# Thread: Points of intersection of 2 curves

1. ## Points of intersection of 2 curves

Find the coordinates of the points of intersection of the curve with parametric equations: $\displaystyle x=8sin^3\theta, y=8cos^3\theta$, where $\displaystyle 0 \leq \theta < \pi$ with the line $\displaystyle y=\sqrt3x-8$

What I have tried is to substitute x and y in $\displaystyle y=\sqrt3x-8$ with those in the parametric equations, but this forms $\displaystyle cos^3\theta=\sqrt3sin^3\theta-1$ which seems hard to solve, unless a graphical calculator is used.

2. ## Re: Points of intersection of 2 curves

Originally Posted by Punch
Find the coordinates of the points of intersection of the curve with parametric equations: $\displaystyle x=8sin^3\theta, y=8cos^3\theta$, where $\displaystyle 0 \leq \theta < \pi$ with the line $\displaystyle y=\sqrt3x-8$

What I have tried is to substitute x and y in $\displaystyle y=\sqrt3x-8$ with those in the parametric equations, but this forms $\displaystyle cos^3\theta=\sqrt3sin^3\theta-1$ which seems hard to solve, unless a graphical calculator is used.
$\displaystyle cos^3\theta=\sqrt3sin^3\theta-1$

$\displaystyle 1 = \sqrt{3} \sin ^3 \theta - \cos ^3 \theta$

since i see $\displaystyle \sqrt{3}$ I think about an angle which has a value include $\displaystyle \sqrt{3}$
we have two angles $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{\pi}{6}$
and I find that $\displaystyle \frac{\pi}{3}$ fit for that

3. ## Re: Points of intersection of 2 curves

Originally Posted by Amer
$\displaystyle cos^3\theta=\sqrt3sin^3\theta-1$

$\displaystyle 1 = \sqrt{3} \sin ^3 \theta - \cos ^3 \theta$

since i see $\displaystyle \sqrt{3}$ I think about an angle which has a value include $\displaystyle \sqrt{3}$
we have two angles $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{\pi}{6}$
and I find that $\displaystyle \frac{\pi}{3}$ fit for that
But when I sub them in x, the x-coordinates i get are 5.196 and 1 whereas the answer is 3.18. In addition, this is not a mathematical approach.

4. ## Re: Points of intersection of 2 curves

Ok
$\displaystyle x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2}$

$\displaystyle y = 8 \cos ^3 \theta \Rightarrow \cos \theta = \frac{\sqrt[3]{y}}{2}$

$\displaystyle \sin ^2 \theta + \cos ^2 \theta = 1 = \frac{\sqrt[3]{x^2}}{4} + \frac{\sqrt[3]{y^2}}{4}$

$\displaystyle 0 \leq x \leq 8 \;\;\; and \;\;\; -8 < y \leq 8$

$\displaystyle y =\sqrt{3} x - 8 \Rightarrow y+8 = \sqrt{3} x\Rightarrow (y+8)^2 = 3x^2$

so now we have to curves see this link which solve these two curves and give x = $\displaystyle 3\sqrt{3}$ which is equal to 5.196

5. ## Re: Points of intersection of 2 curves

Originally Posted by Amer
Ok
$\displaystyle x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2}$

$\displaystyle y = 8 \cos ^3 \theta \Rightarrow \cos \theta = \frac{\sqrt[3]{y}}{2}$

$\displaystyle \sin ^2 \theta + \cos ^2 \theta = 1 = \frac{\sqrt[3]{x^2}}{4} + \frac{\sqrt[3]{y^2}}{4}$

$\displaystyle 0 \leq x \leq 8 \;\;\; and \;\;\; -8 < y \leq 8$

$\displaystyle y =\sqrt{3} x - 8 \Rightarrow y+8 = \sqrt{3} x\Rightarrow (y+8)^2 = 3x^2$

so now we have to curves see this link which solve these two curves and give x = $\displaystyle 3\sqrt{3}$ which is equal to 5.196
Thank you! However, the answer states x=3.18

6. ## Re: Points of intersection of 2 curves

I too get $\displaystyle x=3\sqrt{3}$.

I began by eliminating the parameter θ.

$\displaystyle \frac{x}{8}=\sin^3\theta\:\therefore\:\theta=\sin^ {\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)$

$\displaystyle y=8\cos^3\left(\sin^{\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right) \right)$

$\displaystyle y=8\left(1-\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)^2 \right)^{\small{\frac{3}{2}}}$

$\displaystyle y=8\left(\frac{4-x^{\small{\frac{2}{3}}}}{4}\right)^{\small{\frac{3 }{2}}}=\left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2} }}$

Equating this to the given line, we find:

(1) $\displaystyle \left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2} }}=\sqrt{3}x-8$

Squaring both sides (and being aware that extraneous solutions may be introduced):

$\displaystyle \left(4-x^{\small{\frac{2}{3}}}\right)^3=\left(\sqrt{3}x-8\right)^2$

Expanding gives:

$\displaystyle 64-48x^{\small{\frac{2}{3}}}+12x^{\small{\frac{4}{3}} }-x^2=3x^2-16\sqrt{3}x+64$

Collecting like terms, rearranging and factoring yields:

$\displaystyle 4x^{\small{\frac{2}{3}}}\left(x^{\small{\frac{4}{3 }}}-3x^{\small{\frac{2}{3}}}-4\sqrt{3}x^{\small{\frac{1}{3}}}+12\right)=0$

By substitution into (1), we see x = 0 is an extraneous solution, so we need only consider:

$\displaystyle x^{\small{\frac{4}{3}}}-3x^{\small{\frac{2}{3}}}-4\sqrt{3}x^{\small{\frac{1}{3}}}+12=0$

This equation has two real roots, one of which is extraneous. By substitution into (1), we find the only valid root is:

$\displaystyle x^{\small{\frac{1}{3}}}=3^{\small{\frac{1}{2}}}$

$\displaystyle x=3^{\small{\frac{3}{2}}}=3\sqrt{3}\approx5.196$

7. ## Re: Points of intersection of 2 curves

the graph shows three solutions ...

x = 0 , x = 3.18 , x = 5.196

8. ## Re: Points of intersection of 2 curves

I see now where I erred...

I failed to account for:

$\displaystyle \theta=\pi-\sin^{\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)$

(1) $\displaystyle \pm\left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2} }}=\sqrt{3}x-8$
$\displaystyle x=0$
$\displaystyle x=\left(1-\frac{7}{\sqrt[3]{13+16\sqrt{2}}}+\sqrt[3]{13+16\sqrt{2}}\right)^{\small{\frac{3}{2}}} \approx3.1815$