Video: Slope of a Polar Curve

In this video, we will learn how to find the derivatives of polar curves and the slope of a polar curve.

15:24

Video Transcript

There’ll be times where we might wish to find the slope of the tangent line to a polar curve given in the form π‘Ÿ equals 𝑓 of πœƒ. In this video, we’re going to learn how we can extend our understanding of finding the derivative of parametric equations to help us calculate the derivatives of polar curves and, hence, the slope of a polar curve.

We begin by recalling that we can convert polar coordinates into Cartesian coordinates using the following equations, π‘₯ equals π‘Ÿ cos πœƒ and 𝑦 equals π‘Ÿ sin πœƒ. We’re going to be considering a tangent line to a polar curve with the equation π‘Ÿ is equal to 𝑓 of πœƒ. So, let’s replace π‘Ÿ in these equations with 𝑓 of πœƒ so that π‘₯ is equal to 𝑓 of πœƒ cos πœƒ and 𝑦 is equal to 𝑓 of πœƒ times sin of πœƒ. Notice, we now have what looks a lot like a pair of parametric equations. We have equations for π‘₯ and 𝑦 in terms of a third parameter πœƒ.

We recall that for two functions π‘₯ and 𝑦 in terms of 𝑑, their derivative d𝑦 by dπ‘₯ is d𝑦 by d𝑑 divided by dπ‘₯ by d𝑑. So, we replace 𝑑 with πœƒ, and we see that we can find the slope by calculating d𝑦 by dπ‘₯. It’s d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. But let’s have a look at our expressions for π‘₯ and 𝑦. They’re both actually the product of two functions in πœƒ, so we can use the product rule to find their derivatives.

Remember, the product rule says that the derivative of the product of two differentiable functions 𝑒 and 𝑣 is 𝑒 times d𝑣 by dπ‘₯ plus three times d𝑣 by dπ‘₯. This means dπ‘₯ by dπœƒ is the derivative of 𝑓 times cos πœƒ minus 𝑓 times the derivative of cos πœƒ, which is sin πœƒ. So, that’s 𝑓 prime of πœƒ cos πœƒ minus 𝑓 of πœƒ sin πœƒ. Now, let’s replace 𝑓 of πœƒ with π‘Ÿ and 𝑓 prime of πœƒ with dπ‘Ÿ dπœƒ, and we see that dπ‘₯ by dπœƒ is equal to dπ‘Ÿ dπœƒ cos πœƒ minus π‘Ÿ sin πœƒ.

By performing a similar process for d𝑦 by dπœƒ, we see that it’s equal to dπ‘Ÿ by dπœƒ times sin πœƒ plus π‘Ÿ cos πœƒ. And since we know that d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ, we can say that d𝑦 by dπ‘₯ must be equal to dπ‘Ÿ dπœƒ sin πœƒ plus π‘Ÿ cos πœƒ divided by dπ‘Ÿ dπœƒ cos πœƒ minus π‘Ÿ sin πœƒ. Now, what this also means is that since we find horizontal tangents by finding the points where d𝑦 by dπ‘₯ is equal to zero, we can find horizontal tangents by letting d𝑦 by dπœƒ equal to zero, provided that dπ‘₯ by dπœƒ does not also equal zero. Similarly, we locate vertical tangents at the point where dπ‘₯ by dπœƒ equals zero, assuming that 𝑦 by dπ‘₯ is not equal to zero at these points. Let’s now have a look at the application of these formulae.

Find the slope of the tangent line to the curve π‘Ÿ equals one over πœƒ at πœƒ equals πœ‹.

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 π‘Ÿ equals 𝑓 of πœƒ is d𝑦 by dπ‘₯ equals d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And then, we have specific equations for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ in terms of π‘Ÿ and πœƒ. We can see that our equation is given as π‘Ÿ equals one over πœƒ. So, we’ll begin by working out the value of dπ‘Ÿ dπœƒ.

Let’s start by writing one over πœƒ as πœƒ 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 πœƒ to the power of negative two, which is the same as negative one over πœƒ squared. We’re also given that πœƒ is equal to πœ‹. So, all we need to do is substitute everything we know into our formula. d𝑦 by dπœƒ is equal to negative one over πœƒ squared times sin πœƒ plus one over πœƒ times cos πœƒ.

And then, we obtain dπ‘₯ by dπœƒ to be equal to negative one over πœƒ squared cos πœƒ minus one over πœƒ sin πœƒ. d𝑦 by dπ‘₯ is the quotient of these too. And of course, we could simplify this somewhat by factoring one over πœƒ, for example. However, we’re going to evaluate this when πœƒ is equal to πœ‹, 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œ‹ squared times sin πœ‹ plus one over πœ‹ times cos πœ‹ over negative one over πœ‹ squared cos πœ‹ minus one over πœ‹ sin πœ‹.

And in fact, sin of πœ‹ is zero and cos of πœ‹ is negative one, so this simplifies really nicely to negative one over πœ‹ over one over πœ‹ squared. Due to the nature of dividing fractions, this simplifies further to negative πœ‹ squared over πœ‹, which is simply negative πœ‹. And we found the value of the derivative at πœƒ equals πœ‹. And therefore, the slope of the tangent line to the curve at this point is negative πœ‹.

In our next example, we’ll look at how we can use this formulae to locate horizontal and vertical tangent lines.

Find the points at which π‘Ÿ equals four cos πœƒ has a horizontal or vertical tangent line.

Remember, the formula for the slope of the polar curve π‘Ÿ is equal to 𝑓 of πœƒ is d𝑦 by dπ‘₯ equals d𝑦 by dπœƒ over dπ‘₯ by dπœƒ, where d𝑦 by dπœƒ is equal to dπ‘Ÿ by dπœƒ sin πœƒ plus π‘Ÿ cos πœƒ and dπ‘₯ by dπœƒ is equal to dπ‘Ÿ dπœƒ cos πœƒ minus π‘Ÿ sin πœƒ. 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 dπœƒ equals zero, assuming that dπ‘₯ by dπœƒ does not also equal zero.

Vertical tangents occur where our denominator dπ‘₯ by dπœƒ is equal to zero and the numerator d𝑦 by dπœƒ 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 dπ‘₯ is undefined, in other words, where the denominator dπ‘₯ by dπœƒ is equal to zero, provided that d 𝑦 by dπœƒ does not also equal to zero. So, we’re going to need to work out d𝑦 by dπœƒ and dπ‘₯ by dπœƒ.

We’re told that π‘Ÿ is equal to four cos πœƒ. We quote the general result for the derivative of cos πœƒ as being negative sin πœƒ. And we see that dπ‘Ÿ by dπœƒ must be negative four sin πœƒ. This means d𝑦 by dπœƒ is equal to negative four sin πœƒ times sin πœƒ plus π‘Ÿ times cos πœƒ. And π‘Ÿ is cos four cos πœƒ, so that’s plus four cos πœƒ cos πœƒ. That simplifies to negative four sin squared πœƒ plus four cos squared πœƒ. And we’ll set this equal to zero to find the location of any horizontal tangents.

We begin to solve for πœƒ by dividing through by four and then adding sin squared πœƒ to both sides of the equation. Then, we divide through by cos squared πœƒ. And this is really useful because we know that sin squared πœƒ divided by cos squared πœƒ is equal to tan squared πœƒ. 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 πœƒ to be equal to plus or minus one.

Taking the inverse tan will give us the value of πœƒ. So, πœƒ is equal to the inverse tan of negative one or the inverse tan of one. That gives us values of πœƒ of negative πœ‹ by four and πœ‹ by four. Now, remember, we’re trying to find the points at which these occur, so we do need to substitute our values of πœƒ back into our original function for π‘Ÿ. That gives us π‘Ÿ is equal to four cos of negative πœ‹ by four or four cos of πœ‹ 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œ‹ by four and two root two negative πœ‹ by four. We’re now going to work out the vertical tangent lines. So, we’re going to repeat this process for dπ‘₯ by dπœƒ, substituting π‘Ÿ and dπ‘Ÿ by dπœƒ into our formula for dπ‘₯ by dπœƒ. And we see that it’s equal to negative four sin πœƒ cos πœƒ minus four cos πœƒ sin πœƒ. 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 πœƒ cos πœƒ plus cos πœƒ sin πœƒ, or two sin πœƒ cos πœƒ. Now, actually, we know that sin of two πœƒ is equal to two sin πœƒ cos πœƒ, so we can solve this by setting zero equal to sin of two πœƒ. So, two πœƒ 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 πœƒ must be equal to πœ‹. And since πœƒ is greater than negative πœ‹ and less than or equal to πœ‹, these are actually the only options we consider.

Dividing through by two, we obtain πœƒ to be equal to zero and πœ‹ by two. Remember, we substitute this back into our original equation for π‘Ÿ. And we get that π‘Ÿ equals four cos πœƒ or four cos πœ‹ by two, which gives us values of π‘Ÿ as four and zero. And we found the points at which π‘Ÿ equals four cos πœƒ has horizontal and vertical tangent lines. The horizontal tangent lines are at the point two root two, πœ‹ by four and two root two, negative πœ‹ by four and the vertical tangent lines are at the points four, zero and zero, πœ‹ by two.

We’ll now consider how we can find the slopes of the tangent lines to a polar curve at the tips of the leaves.

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.

We can also use the techniques outlined in this video to find an equation for the tangent or normal to a polar curve at a point. Let’s see what that might look like.

Find the equation of the tangent line to the polar curve with equation π‘Ÿ equals cos two πœƒ at πœƒ is equal to πœ‹ by four.

Remember, the equation for a straight line is given by 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. Here, π‘š is the slope, whereas π‘₯ one, 𝑦 one are the point where the tangent line hits the curve. So, we’re going to need to begin by working out the slope of our tangent line. The formula we can use is d𝑦 by dπ‘₯ equals dπ‘Ÿ by dπœƒ sin πœƒ plus π‘Ÿ cos πœƒ over dπ‘Ÿ dπœƒ cos πœƒ minus π‘Ÿ sin πœƒ.

Well, our π‘Ÿ is equal to cos two πœƒ, so dπ‘Ÿ by dπœƒ must be equal to negative two sin of two πœƒ. We can substitute this into our equation for d𝑦 by dπ‘₯, and we get 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. Remember, we’re looking to find the slope when πœƒ is equal to πœ‹ by four. So, let’s evaluate this when πœƒ is equal to πœ‹ by four. When we replace πœƒ with πœ‹ by four, we find d𝑦 by dπ‘₯ to be equal to one. So, we know the slope of the curve, but what about π‘₯ one, 𝑦 one?

Well, we use the fact that π‘₯ is equal to π‘Ÿ cos πœƒ and 𝑦 is equal to π‘Ÿ sin πœƒ. And since π‘Ÿ is equal to cos two πœƒ, we find that π‘₯ is equal to cos two πœƒ cos πœƒ and 𝑦 is equal to cos two πœƒ sin πœƒ. Then, when πœƒ is equal to πœ‹ by four, we find that π‘₯ is equal to cos of two times πœ‹ by four times cos of πœ‹ by four, which is simply equal to zero. Similarly, 𝑦 is equal to cos of two times πœ‹ by four times sin of πœ‹ by four, which is also zero. So, the equation of the tangent line to our curve at πœƒ is equal to πœ‹ by four is 𝑦 minus zero equals one times π‘₯ minus zero, which is simply 𝑦 equals π‘₯.

In this video, we’ve seen that we can use the formula d𝑦 by dπ‘₯ equals d𝑦 by dπœƒ over dπ‘₯ by dπœƒ to find the derivative of a polar curve given in the form π‘Ÿ equals 𝑓 of πœƒ, where d𝑦 by dπœƒ and dπ‘₯ by dπœƒ are as shown. By setting d𝑦 by dπœƒ equals zero, we can establish the existence of horizontal tangents. And by setting the dπ‘₯ by dπœƒ equal to zero, we find the vertical tangents. We can also combine this formula with the equation of a straight line to help us find the equation of the tangent and normal to a curve.

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