Find the slope of the tangent line
to the curve 𝑟 equals one over 𝜃 at 𝜃 equals 𝜋.
Remember, when finding the slope,
we need to evaluate the derivative of that curve at a specific point. The formula we’re interested in for
a polar curve of the form 𝑟 equals 𝑓 of 𝜃 is d𝑦 by d𝑥 equals d𝑦 by d𝜃 divided
by d𝑥 by d𝜃. And then, we have specific
equations for d𝑦 by d𝜃 and d𝑥 by d𝜃 in terms of 𝑟 and 𝜃. We can see that our equation is
given as 𝑟 equals one over 𝜃. So, we’ll begin by working out the
value of d𝑟 d𝜃.
Let’s start by writing one over 𝜃
as 𝜃 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, that’s negative one times 𝜃 to
the power of negative two, which is the same as negative one over 𝜃 squared. We’re also given that 𝜃 is equal
to 𝜋. So, all we need to do is substitute
everything we know into our formula. d𝑦 by d𝜃 is equal to negative one over 𝜃
squared times sin 𝜃 plus one over 𝜃 times cos 𝜃.
And then, we obtain d𝑥 by d𝜃 to
be equal to negative one over 𝜃 squared cos 𝜃 minus one over 𝜃 sin 𝜃. d𝑦 by d𝑥
is the quotient of these too. And of course, we could simplify
this somewhat by factoring one over 𝜃, for example. However, we’re going to evaluate
this when 𝜃 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 𝜃 equals 𝜋. And therefore, the slope of the
tangent line to the curve at this point is negative 𝜋.