A certain lunar crater is shaped approximately as a paraboloid (see figure). The cross sections of the crater parallel to the lunar surface are circular, with a maximum radius of 60 meters. The depth of the crater is 9 meters. Find an equation for the surface of the crater. [Write your equation such that the bottom of the crater is located at (x, y, z) = (0, 0, 0). Assume the z-axis is perpendicular to the lunar surface.]

Question

Quizplus · Accepted Answer

Since the cross sections are circular, we know they are of the form $x^2+y^2=r^2$ for some $r$ less than or equal to $60$. We also know the depth is $9$, so we can use this information to find the equation of the paraboloid.
At the maximum radius, $r=60$, the depth is $9$, so the equation of the paraboloid at this point is $z=9$. At the center of the crater, $r=0$, the depth is $0$, so the equation of the paraboloid at this point is $z=0$.
So, we can use these two points to find the equation of the paraboloid.
We can write the equation of the cross sections as $z=a(x^2+y^2)$, where $a$ is some constant. At $r=60$, we have $9=a(60^2)$, so $a=9/3600=1/400$. Therefore, the equation of the paraboloid is $z=\frac{1}{400}(x^2+y^2)$. Substituting $x^2+y^2=r^2$, we get $z=\frac{1}{400}(r^2)$.
Therefore, the equation of the surface of the crater is $z=\frac{1}{400}(x^2+y^2)-9$. Since we want the bottom of the crater to be at $(0,0,0)$, we can shift the whole equation up by $9$ to get $z=\frac{1}{400}(x^2+y^2)$ as given in choice $\boxed{	extbf{(B)}}$.

A Certain Lunar Crater Is Shaped Approximately as a Paraboloid