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

Find the slope of the tangent line to the polar curve π‘Ÿ = cos 2πœƒ at the point πœƒ = πœ‹/6.

02:38

Video Transcript

Find the slope of the tangent line to the polar curve π‘Ÿ equals cos two πœƒ at the point where πœƒ equals πœ‹ 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 π‘Ÿ equals 𝑓 of πœƒ, it’s some function of πœƒ, d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. We use the formula d𝑦 by dπœƒ equals dπ‘Ÿ by dπœƒ times sin πœƒ plus π‘Ÿ cos πœƒ and dπ‘₯ by dπœƒ equals dπ‘Ÿ by dπœƒ times cos πœƒ minus π‘Ÿ sin πœƒ.

Now, if we go back to our question, we see that π‘Ÿ is defined as cos of two πœƒ. And since our equations for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ are in terms of dπ‘Ÿ by dπœƒ and π‘Ÿ, we’re going to need to begin by calculating dπ‘Ÿ by dπœƒ. To do so, we quote the general result that the derivative of cos of π‘Žπ‘₯ is equal to negative π‘Ž sin of π‘Žπ‘₯. And that’s great, because that means dπ‘Ÿ by dπœƒ is negative two sin of two πœƒ.

Let’s now work out d𝑦 by dπœƒ and dπ‘₯ by dπœƒ. d𝑦 by dπœƒ is dπ‘Ÿ by dπœƒ times sin πœƒ. So, that’s negative two sin two πœƒ sin πœƒ plus π‘Ÿ times cos πœƒ. And in this case, that’s cos two πœƒ times cos πœƒ. We then use the formula for dπ‘₯ by dπœƒ. It’s dπ‘Ÿ dπœƒ times cos πœƒ. So, that’s negative two sin two πœƒ times cos πœƒ. We then subtract π‘Ÿ times sin πœƒ. So, here, we subtract cos two πœƒ sin πœƒ. d𝑦 by dπ‘₯ is the quotient of these. It’s d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. So, it’s negative two sin two πœƒ sin πœƒ plus cos two πœƒ cos πœƒ all over negative two sin two πœƒ cos πœƒ minus cos two πœƒ sin πœƒ.

Now, remember, we’re looking to work out the slope of the tangent line to our curve at the point where πœƒ is equal to πœ‹ by six. We’ll achieve this by substituting πœƒ equals πœ‹ 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 π‘Ÿ equals cos two πœƒ at the point where πœƒ is equal to πœ‹ by six is root three over seven.

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