 # Calculus III: Three Dimensional Coordinate Systems (Level 9 of 10) | Sphere Examples III

Three-Dimensional Coordinate systems (Level 9) In the previous video we covered intermediate
level examples that required the use of the equation of a sphere. Now we are going to
go over 2 more challenging examples. Let’s go over the first example,
Find an equation of the largest sphere with center (5,4,9) that is contained in the first
octant. This examples builds upon the problems from
the previous video, the problem is asking us to find the largest sphere with center
(5,4,9) that is contained in the first octant. How can we figure this one out? Let’s first think
about an easier problem that is related to this one, say we have a random circle that
is placed in the first quadrant of a two dimensional coordinate system. If we were trying to find
the largest circle that is contained in this quadrant, one way we can solve this problem
is by expanding the circle until it literally hits one of the two axis if we allow this
to happen the circle will actually be tangent to one of the coordinate axis, it turns out
that we will actually hit the coordinate axis this is closest to the center. In the three
dimensional case if we allow this sphere to expand as if we were inflating a ball or a
balloon it will end up hitting one of the coordinate planes, similar to the two dimensional
case the sphere will end up touching the coordinate plane that is closest to its center. Lets
actually find the distance from the center of the sphere to the each of the 3 coordinate
planes. Recall from the previous video that since all 3 coordinate planes are perpendicular
to one of the coordinate axis, finding the length reduces to just measuring the distance
from the center to each of the coordinate planes along the direction of a single coordinate
axis. This means that the xy-plane is 9 units away from the center, the yz-plane is 5 units
away from the center and the xz-plane is 4 units away from the center, If we were to
let this sphere expand it will end up hitting the xz coordinate plane, if the sphere is
allowed to expand beyond this plane it will seize to be contained in the first octant,
so the radius of the sphere is equal to 4, the largest sphere that is contained in the
first octant with the given center is represented by the equation the quantity (x-5) squared plus the quantity (y-4) squared plus the quantity (z-9) squared equals 16. The hard part of this problem was to understand the idea
of containing the sphere in the first octant it turns out that the shortest distance from
the center to the coordinate plane limits how much the sphere will be able to expand
and at the same time be constrained in the first octant.
Alright, lets try the next example. Find an equation of the sphere with center
in the xz-plane and passing through the points P, Q, and R
Notice that in this problem we are not provided with the center or the radius of
the sphere, nor are we provided with the end points of the diameter, we are dealing with
a completely different problem not addressed in the previous videos. We are asked to find
the equation of a sphere with its center located in the xz-plane, all we know from this specific
information is that the center of the sphere will have coordinates equal to (h,0,l). Since
we don’t know what the x and z coordinates of the center are equal to we leave them as
h and l, in addition, we also know that k is equal to zero because any point located
in the xz plane has 0 as its y coordinate. Next we are given 3 distinct points that are
located on the surface of this sphere, we need to somehow use these points to find the
value for h, l, and the radius of the sphere. How do we accomplish this? What kind of geometric
relations can we construct with the given information? We have an expression for the
center of the sphere and we also have 3 points that are located on the surface of
the sphere, with this information we can generate a relation for the radius of the sphere by
using the center and one of these points, it turns out that we can actually generate
3 relations that represent the radius of the sphere, it can be seen that the radius is
equal to the length between point P and C which is also equal to the length between
point Q and C which in turn is also equal to the length between point R and C. Since
all of these relations represent the radius of the sphere we can equate all three expressions
together, this is similar to the law of sines that you learned in your precaluclus or math
analysis recall that you can equate any two relations with each other in order to solve a specific
expression. By using the distance formula in three dimensions and using point P as point
1 and the center C as point 2 we obtain the following expression for the radius, in the
same manner we can find an expression for the distance between point Q and point C as
follows, simplifying the expression we obtain the following, and finally the distance between
point R and the center C is expressed as follows, simplifying the expression we obtain the following,
since all of these expressions represent the radius of the sphere we can equate all 3 expressions
with each other. We can make the expressions more user friendly by getting rid of the radical
so we go ahead and square all three expressions as follows. Next it’s just a matter of solving the system
of equations, we need to solve for two distinct variables in this case h and l, and we can
also generate three equations from the given relations, this is more than enough to solve
the system of equations. Let’s start by equating the first expression with the third expression,
Lets avoid using the second expression for now since it contains two binomials squared
we want to work with the simpler expressions first, notice that when we equate the first
and third expression we can eliminate the h squared term since it appears on both sides
of the equation, next lets go ahead and expand the binomial square, doing that we have the
following, notice how we can also eliminate the l squared term since it appears on both
sides of the equation, next we go ahead and add like terms moving the constant on one
side of the equation and the terms with variables on the other side, doing that we have the following,
finally we go ahead and solve for l, doing that we obtain l=12, now we need equate
two distinct expressions and substitute this value for l and solve for h, lets equate the
first and second equations together, next lets go ahead replace the value of l with
12 and simplify the expressions, after that lets go ahead and expand the binomial square,
notice that the h squared terms cancel out since it appears on both sides, then it’s
just a matter of getting all the terms with the variables to one side and the constants
to the other side and simplify the expression, lastly we go ahead and solve for h, doing
that we obtain h is equal to -7, we now have the center of our sphere, to find the radius
its just a matter of substituting the values for h and l into any of the three line segments
since all three of them represent the radius of the sphere, let’s go ahead and substitute
them into the expression representing line segment PC, simplifying the expression we
obtain the square root of 257, so the equation of the sphere that is located in the xz-plane
and that passes through points P,Q, and R is represented by the equation the quantity (x+7) squared
plus y-squared plus the quantity (z-12) squared equals 257
Alright, in our final video on three dimensional coordinate systems we will go over how to
graph equations whose domain are restricted to a given interval.

## 13 thoughts on “Calculus III: Three Dimensional Coordinate Systems (Level 9 of 10) | Sphere Examples III”

1. Hyun Yong Seung says:

Hey Mathfortress, thanks for all these videos. I've definitely learned a lot from watching these videos. I'll make sure to spread the word to my friends about your channel! 🙂

2. Math Fortress says:

3. Nafisatzc says:

i cannot believe that i understood this problem in such a short period of time. Thanks a lot! U r awesome in explaining.

4. Marius says:

Easily on par with Khan Academy!

5. Marius says:

Quality of the explanation that is.

6. Omar Tayyab says:

the question says that its located in xz plane. how can we conclude that y=0. We can only say that y is constant!. Pls explain

7. J Sharp says:

This is great-I'll spread the word!!!

8. Ayy Foo It's Modelo Time says:

looking for angles (vertexes in 3d spaces) and transformations in 3d space 🙁

9. James Ho says:

This is easy to understand. great video series.

10. Ali Abdu says:

Thank you..

11. PAY says:

Extremely helpful! Thank you so much!

12. Seraj Sersawi says:

13. Jossy Beth says: