How to find rate of change of radius of cross sections of sphere as a function of its y-coordinate
By Sebastian Wright
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Say I have a sphere with center at $C(0,0,5)$ and a radius of $r = 2$. If we take horizontal cross section of the sphere that passes through the center, it'll have a radius of 2. But if we took a horizontal cross section of the sphere at a height of say h from the center, what would be the radius of the sphere? And how do we determine the rate at which the radius changes with respect to y?
I'm not even sure where to start with this. If we take the equation of a sphere $x^2 + y^2 + z^2 = r^2$, and plug in a value, h, for y, then we're still left with three constants, x, z, and r. I'm not sure how to proceed. I would appreciate any help you guys could give.
Thank You
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$\begingroup$If the radius of the sphere is $R$ and the distance between the center and the crossing plane $h$, then $$r^2 = R^2 - h^2$$
$\endgroup$ 2 $\begingroup$Note that the radius of such a cross-section will be identical to that of the corresponding cross section of the sphere of radius $2$ centered at the origin. Thus, there's no particular reason why we should consider the actual center.
Note next that such a cross section (of the sphere centered at the origin) will be centered on the $y$-axis, say at the point $y=h,$ so that our equation becomes $$(x-0)^2+(z-0)^2=r^2,$$ or more simply, $$x^2+z^2=r^2.$$ However, points on this circle will also be points on our origin-centered sphere $$x^2+y^2+z^2=4,$$ and since each such point has a $y$-value of $h,$ then we have $$x^2+h^2+z^2=4,$$ so that $$r^2+h^2=4,$$ whence $$r=\sqrt{4-h^2}.$$
To find the rate of change with respect to $y,$ simply find $\frac{dr}{dh}.$
If we'd wanted to keep the actual center, then we'd instead consider the equations $$x^2+(z-5)^2=r^2$$ and $$x^2+y^2+(z-5)^2=4,$$ from which we'd draw the same conclusions.
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