Question Video: Finding the Slope of the Tangent to a Polar Curve at a Given Point | Nagwa Question Video: Finding the Slope of the Tangent to a Polar Curve at a Given Point | Nagwa

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Question Video: Finding the Slope of the Tangent to a Polar Curve at a Given Point Mathematics

Find the slope of the tangent line to the polar curve 𝑟 = cos 2𝜃 at the point 𝜃 = 𝜋/6.

02:38

Video Transcript

Find the slope of the tangent line to the polar curve 𝑟 equals cos two 𝜃 at the point where 𝜃 equals 𝜋 by six.

Remember, to find the slope of the tangent line to a curve at a given point, we need to begin by calculating the derivative. For a polar curve of the form 𝑟 equals 𝑓 of 𝜃, it’s some function of 𝜃, d𝑦 by d𝑥 is equal to d𝑦 by d𝜃 divided by d𝑥 by d𝜃. We use the formula d𝑦 by d𝜃 equals d𝑟 by d𝜃 times sin 𝜃 plus 𝑟 cos 𝜃 and d𝑥 by d𝜃 equals d𝑟 by d𝜃 times cos 𝜃 minus 𝑟 sin 𝜃.

Now, if we go back to our question, we see that 𝑟 is defined as cos of two 𝜃. And since our equations for d𝑦 by d𝜃 and d𝑥 by d𝜃 are in terms of d𝑟 by d𝜃 and 𝑟, we’re going to need to begin by calculating d𝑟 by d𝜃. To do so, we quote the general result that the derivative of cos of 𝑎𝑥 is equal to negative 𝑎 sin of 𝑎𝑥. And that’s great, because that means d𝑟 by d𝜃 is negative two sin of two 𝜃.

Let’s now work out d𝑦 by d𝜃 and d𝑥 by d𝜃. d𝑦 by d𝜃 is d𝑟 by d𝜃 times sin 𝜃. So, that’s negative two sin two 𝜃 sin 𝜃 plus 𝑟 times cos 𝜃. And in this case, that’s cos two 𝜃 times cos 𝜃. We then use the formula for d𝑥 by d𝜃. It’s d𝑟 d𝜃 times cos 𝜃. So, that’s negative two sin two 𝜃 times cos 𝜃. We then subtract 𝑟 times sin 𝜃. So, here, we subtract cos two 𝜃 sin 𝜃. d𝑦 by d𝑥 is the quotient of these. It’s d𝑦 by d𝜃 divided by d𝑥 by d𝜃. So, it’s negative two sin two 𝜃 sin 𝜃 plus cos two 𝜃 cos 𝜃 all over negative two sin two 𝜃 cos 𝜃 minus cos two 𝜃 sin 𝜃.

Now, remember, we’re looking to work out the slope of the tangent line to our curve at the point where 𝜃 is equal to 𝜋 by six. We’ll achieve this by substituting 𝜃 equals 𝜋 by six into our expression for the derivative. And this gives us the expression shown. When we evaluate the numerator, we get negative root three over two plus root three over four. And the denominator, we get negative three over two minus one-quarter. The quotient of these is root three over seven. And so, we see the slope of the tangent line to the polar curve 𝑟 equals cos two 𝜃 at the point where 𝜃 is equal to 𝜋 by six is root three over seven.

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