Video Transcript
Find the points at which 𝑟 equals
four cos 𝜃 has a horizontal or vertical tangent line.
Remember, the formula for the slope
of the polar curve 𝑟 is equal to 𝑓 of 𝜃 is d𝑦 by d𝑥 equals d𝑦 by d𝜃 over d𝑥
by d𝜃, where d𝑦 by d𝜃 is equal to d𝑟 by d𝜃 sin 𝜃 plus 𝑟 cos 𝜃 and d𝑥 by d𝜃
is equal to d𝑟 d𝜃 cos 𝜃 minus 𝑟 sin 𝜃. Horizontal tangents will occur when
the derivative is equal to zero, in other words, where the numerator is equal to
zero. So, we locate horizontal tangents
by finding the points where d𝑦 by d𝜃 equals zero, assuming that d𝑥 by d𝜃 does
not also equal zero.
Vertical tangents occur where our
denominator d𝑥 by d𝜃 is equal to zero and the numerator d𝑦 by d𝜃 is not equal to
zero. In some sense, we can think of our
gradient as positive or negative infinity here. But strictly speaking, we say that
d𝑦 by d𝑥 is undefined, in other words, where the denominator d𝑥 by d𝜃 is equal
to zero, provided that d 𝑦 by d𝜃 does not also equal to zero. So, we’re going to need to work out
d𝑦 by d𝜃 and d𝑥 by d𝜃.
We’re told that 𝑟 is equal to four
cos 𝜃. We quote the general result for the
derivative of cos 𝜃 as being negative sin 𝜃. And we see that d𝑟 by d𝜃 must be
negative four sin 𝜃. This means d𝑦 by d𝜃 is equal to
negative four sin 𝜃 times sin 𝜃 plus 𝑟 times cos 𝜃. And 𝑟 is cos four cos 𝜃, so
that’s plus four cos 𝜃 cos 𝜃. That simplifies to negative four
sin squared 𝜃 plus four cos squared 𝜃. And we’ll set this equal to zero to
find the location of any horizontal tangents.
We begin to solve for 𝜃 by
dividing through by four and then adding sin squared 𝜃 to both sides of the
equation. Then, we divide through by cos
squared 𝜃. And this is really useful because
we know that sin squared 𝜃 divided by cos squared 𝜃 is equal to tan squared
𝜃. By finding the square root of both
sides of this equation, remembering to take by the positive and negative square root
of one, we obtain tan 𝜃 to be equal to plus or minus one.
Taking the inverse tan will give us
the value of 𝜃. So, 𝜃 is equal to the inverse tan
of negative one or the inverse tan of one. That gives us values of 𝜃 of
negative 𝜋 by four and 𝜋 by four. Now, remember, we’re trying to find
the points at which these occur, so we do need to substitute our values of 𝜃 back
into our original function for 𝑟. That gives us 𝑟 is equal to four
cos of negative 𝜋 by four or four cos of 𝜋 by four, which is two root two in both
cases.
And so, we see that we have
horizontal tangent lines at the points two root two 𝜋 by four and two root two
negative 𝜋 by four. We’re now going to work out the
vertical tangent lines. So, we’re going to repeat this
process for d𝑥 by d𝜃, substituting 𝑟 and d𝑟 by d𝜃 into our formula for d𝑥 by
d𝜃. And we see that it’s equal to
negative four sin 𝜃 cos 𝜃 minus four cos 𝜃 sin 𝜃. We set this equal to zero, and we
see that we can divide through by negative four.
So, we obtain zero to be equal to
sin 𝜃 cos 𝜃 plus cos 𝜃 sin 𝜃, or two sin 𝜃 cos 𝜃. Now, actually, we know that sin of
two 𝜃 is equal to two sin 𝜃 cos 𝜃, so we can solve this by setting zero equal to
sin of two 𝜃. So, two 𝜃 is equal to the inverse
tan of zero. Now, sin is periodic, and we need
to consider that this is also the case for two 𝜃 must be equal to 𝜋. And since 𝜃 is greater than
negative 𝜋 and less than or equal to 𝜋, these are actually the only options we
consider.
Dividing through by two, we obtain
𝜃 to be equal to zero and 𝜋 by two. Remember, we substitute this back
into our original equation for 𝑟. And we get that 𝑟 equals four cos
𝜃 or four cos 𝜋 by two, which gives us values of 𝑟 as four and zero. And we found the points at which 𝑟
equals four cos 𝜃 has horizontal and vertical tangent lines. The horizontal tangent lines are at
the point two root two, 𝜋 by four and two root two, negative 𝜋 by four and the
vertical tangent lines are at the points four, zero and zero, 𝜋 by two.
