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Chapter 10: Problem 90

Eliminate the parameter: \(x=\cos ^{3} t\) and \(y=\sin ^{3} t\)

Short Answer

Expert verified
The equation after eliminating the parameter 't' is \(x^2 + y^2 - 3xy = 0\).

Step by step solution

01

Express sin t and cos t in terms of x and y

It is given that \(x = \cos^{3}t\) and \(y = \sin^{3}t\). We can express \(\cos t\) and \(\sin t\) in terms of x and y by taking the cube root of both sides of the given equations. So, we have \(\cos t = x^{1/3}\) and \(\sin t = y^{1/3}\).
02

Use the Pythagorean identity

Now we substitute \(\cos t = x^{1/3}\) and \(\sin t = y^{1/3}\) into the Pythagorean identity \(\sin^{2}t + \cos^{2}t = 1\) to obtain the equation \((x^{1/3})^{2} + (y^{1/3})^{2} = 1\). Simplifying this gives \(x^{2/3} + y^{2/3} = 1\).
03

Final simplification

The equation as it stands is correct but quite complex and can be made simpler. We can simplify the equation further by cubing both sides of the equation. This gives the final answer as \(x^2 + y^2 - 3xy = 0\).

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Most popular questions from this chapter

The equation \(3 x^{2}-2 \sqrt{3} x y+y^{2}+2 x+2 \sqrt{3} y=0\) is in a he form \(A x^{2}+B x y+C y^{2}+D x+E y+F=0 .\) Use the equation to determine the value of \(B^{2}-4 A C\)

Describe a strategy for graphing \(r=\frac{1}{1+\sin \theta}\)

Describe how to locate the foci for \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

In Exercises \(61-66,\) find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. $$ \left\\{\begin{array}{r} {\frac{x^{2}}{25}+\frac{y^{2}}{9}=1} \\ {y=3} \end{array}\right. $$

Exercises 105鈥�107 will help you prepare for the material covered in the next section. Simplify and write the equation in standard form in terms of \(x^{\prime}\) and \(y^{\prime}\) $$ \left[\frac{\sqrt{2}}{2}\left(x^{\prime}-y^{\prime}\right)\right]\left[\frac{\sqrt{2}}{2}\left(x^{\prime}+y^{\prime}\right)\right]=1 $$
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