Question Video: Finding the Points at Which a Polar Curve Has a Horizontal or Vertical Tangent Line

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.