{\sqrt 3}{2} \cos t {/eq} is a tilted ellipse. What are the dimensions of the rectangular box containing the ellipse?

Parametric Curves


A parametric curve given as {eq}\displaystyle x=x(t), y=y(t), a\leq t\leq b, {/eq} is a curve whose points {eq}\displaystyle (x,y), {/eq} are obtained by giving values to the parameter {eq}\displaystyle t. {/eq}

A parametric curve has a direction, which is given by the trace on the curve when the values of t are increasing.

The slope of the tangent line to the parametric curve given above is calculated as

{eq}\displaystyle \frac{dy}{dx}(t)=\frac{y'(t)}{x'(t)}, {/eq} where {eq}\displaystyle '=\frac{d}{dt}, {/eq}

and the points with horizontal tangent line are the points such that {eq}\displaystyle \frac{y'(t)}{x'(t)}=0. {/eq}

Answer and Explanation:


To determine the box containing the ellipse {eq}\displaystyle \left(\sqrt{3}\sin t+\frac{1}{2}\cos t , -\sin t+\frac{\sqrt{3}}{2}\cos t\right) {/eq} we will find the points where the ellipse admits horizontal and vertical tangent lines.

To find the horizontal tangent lines, we solve the equation corresponding to zero slope,

{eq}\displaystyle \begin{align} \frac{y'(t)}{x'(t)}&=0\iff y'(t)=0\\ &\iff -\cos t-\frac{\sqrt{3}}{2}\sin t =0\\ & \tan t=-\frac{2\sqrt{3}}{3} \implies t=-\tan^{-1}\left(\frac{2\sqrt{3}}{3}\right) \text{ and } t=\pi-\tan^{-1}\left(\frac{2\sqrt{3}}{3}\right). \end{align} {/eq}

To calculate the exact value of the y coordinate of the point with horizontal tangent line, we will obtain {eq}\displaystyle y=-\sin t+\frac{\sqrt{3}}{2}\cos t {/eq}

by substituting {eq}\displaystyle \cos t=-\frac{\sqrt{3}}{2}\sin t: y=-\frac{7}{4}\sin t, {/eq} where {eq}\displaystyle \tan t=-\frac{2\sqrt{3}}{3} {/eq}

Because that tangent above is negative on the domain of its inverse, implies the sine function is negative and cosine positive,

so

{eq}\displaystyle \tan t=-\frac{2\sqrt{3}}{3}=\frac{\sin t}{\sqrt{1-\sin^2 t}} \implies \frac{7}{3}\sin^2 t=\frac{4}{3}\implies \sin t=-\frac{2}{\sqrt{7}}. {/eq}

Therefore, the y coordinate is {eq}\displaystyle y=-\frac{7}{4}\sin t=\frac{\sqrt{7}}{2}. {/eq}

Because the center of the ellipse is the origin, the other horizontal line is {eq}\displaystyle y=-\frac{\sqrt{7}}{2} \implies \boxed{ \text{ one side of the box is }\sqrt{7}}. {/eq}


Similarly, we find the vertical tangent lines, by solving the equation corresponding to undefined slope,

{eq}\displaystyle \begin{align} x'(t)=0&\iff \sqrt{3}\cos t -\frac{1}{2}\sin t =0\\ & \tan t=2\sqrt{3} \implies t=\tan^{-1}\left(2\sqrt{3}\right) \text{ and } t=\pi+\tan^{-1}\left(2\sqrt{3}\right). \end{align} {/eq}

To calculate the exact value of the x coordinate of the point with vertical tangent line, we will obtain {eq}\displaystyle x=\sqrt{3}\sin t+\frac{1}{2}\cos t {/eq}

by substituting {eq}\displaystyle \sin t=2\sqrt{3}\cos t: x=\frac{13}{2}\cos t, {/eq} where {eq}\displaystyle \tan t=2\sqrt{3} {/eq}

Because that tangent above is positive on the domain of its inverse, implies the sine and cosine functions are both positive,

so

{eq}\displaystyle \tan t=2\sqrt{3}=\frac{\sqrt{1-\cos^2 t}}{\cos t} \implies \cos^2 t=\frac{1}{13}\implies \sin t=\frac{1}{\sqrt{13}}. {/eq}

Therefore, the x coordinate is {eq}\displaystyle x=\frac{1}{13}\cos t=\frac{\sqrt{13}}{2}. {/eq}

Because the center of the ellipse is the origin, the other vertical line is {eq}\displaystyle x=-\frac{\sqrt{13}}{2} \implies \boxed{ \text{the other side of the box is }\sqrt{13}}. {/eq}

So, the dimensions of the box are {eq}\displaystyle \sqrt{7} \text{ and } \sqrt{13}. {/eq}


Learn more about this topic:

Loading...
Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
4.8K

Related to this Question

Explore our homework questions and answers library