Three-dimensional coordinate systems (level 7)
in the previous video we derived the equation of a sphere
by using the definition of a sphere and the distance formula
in three dimensions. Now we're going to go over three basic
examples that require the use of this equation. Let's go over an example
where the radius and center of the sphere are provided to us
Find the equation of the sphere with radius 5 and center (2,-1,3).
This problem is pretty straightforward
all we need to do is to use the equation of a sphere and identify the values
for h,k, l, and r.
Recall that the constants
h, k, and l represent the coordinates of the center of the sphere
notice that in this problem the center is given to us
the coordinates to the center of the sphere are two, negative one,
three
so we go ahead and substituted these values into the equation
as follows: be careful with the value of k
since its equal to a negative number it will end up being a positive number
when we substitute it into the equation
for the most part the values of h, k and l usually have the opposite sign
than the one that is expressed in the actual equation of the sphere
having substitute it the values for h,k and l we go ahead and find the measurement
the radius, notice that this value is also given to us
the radius is just equal to 5, so substituting this value into the equation
we obtain five squared which is equal to 25
a common mistake that students make is forgetting to square the value of the
radius
and just make it equal to 5 without squaring the expression
make sure you square the expression since its actually r-squared
and not just r, so the final equation of the sphere
is equal to the quantity (x-2) squared plus the quanitty (y+1) squared plus
the quantity (z-3) squared equals 25
the graph of the equation is represented as follows:
it's literally a sphere centered at two negative one, three
with radius equal to 5. Alright,
let's try a more challenging problem, let's go over a problem
where the center and a point on the sphere are provided to us
find an equation of the sphere that passes through the point
(4,3,-1) and has center (3,8,1)
this problem is similar to the previous example
all we need to do is to find the center and radius of the sphere
and substitute the values for h,k, l and the radius
into the equation, the center is already provided to us
so the values of h,k, and l are three, eight, one respectively
now, the only difference is that the radius
is not directly provided to us we actually have to figure it out
from the given information, notice that we are given the center
and a point that passes through the surface of the sphere remember
that the radius of a sphere is the line segment from the center
to any point on the surface of the sphere we know the center
and we also have a point on the surface this means
that we can use the distance formula to find the radius formed by these two points
so using the coordinates (4,3,-1) as point 1
and the coordinates (3,8,1) as point 2 we go ahead and find the difference
of each points coordinates and square them
doing that we have the following
next we go ahead and simplify the expression obtaining the square root of 30
having found the value for the radius we go ahead and substitute it
into the equation, then its just a matter of squaring it
doing that we obtain the final equation of the sphere
equal to the quanitty (x-3) squared
plus the quantity (y-8) squared plus the quantity (z-1) squared
equals 30. Alright, now lets try an example
where the endpoints of the diameter are provided to us
find an equation of the sphere with points (7,-1,-2)
and (3,4,6) as the endpoints of a
diameter
this problem is similar to the last two examples in order to generate the
equation of a sphere
we need the center and radius of the sphere, the only problem is that we are not
directly provided
with the center or radius we actually need to find them
let's find the center of the sphere first, notice that we are provided with
the endpoints of the diameter
recall from elementary geometry that the diameter is a line segment
that passes through the center of the circle in this case the line segment passes through the
center of the sphere
so how can we find the coordinates of the center of the diameter
we can find the coordinates of the center by finding the midpoint of the diameter
in other words we use the midpoint formula in three dimensions
so we essentially find the average of the two points x, y, and z coordinates
simplifying the numerical expression we have that the center of the sphere
has coordinates (5,3/2,2) having found the center
of the sphere we now have the values for h, k and l all we have to do is to find the
measurement
of the radius of the sphere we can find the radius in a couple ways
we can find the distance from the center of the sphere to any of the endpoints of the
diameter
either P or Q since the line segment from the center
to any of these points represents the radius, alternatively
we can find the distance from point P and Q this will give us the length of the
diameter
of the sphere recall from elementary geometry
that the diameter is equal to twice the radius so finding the length from point P
to point Q
we would actually find twice the length of the radius
let's go ahead and find this measurement by using the distance formula
in three-dimensions using point P as point 1
and point Q as point 2, we go ahead and find the difference
of each of the points x, y and z coordinates and square them
doing that we obtain the following, next we go ahead and simplify expression
obtaining the square root of 105
next is just a matter of solving for the radius, remember
the distance we just found is the length of the diameter
of the sphere which is equal to twice the radius
dividing both sides by two we have that the radius
is equal to the square root of 105 over 2
having found the radius we go ahead and substitute this value
into the equation of the sphere, remember we need to square the radius
recall that the way we square a fraction is by squaring the numerator
and denominator as follows, doing that we obtain the final equation of the
sphere
equal to, the quantity (x-5) squared plus
the quantity (y-3/2) squared plus the quantity (z-2) squared
equals 105 over 4. Alright,
in our next video we will continue going over examples that require the use of the
equation of a sphere.