### Video Transcript

If π equals negative π’ minus π£,
then the polar form of π is blank. (A) Root two, π over four, (B)
root two, three π over four, (C) root two, five π over four, (D) root two, seven
π over four.

All right, so given this vector π
in rectangular form, we want to solve for its polar form. Another way we can write π in
rectangular form is to express it in terms of its π₯- and π¦-components like
this. And now letβs recall that for a
vector expressed in polar form, we give it not by its π₯-, π¦-components, but rather
by π and π. Here, π is equal to the square
root of π₯ squared plus π¦ squared, and the tan of π is equal to π¦ divided by
π₯. Using these relationships, we then
have the ability to convert π from its rectangular form as given to its polar
form.

That polar form, as weβve seen, is
defined by a radial distance and an angle. And π, we know, is equal to the
square root of the π₯-component squared plus the π¦-component squared. This is equal to the square root of
one plus one or the square root of two, so weβll substitute this result in for π in
our polar form of π. But as we look again at our answer
options, notice that this doesnβt narrow down the list. All four choices had the same
π-value of the square root of two. So letβs move on to calculating the
angle π of our vector. As we do this, it will be helpful
to sketch our vector on a coordinate plane.

Letβs say that each of these tick
marks represents one unit of distance. And since our vector has
rectangular components of negative one and negative one, if the tail of the vector
was at the origin, then the vector would look like this. And the angle π, defining its
direction, would be measured from the positive π₯-axis to the vector. We see then that π will be greater
than π radians but less than three-halves π. We can now use this relationship
here to solve for it. We know that the components of our
vector π₯ and π¦ are both negative one. Here weβve taken the inverse tan of
both sides of our tan of π equation, meaning that π equals the arc or inverse tan
of negative one divided by negative one.

Hereβs whatβs interesting,
though. If we evaluate this inverse tangent
on our calculator, the result we get is exactly π divided by four. But looking at our sketch of our
vector, we know that that canβt be the correct angle for π. Here, we need to recall the rule
that when we calculate an inverse tangent with a negative π₯-value in the fraction,
then to correctly solve for the angle π measured relative to the positive π₯-axis,
we need to add π radians to our result. Itβs by doing this that we avoid a
possible error in calculating π. This potential error can be traced
back to the tangent function.

But suffice to say whenever we
calculate the inverse tangent of a fraction with a negative π₯-value, that is,
whenever our vectorβs in the second or third quadrants, weβll need to add π radians
or 180 degrees, as the case may be, to properly solve for the angle relative to the
positive π₯-axis. π over four plus π is equal to
five over four times π. And substituting this value in for
π in our polar form of π, we see that in polar form π is equal to the square root
of two, five π over four. We find that listed as option (C)
among our answer choices. So, to complete our sentence, if π
equals negative π’ minus π£, then the polar form of π is square root of two, five
π over four.