Video Transcript
Two points in a plane have polar
coordinates π sub one 2.500 meters, π over six and π sub two 3.800 meters, two π
over three. Determine the Cartesian coordinates
of π sub one. Determine the Cartesian coordinates
of π sub two. Determine the distance between the
points, to the nearest centimeter.
We can call the Cartesian
coordinates of point π sub one π₯ one, π¦ one and the Cartesian coordinates of π
sub two π₯ two, π¦ two. The distance between these two
points weβll call capital π·. As starting information in this
exercise, weβre given two points, π sub one and π sub two, in their polar
coordinate setup. This means that the first
coordinate in each pair is the radial distance. Weβll call that π sub one for π
sub one and π sub two for π sub two. The second coordinate in the pair
is the angular coordinate. In π sub one, weβll call that
value π sub one. And in π sub two, weβll call it π
sub two.
We can recall the coordinate
conversion relationships between polar coordinates and Cartesian or π₯-, π¦-,
π§-coordinates. In our two-dimensional setup, if we
take the polar coordinate π and multiply it by the cosine of the angular polar
coordinate π, then weβll get the Cartesian coordinate π₯. Similarly, if we take that
coordinate π and multiply it by the sine of the angle π, weβll get the Cartesian
coordinate π¦.
This means that the Cartesian
coordinates of the first point π₯ one, π¦ one are equal to π sub one cos π sub one
and π sub one sin π sub one. When we plug these values in, using
2.500 meters for π sub one and π over six for π sub one, we find a result of
2.165 meters in the π₯-direction and 1.250 meters in the π¦. These are the Cartesian coordinates
of the point π sub one.
Likewise, for π₯ two, π¦ two, where
weβll use polar values π sub two and π sub two. Using a value of 3.800 meters for
π sub two and two π over three for π sub two, we find a result of negative 1.900
meters in the π₯-direction and 3.291 meters in the π¦-direction. These are the Cartesian coordinates
of the polar point π sub two.
Finally, we want to solve for the
distance π· between these two points. That distance, mathematically, is
equal to the square root of the change in π₯ squared plus the change in π¦
squared. In our case then, π· is equal to
the square root of π₯ two minus π₯ one quantity squared plus π¦ two minus π¦ one
quantity squared. When we plug in our values for π₯
two, π₯ one, π¦ two, and π¦ one and then enter this expression on our calculator, we
find that π· is 4.55 meters. Thatβs the distance between these
two points, to the nearest centimeter.