Video: Finding the Slope of a Tangent Line to a Polar Curve

Find the slopes of the tangent lines to π‘Ÿ = 2 sin (3πœƒ) at the tips of the leaves.

03:26

Video Transcript

Find the slopes of the tangent lines to π‘Ÿ equals two sin three πœƒ at the tips of the leaves.

Remember the leaves in our polar curves are the repeating loops. And here, we recall that the graph of π‘Ÿ equals sin of π‘›πœƒ β€” and in fact, multiples of sin π‘›πœƒ β€” has 𝑛 loops when 𝑛 is odd, and two 𝑛 loops when 𝑛 is an even integer. In our case, 𝑛 is three; it’s odd. So, the graph of π‘Ÿ equals two sin of three πœƒ is going to have three loops, or petals. So, how do we find the loops? Well, we begin by finding the two smallest non-negative values of πœƒ for which two sin of three πœƒ equals zero. These will describe the beginning and end point of each of our loops, of course, when π‘Ÿ is equal to zero.

By dividing through by two, we see that sin of three πœƒ equals zero. And then, we take the inverse sin of zero, and we see that three πœƒ is equal to zero or, alternatively, πœ‹. This means that πœƒ must be equal to zero or πœ‹ by three. One leaf is, therefore, described by letting πœƒ go from zero to πœ‹ by three. That might look a little something like this. This means the tip of its petal will occur when πœƒ is exactly halfway between πœƒ equals zero and πœƒ equals πœ‹ by three. That’s πœƒ equals πœ‹ by six.

Since we have three loops, we continue by adding a third of a full turn. That’s two πœ‹ by three. So, we find that πœƒ is equal to πœ‹ by six, πœ‹ by six plus two πœ‹ over three, which is five πœ‹ by six, and then we also subtract two πœ‹ by three from πœ‹ by six. And that gives us negative πœ‹ by two. It’s important to substitute these values into our original equation for π‘Ÿ, and we find that π‘Ÿ is equal to two throughout. We can say that the tips of the leaves are at two, πœ‹ by six; two, five πœ‹ by six; and two, negative πœ‹ by two.

Now, we can alternatively redefine our final point as negative two, πœ‹ by two. And this is because a negative value of π‘Ÿ takes us through the origin and out the other side at exactly the same distance. So, these are actually the exact same points. Remember, we need to find the slopes of the tangent lines to our curve at these points. And we recall that we can find the derivative and, therefore, the slope by using the given formula. The derivative of sin of three πœƒ is three cos of three πœƒ. So, dπ‘Ÿ by dπœƒ is six cos of three πœƒ.

We can substitute this into our formula for d𝑦 by dπ‘₯. And we see that it’s equal to six cos of three πœƒ sin πœƒ plus two sin of three πœƒ cos πœƒ over six cos three πœƒ cos πœƒ minus two sin of three πœƒ sin πœƒ. We’ll evaluate this at the tip of our first leaf by substituting πœƒ equals πœ‹ by six. And that gives us a slope of negative root three. We then substitute in πœƒ is equal to five πœ‹ by six, and we find the slope here is root three. And finally, we substitute in πœƒ equals πœ‹ by two, which gives us a slope of zero.

And it’s important to realise that have we substituted the πœƒ equals negative πœ‹ by two, we would have got the same answer. So, we find the slope at the first tip is a negative root three, and that’s at the point two, πœ‹ by six. The second leaf tip is at two, five πœ‹ by six. And the slope there is root three. And our third is at negative two, πœ‹ over two or two, negative πœ‹ over two. And the slope there is zero.

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