# Points of intersection of 2 curves

• Dec 9th 2011, 12:45 AM
Punch
Points of intersection of 2 curves
Find the coordinates of the points of intersection of the curve with parametric equations: $x=8sin^3\theta, y=8cos^3\theta$, where $0 \leq \theta < \pi$ with the line $y=\sqrt3x-8$

What I have tried is to substitute x and y in $y=\sqrt3x-8$ with those in the parametric equations, but this forms $cos^3\theta=\sqrt3sin^3\theta-1$ which seems hard to solve, unless a graphical calculator is used.
• Dec 9th 2011, 08:46 PM
Amer
Re: Points of intersection of 2 curves
Quote:

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

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

$cos^3\theta=\sqrt3sin^3\theta-1$

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

since i see $\sqrt{3}$ I think about an angle which has a value include $\sqrt{3}$
we have two angles $\frac{\pi}{3}$ and $\frac{\pi}{6}$
and I find that $\frac{\pi}{3}$ fit for that
• Dec 9th 2011, 11:17 PM
Punch
Re: Points of intersection of 2 curves
Quote:

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

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

since i see $\sqrt{3}$ I think about an angle which has a value include $\sqrt{3}$
we have two angles $\frac{\pi}{3}$ and $\frac{\pi}{6}$
and I find that $\frac{\pi}{3}$ 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.
• Dec 10th 2011, 01:38 AM
Amer
Re: Points of intersection of 2 curves
Ok
$x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2}$

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

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

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

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

so now we have to curves see this link which solve these two curves and give x = $3\sqrt{3}$ which is equal to 5.196
• Dec 10th 2011, 03:25 AM
Punch
Re: Points of intersection of 2 curves
Quote:

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

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

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

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

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

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

Thank you! However, the answer states x=3.18
• Dec 22nd 2011, 01:19 PM
MarkFL
Re: Points of intersection of 2 curves
I too get $x=3\sqrt{3}$.

I began by eliminating the parameter θ.

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

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

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

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

Equating this to the given line, we find:

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

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

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

Expanding gives:

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

Collecting like terms, rearranging and factoring yields:

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

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

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

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

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

$x=3^{\small{\frac{3}{2}}}=3\sqrt{3}\approx5.196$
• Dec 22nd 2011, 02:26 PM
skeeter
Re: Points of intersection of 2 curves
the graph shows three solutions ...

x = 0 , x = 3.18 , x = 5.196
• Dec 22nd 2011, 03:33 PM
MarkFL
Re: Points of intersection of 2 curves
I see now where I erred...

I failed to account for:

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

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