# Question Video: Finding the Slope of the Tangent to a Polar Curve at a Certain Point Mathematics • Higher Education

Find the slope of the tangent line to the curve 𝑟 = 1/𝜃 at 𝜃 = 𝜋.

02:09

### Video Transcript

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.

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