### Video Transcript

Find the slope of the tangent line
to the curve π equals one over π at π equals π.

Remember, when finding the slope,
we need to evaluate the derivative of that curve at a specific point. The formula weβre interested in for
a polar curve of the form π equals π of π is dπ¦ by dπ₯ equals dπ¦ by dπ divided
by dπ₯ by dπ. And then, we have specific
equations for dπ¦ by dπ and dπ₯ by dπ in terms of π and π. We can see that our equation is
given as π equals one over π. So, weβll begin by working out the
value of dπ dπ.

Letβs start by writing one over π
as π to the power of negative one. And of course, to differentiate a
simple polynomial terms like this one, we multiply by the exponent and then reduce
that exponent by one. So, thatβs negative one times π to
the power of negative two, which is the same as negative one over π squared. Weβre also given that π is equal
to π. So, all we need to do is substitute
everything we know into our formula. dπ¦ by dπ is equal to negative one over π
squared times sin π plus one over π times cos π.

And then, we obtain dπ₯ by dπ to
be equal to negative one over π squared cos π minus one over π sin π. dπ¦ by dπ₯
is the quotient of these too. And of course, we could simplify
this somewhat by factoring one over π, for example. However, weβre going to evaluate
this when π is equal to π, so we may as well go ahead and substitute that in. When we do, we find that the slope
is equal to negative one over π squared times sin π plus one over π times cos π
over negative one over π squared cos π minus one over π sin π.

And in fact, sin of π is zero and
cos of π is negative one, so this simplifies really nicely to negative one over π
over one over π squared. Due to the nature of dividing
fractions, this simplifies further to negative π squared over π, which is simply
negative π. And we found the value of the
derivative at π equals π. And therefore, the slope of the
tangent line to the curve at this point is negative π.