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I have come to a problem in a multivariable calculus book that I'm having trouble with.

The problem statement is :

"Find a constant $c$ such that for any point of intersection of the two spheres $(x-c)^{2} + y^{2} + z^{2} = 3$ and $x^{2} + (y-1)^{2} + z^{2} = 1$, the corresponding tangent planes will be perpendicular to each other."

So I define two functions : \begin{align} f(x,y,z) & = (x-c)^{2} + y^{2} + z^{2} - 3 \\ g(x,y,z) & = x^{2} + (y-1)^{2} + z^{2} - 1 \end{align} We denote the partial derivative with respect to $x$ as $f_{x}$ and so on... We see : \begin{align} f_{x}(x,y,z) & = 2(x-c) = 2x - 2c\\ f_{y}(x,y,z) & = 2y\\ f_{z}(x,y,z) & = 2z \end{align} and : \begin{align} g_{x}(x,y,z) & = 2x \\ g_{y}(x,y,z) & = 2(y-1) = 2y - 2 \\ g_{z}(x,y,z) & = 2z \end{align} I assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. So for every intersection point (x,y,z) we have : \begin{equation} \bigtriangledown f(x,y,z) \cdot \bigtriangledown g(x,y,z) = 0 \end{equation} So : \begin{align} \require{cancel} (2x - 2c, 2y, 2z) \cdot (2x, 2y-2, 2z) & = 0\\ (2x-2c)2x + (2y)(2y) - 2(2y) + 4z^{2} & = 0\\ 4x^{2} - 4xc + 4y^{2} - 4y + 4z^{2} & = 0\\ 4(x^{2} + y^{2} + z^{2}) - 4(xc + y) & = 0\\ \cancel{4} \left[ (x^{2}+y^{2}+z^{2}) - (xc + y) \right] & = 0\\ (x^{2}+y^{2}+z^{2}) - (xc + y) & = 0 \\ x^{2} + y^{2} + z^{2} & = xc + y \end{align} We also see : \begin{align} \require{cancel} f(x,y,z) & = x^{2} - 2xc + c^{2} + y^{2} + z^{2} - 3 \\ & = (x^{2} + y^{2} + z^{2}) + (c^{2} - 2xc - 3) \\ g(x,y,z) & = x^{2} + y^{2} - 2y + \cancel{1} + z^{2} - \cancel{1} \\ & = (x^{2}+y^{2}+z^{2}) - 2y \end{align} and : \begin{align} f(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = -(c^{2}-2xc-3) = -c^{2} + 2xc + 3\\ g(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = 2y \end{align} So we have : \begin{equation} xc + y = -c^{2} + 2xc + 3 = 2y \end{equation} It is here that I am stuck. It seems that the goal here would be to obtain an expression for $c$ that doesn't include $x$ or $y$, but I do not know how to obtain it.

Can someone help with this ?

scipio
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2 Answers2

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Because of the symmetry of spheres with respect to rotation, the statement “for any point of intersection between two spheres $A$ and $B$, the corresponding tangent planes are perpendicular” is equivalent to “between two spheres $A$ and $B$, there is a point of intersection at which the corresponding tangent planes are perpendicular.”

This means that we can reduce this problem to two dimensions by looking at the intersection of the spheres with a plane that passes through the center of both. Doing so we get two intersecting circles, one with radius $1$, the other with radius $\sqrt{3}$.

Rather than look at the two circles actually created by this, for now we will look at the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{3-(x-d)^2}$. These intersect at $x=\frac{d^2-2}{2d}$, and for their tangent lines to be perpendicular, their derivatives must be perpindicular, so we have $$0=\frac{\sqrt{3-(x-d)^2}}{x-d}+\frac{x}{\sqrt{1-x^2}}$$

Substituting for $x$ and solving, we get $d=\pm 2$. Because of this, we know that the centers of the spheres in the problem must be a distance of 2 apart, which is equivalent to $c^2+1^2=2^2 \to c=\pm \sqrt{3}$.

iosce
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Math is beautiful in that a good problem gives you everything that you need to solve it, and nothing more. Taking this into consideration, let's see what we're given. We have: $$ (x-c)^2 + y^2 + z^2 = 3 $$ $$ x^2 + (y-1)^2 + z^2 = 1 $$ So, we can define two functions: $$ f(x,y,z) = (x-c)^2 + y^2 + z^2 - 3 $$ $$ g(x,y,z) = x^2 + (y-1)^2 + z^2 - 1 $$ $$ $$ Taking the partial derivatives, we have:

$$ \frac{\partial f}{\partial x} = 2x-2c , \frac{\partial f}{\partial y} = 2y , \frac{\partial f}{\partial z} = 2z $$

$$ \frac{\partial g}{\partial x} = 2x , \frac{\partial g}{\partial y} = 2y-2 , \frac{\partial g}{\partial z} = 2z $$

Thus, $$ \nabla f = (2x-2c,2y,2z) $$ $$ \nabla g = (2x,2y-2,2z) $$ You are correct to assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. Thus, we have: $$ \nabla f(x,y,z) \cdot \nabla g(x,y,z) = 0 $$

$$ (2x-2c,2y,2z) \cdot (2x,2y-2,2z) = 0 $$ $$ 4x^2 - 4xc + 4y^2 - 4y + 4z^2 = 0 $$

Completing the square, we have: $$ 4(x^2 - cx + \frac{1}{4} c^2) - c^2 + 4(y^2 - y + \frac{1}{4}) - 1 + 4z^2 = 0 $$

$$ 4(x - \frac{1}{2} c)^2 + 4(y - \frac{1}{2})^2 + 4z^2 = 1 + c^2 $$

Adding $ 1 + c^2 $ to both sides and simplifying, we have:

$$ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $$

$$ $$

Now we need to look back at what we're given. We know that $$ (x-c)^2 + y^2 + z^2 = 3 $$ $$ x^2 + (y-1)^2 + z^2 = 1 $$

So, multiplying both equations by $ 2 $, we have: $$ 2((x-c)^2 + y^2 + z^2) = 2(3) \implies 2x^2 - 4cx + 2c^2 + 2y^2 + 2z^2 = 6 $$ $$ 2(x^2 + (y-1)^2 + z^2) = 2(1) \implies 2x^2 + 2 - 4y + 2y^2 + 2z^2 = 2 $$

Substituting these equations into $ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $ , we can see that this simplifies to: $$ 2c^2 + 2 = 8 $$ $$ 2c^2 = 6 $$ $$ c = \pm \sqrt3 $$

Pdogg
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