A particular curve is represented parametrically by x = -2sec3t, \; y=2tan3t, \; t \in...


A particular curve is represented parametrically by

{eq}x = -2\sec3t, \; y=2\tan3t, \; t \in (-\frac{\pi}{6}, \frac{\pi}{6}) {/eq}.

(1) Where in the Cartesian plane is this curve located?

A. the right half-plane only

B. the left-half plane only

C. both left and right half-planes

(2) What is the corresponding Cartesian equation for this curve (the equation in x and y only)?

Cartesian equation: _____ = 0.

Write your answer as an expression in x and y which is the left-hand side of the equation when the right-hand side is set equal to zero. For example, the equation of the parabola {eq}y = x^2 +1 {/eq} can be rewritten as {eq}y - x^2 -1=0 {/eq}, so your answer would be {eq}y - x^2 -1 {/eq}. Write the equation of the full curve, even if only part of the curve is given by the parametrization.

(3) Give the range of y-values for this curve (both inf and -inf are possible answers.)

_____ < y < _____ .

Cartesian Form of a Parametric Curve

To obtain a Cartesian form from a parametric curve {eq}\displaystyle x(t), y(t), t\in\mathbf{R}, {/eq}

we eliminate the parameter {eq}\displaystyle t, {/eq} to obtain a relationship between {eq}\displaystyle x \text{ and } y, {/eq} only.

When eliminating the parameter, we may need to use some trigonometric identities, like

{eq}\displaystyle \sin^2 t+\cos^2 =1 \text{ or } \sec^2 t=1+\tan^2 t. {/eq}

To obtain the range of one of the Cartesian variables, we will determine the values of the variable in question when parameter t takes values from its domain.

For example, the range of {eq}\displaystyle x(t)=\sin t {/eq} for any value of t is {eq}\displaystyle [-1,1], {/eq}

because the function sine is bounded between -1 and 1.

Answer and Explanation:

(1) The parametric curve {eq}\displaystyle x = -2\sec (3t), \; y=2\tan (3t), \; t \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) {/eq}

is a curve whose x coordinate is negative because {eq}\displaystyle \sec (3t)>0\text{ on }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), {/eq}

and y coordinate could be positive or negative, because {eq}\displaystyle \tan (3t)>0 \text{ or } \tan (3t)<0 \text{ on }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), {/eq}

therefore the curve {eq}\displaystyle \boxed{\text{ is in both left and right half-planes (option C)}}. {/eq}

(2) For the Cartesian form of the curve, we will use the trigonometric identity {eq}\displaystyle \sec^2 x=1+\tan^2 x. {/eq}

Therefore, {eq}\displaystyle \tan^2(3t)=\frac{y^2}{4} \text{ and } \sec^2(3t)=\frac{x^2}{4}\implies \frac{x^2}{4}=1+\frac{y^2}{4}\iff x^2 =4+y^2. {/eq}

Knowing that {eq}\displaystyle x<0, {/eq} then the Cartesian form is {eq}\displaystyle x=-\sqrt{4+y^2} \iff \boxed{ x+\sqrt{4+y^2}=0 }. {/eq}

(3) The range of the variable y is the entire real axis, because there is no restriction on y in the parametrization obtained in (2),

{eq}\displaystyle \boxed{-\infty \leq y\leq \infty }. {/eq}

Learn more about this topic:

Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3

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