MATH200 December 2011
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Question 04

Find the radius of the largest sphere centred at the origin that can be inscribed inside (that is, enclosed inside) the ellipsoid
 $\displaystyle 2(x+1)^{2}+y^{2}+2(z1)^{2}=8.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

If the sphere with radius in question $\displaystyle R$ is enclosed by the ellipse, and the radius is maximal, then the surface of the ellipse and the sphere just touch in only one point.
What must hold for the tangent planes to the surface $\displaystyle 2(x+1)^{2}+y^{2}+2(z1)^{2}=8$
and the sphere with radius $\displaystyle R$ at the points where the surface and the sphere touch?

Hint 2

In the points where the ellipsiod $\displaystyle 2(x+1)^{2}+y^{2}+2(z1)^{2}=8$ and the sphere $\displaystyle x^{2}+y^{2}+z^{2}=R^{2}$ touch in only one point, the surfaces are parallel. Hence, the normal vector to the surfaces are parallel in these points, too.
You find the normal vectors by regarding the ellipsiod and the sphere as level sets of two functions
$\displaystyle f,\ g:\mathbb {R} ^{3}\rightarrow \mathbb {R}$. Then the gradients of these functions are the normal vectors.

Hint 3

The function $\displaystyle f,\ g:\mathbb {R} ^{3}\rightarrow \mathbb {R}$ are
$\displaystyle {\begin{aligned}f(x,y,z)&=2(x+1)^{2}+y^{2}+2(z1)^{2}\\g(x,y,z)&=x^{2}+y^{2}+z^{2}\end{aligned}}$
Hence, $\displaystyle \nabla f={\begin{pmatrix}4(x+1)\\2y\\4(z1)\end{pmatrix}}$
and $\displaystyle \nabla g={\begin{pmatrix}2x\\2y\\2z\end{pmatrix}}$
To be parallel, $\displaystyle \nabla f$ and $\displaystyle \nabla g$
must satisfy
$\displaystyle \nabla f=\lambda \nabla g$
for a $\displaystyle \lambda \in \mathbb {R}$.

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Solution

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If the sphere with radius in question $\displaystyle R$ is enclosed by the ellipse, and the radius is maximal, then the surface of the ellipse and the sphere just touch in one point.
The tangent planes to the surface $\displaystyle 2(x+1)^{2}+y^{2}+2(z1)^{2}=8$
and the sphere with radius $\displaystyle R$ are parallel at the points where the surface and the sphere touch.
Hence the normal vectors to the tangent planes are also parallel.
Regarding the surface equation and the sphere equation as level sets of functions $f,g:\mathbb {R} ^{3}\rightarrow \mathbb {R}$,
then the normal vectors to these tangent planes are given by the gradients of the functions $\displaystyle f$ and $\displaystyle g$.
The sphere equation with the origin as center and radius $\displaystyle R$ is $\displaystyle x^{2}+y^{2}+z^{2}=R^{2}$. Then $\displaystyle f(x,y,z)=2(x+1)^{2}+y^{2}+2(z1)^{2}$ and $\displaystyle g(x,y,z)=x^{2}+y^{2}+z^{2}$. We are looking for $\displaystyle x,y,z,\lambda$ such that the gradients are parallel,
${\begin{aligned}\nabla f&=\lambda \nabla g\\{\begin{pmatrix}4(x+1)\\2y\\4(z1)\end{pmatrix}}&=\lambda {\begin{pmatrix}2x\\2y\\2z\end{pmatrix}}\end{aligned}}$
The second line yields that $\displaystyle \lambda =1$ or $\displaystyle y=0$.
 $\displaystyle \lambda =1$ leads to $\displaystyle x=2,z=2$ and using the surface equation we obtain the points $\displaystyle p_{1}=(2,2,2),\ \ p_{2}=(2,2,2)$
 $\displaystyle y=0$ leads to $\displaystyle x=z$ and using the surface equation we obtain the points $\displaystyle p_{3}=({\sqrt {2}}1,0,{\sqrt {2}}+1),\ \ p_{4}=({\sqrt {2}}1,0,{\sqrt {2}}+1)$.
These four points are candidates for where the ellipse and the sphere touch. Since we are searching for the radius $\displaystyle R$ such that the sqhere is enclosed in the ellipse, we need the point $\displaystyle p_{i}$ with the shortest distance to the origin.
 ${\begin{aligned}p_{1}&={\sqrt {12}}\\p_{2}&={\sqrt {12}}\\p_{3}&={\sqrt {64{\sqrt {2}}}}\\p_{4}&={\sqrt {6+4{\sqrt {2}}}}\end{aligned}}$
Since $\displaystyle {\sqrt {64{\sqrt {2}}}}$ is the smallest of these numbers, this is the radius $\displaystyle R$ of the sqhere.

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