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Given a set of parametric equations x = 3 \cos 2t, y = 1 + cos^2 2t . (a) What would be the...

Question:

Given a set of parametric equations {eq}x = 3 \cos 2t, y = 1 + cos^2 2t {/eq}.

(a) What would be the range of values on t that would retrace the path of the curve once?

(b) Find the corresponding Cartesian equation, including the appropriate limits on the values of x and y.

(c) Sketch the curve, using arrows.

Parametric Curves


A curve given parametrically as {eq}\displaystyle x=f(t), y=g(t), a\leq t\leq b, {/eq}

where {eq}\displaystyle f(t), g(t) {/eq} are trigonometric functions,

has the parameter t limited to the period of each trigonometric function.

The period of a trigonometric function {eq}\displaystyle \sin (\omega t) {/eq} is {eq}\displaystyle \frac{2\pi}{\omega}. {/eq}

If the trigonometric function is squared, {eq}\displaystyle \sin^2 (\omega t), {/eq} then the period is {eq}\displaystyle \frac{\pi}{\omega}. {/eq}

Answer and Explanation:


(a) To find the range of values of t such that the parametric curve {eq}\displaystyle x=3\cos(2t), y=1+\cos^2(2t), {/eq} is traced once,

we will analyze the periods of the two functions.

Cosine function {eq}\displaystyle \cos(2t) {/eq} has the period {eq}\displaystyle \pi {/eq} and because the y coordinate is a square of cosine, therefore the period is {eq}\displaystyle \frac{\pi}{2}. {/eq}

So, the range of t for one trace is {eq}\displaystyle \boxed{\left[0, \frac{\pi}{2}\right]} {/eq}

(b) To find the Cartesian form of the parametric curve, we will eliminate t, or the cosine function

{eq}\displaystyle \cos(2t)=\frac{x}{3} \implies y=1+\frac{x^2}{9}. {/eq}

The domain of x is given by the range of the function {eq}\displaystyle x=3\cos (2t) \text{ which is } \left[-3,3\right], {/eq} because {eq}\displaystyle -1\leq \cos(2t)\leq 1. {/eq}

The domain of y is given by the range of the function {eq}\displaystyle y=1+\cos^2 (2t) \text{ which is } \left[1,2\right], {/eq} because {eq}\displaystyle 0\leq \cos^2(2t)\leq 1. {/eq}

So, the Cartesian form is {eq}\displaystyle \boxed{y=1+\frac{x^2}{9}, \text{ for }-3\leq x\leq 3 \text{ and } 1\leq y\leq 2}. {/eq}

(c) The graph of the parametric curve with the clockwise direction from the point {eq}\displaystyle t=0: (3,2) {/eq} to the end point {eq}\displaystyle t=\frac{\pi}{2}: (-3,2) {/eq} is shown below by the green curve.


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Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
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