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

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

06:23

Video Transcript

Find the slope of the tangent line to the curve π‘Ÿ is equal to the cos of πœƒ at πœƒ is equal to πœ‹ by six.

We’re given a curve defined by a polar equation. We need to determine the slope of the tangent lines to this polar curve when πœƒ is equal to πœ‹ by six. We know to find the slope of a tangent line to a curve, we need to find an expression for d𝑦 by dπ‘₯. That’s the rate of change of 𝑦 with respect to π‘₯. Normally, we would do this by differentiating our expression for 𝑦 with respect to π‘₯. However, in this case, we’re given a polar curve. This means we’re going to need to use our formula for finding d𝑦 by dπ‘₯ for polar curves.

We recall by using the chain rule and the inverse function theorem, we get that d𝑦 by dπ‘₯ will be equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And this is provided our denominator of dπ‘₯ by dπœƒ is not equal to zero. So to find an expression for d𝑦 by dπ‘₯, we first need to find an expression for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ. But we’re not given π‘₯ and 𝑦 in terms of πœƒ, so we’re going to need to find expressions for these. To find these expressions, we need to recall the standard equations for polar coordinates: 𝑦 is equal to π‘Ÿ times the sin of πœƒ and π‘₯ is equal to π‘Ÿ times the cos of πœƒ. And we know, for the polar curve given to us in the question, π‘Ÿ is equal to the cos of πœƒ, so we can substitute this in.

Substituting π‘Ÿ is equal to the cos of πœƒ into these equations, we get that 𝑦 is equal to the cos of πœƒ times the sin of πœƒ and π‘₯ is equal to the cos of πœƒ multiplied by the cos of πœƒ. And, of course, we can simplify our expression for π‘₯ to give us the cos squared of πœƒ. We now need to find an expression for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ. Let’s start with d𝑦 by dπœƒ. There’s a couple of different methods we could use to differentiate 𝑦 with respect to πœƒ. For example, this is the product of two differentiable functions, so we could use the product rule.

However, there’s another method. We could also use the double-angle formula for sine. Recall this tells us the sin of two πœƒ is equivalent to two times the sin of πœƒ multiplied by the cos of πœƒ. If we then divide this equivalence through by two, we can get an expression for the sin of πœƒ multiplied by the cos of πœƒ. So the sin of πœƒ multiplied by the cos of πœƒ is one-half times the sin of two πœƒ, so we can use this to rewrite our equation for 𝑦. Now, we can use this to find an expression for the d𝑦 by dπœƒ. That’s the derivative of one-half times the sin of two πœƒ with respect to πœƒ.

And we can evaluate this derivative by recalling one of our standard trigonometric derivative results. For any real constant π‘Ž, the derivative of the sin of π‘Žπœƒ with respect to πœƒ is equal to π‘Ž times the cos of π‘Žπœƒ. In our case, the value of π‘Ž is two. So when we differentiate this, we get one-half multiplied by two cos of two πœƒ. And we can simplify this since one-half multiplied by two is equal to one. So we found our expression for d𝑦 by dπœƒ; it’s the cos of two πœƒ.

We now want to find an expression for dπ‘₯ by dπœƒ. Once again, there’re several different ways we could do this. For example, we could use the product rule, the chain rule, or the general power rule. However, just like we did above, we can also simplify this by using a double-angle formula, this time the double-angle formula for the cosine. We have the cos of two πœƒ is equivalent to two cos squared of πœƒ minus one. We want to rearrange this for the cos of πœƒ, so we need to add one to both sides of this equivalence and then divide through by two. Doing this gives us the cos squared of πœƒ is equivalent to the cos of two πœƒ minus one all divided by two. We can then use this to rewrite our expression for π‘₯ to make it easier to differentiate.

We’re now already to find an expression for dπ‘₯ by dπœƒ. That’s the derivative of the cos of two πœƒ minus one all divided by two with respect to πœƒ. Once again, we can differentiate this by using one of our standard trigonometric derivative results. For any real constant π‘Ž, the derivative of the cos of π‘Žπœƒ with respect to πœƒ is equal to negative π‘Ž times the sin of π‘Žπœƒ. Once again, our value of π‘Ž is two, so this gives us negative two times the sin of two πœƒ divided by two. And we know the derivative of the constant negative one-half is equal to zero. And we can simplify this since two divided by two is equal to one. So we’ve shown dπ‘₯ by dπœƒ is equal to negative the sin of two πœƒ.

Now that we’ve found expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ, we can use our formula to find an expression for d𝑦 by dπ‘₯. Substituting in d𝑦 by dπœƒ is the cos of two πœƒ and dπ‘₯ by dπœƒ is negative the sin of two πœƒ, we get d𝑦 by dπ‘₯ is equal to the cos of two πœƒ divided by negative the sin of two πœƒ. And we can simplify this. First, we’ll take the constant of negative one outside of our fraction. And if we wanted, we could simplify this even further. We know from the definition of cotangent that the cot of two πœƒ will be equivalent to the cos of two πœƒ divided by the sin of two πœƒ. So we could write d𝑦 by dπ‘₯ as negative the cot of two πœƒ.

However, this is not necessary. Instead, we know we need to find the slope of our tangent line when πœƒ is equal to πœ‹ by six, so we’ll just substitute πœƒ is equal to πœ‹ by six into this expression. Substituting in πœƒ is equal to πœ‹ by six, we get the slope of our tangent line to the curve π‘Ÿ is equal to the cos of πœƒ when πœƒ is equal to πœ‹ by six is given by negative the cos of two times πœ‹ by six all divided by the sin of two times πœ‹ by six. To evaluate this expression, we’ll start by simplifying our arguments. Two times πœ‹ by six is equal to πœ‹ by three. So this gives us negative the cos of πœ‹ by three divided by the sin of πœ‹ by three. And these are both standard angles, so we should know these results.

First, in our numerator, the cos of πœ‹ by three is equal to one-half. Next, in our denominator, the sin of πœ‹ by three is equal to root three divided by two. So we now have negative one-half divided by root three over two. And there’re several different ways we could simplify this. We’ll multiply both our numerator and our denominator by root three over two. In our numerator, we have one-half multiplied by root three over two is root three divided by four. And then in our denominator, we have root three over two times root three over two is three-quarters. Finally, the last thing we’ll do is multiply both our numerator and our denominator through by four. And this leaves us with our final answer of negative root three divided by three.

In this question, we were asked to find the slope of a tangent line to our polar curve at a point. We were able to use our formula to find this answer. We were able to show the slope of the tangent line to the curve π‘Ÿ is equal to the cos of πœƒ at πœƒ is equal to πœ‹ by six is equal to negative square root of three divided by three.

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