Question Video: Finding the Slope of the Tangent to a Polar Curve at a Certain Point Mathematics • Higher Education

Video Transcript

Find the slope of the tangent line to the curve 𝑟 is equal to two minus the sin of three 𝜃 at 𝜃 is equal to five 𝜋 by four.

We’re given a curve defined by a polar equation. And we need to determine the slope of the tangent line to this polar curve when 𝜃 is equal to five 𝜋 by four. To do this, we need to start by recalling the slope of our tangent lines will be given by d𝑦 by d𝑥, the rate of change of 𝑦 with respect to 𝑥. However, in this case, we’re not given 𝑦 as a function in 𝑥; instead this time, we’re given a polar curve. So, we’re going to need to use our formula for finding d𝑦 by d𝑥 for polar curves.

And we recall by applying the chain rule and inverse function theorem, we get that d𝑦 by d𝑥 will be equal to d𝑦 by d𝜃 divided by d𝑥 by d𝜃. And this will work provided our denominator d𝑥 by d𝜃 is not equal to zero. So, to find the slope of our tangent line, we’re going to first need to find expressions for d𝑦 by d𝜃 and d𝑥 by d𝜃. This means the first thing we’re going to need to do is write 𝑦 and 𝑥 in terms of 𝜃. And to do this, we’re going to need to recall our standard polar equations: 𝑦 is equal to 𝑟 times the sin of 𝜃 and 𝑥 is equal to 𝑟 times the cos of 𝜃. And remember, for our polar curve, 𝑟 is equal to two minus the sin of three 𝜃. So, we can substitute this into our equations.

So by substituting 𝑟 is equal to two minus the sin of three 𝜃, we get that 𝑦 is equal to two minus the sin of three 𝜃 multiplied by the sin of 𝜃 and 𝑥 is equal to two minus the sin of three 𝜃 multiplied by the cos of 𝜃. So, now, we have 𝑦 and 𝑥 written in terms of 𝜃, so we can differentiate these to find expressions for d𝑥 by d𝜃 and d𝑦 by d𝜃. Let’s start by finding an expression for d𝑦 by d𝜃.

The first thing we’ll do is distribute the sin of 𝜃 over the parentheses in our equation for 𝑦. We get 𝑦 is equal to two sin 𝜃 minus sin three 𝜃 multiplied by sin 𝜃. And now to find d𝑦 by d𝜃, we would need to differentiate this term by term. However, we see this means we need to differentiate the sin of three 𝜃 multiplied by the sin of 𝜃. And since this is the product of two differentiable functions, we’ll do this by using the product rule. So, we’ll start by recalling the product rule.

The product rule tells us the derivative of the product of two differentiable functions 𝑢 of 𝜃 times 𝑣 of 𝜃 with respect to 𝜃 is equal to 𝑢 prime of 𝜃 times 𝑣 of 𝜃 plus 𝑣 prime of 𝜃 times 𝑢 of 𝜃. In our case, we’ll set 𝑢 of 𝜃 to be the sin of three 𝜃 and 𝑣 of 𝜃 to be the sin of 𝜃. Now, to use the product rule, we’re going to need to find expressions for 𝑢 prime and 𝑣 prime.

Let’s start with 𝑢 prime. That’s the derivative of the sin of three 𝜃 with respect to 𝜃. And to evaluate this, we need to recall for any real constant 𝑎, the derivative of the sin of 𝑎 𝜃 with respect to 𝜃 is equal to 𝑎 times the cos of 𝑎𝜃. In our case, the value of 𝑎 is equal to three. So, we get that 𝑢 prime of 𝜃 is equal to three times the cos of three 𝜃. We now need to find an expression for 𝑣 prime of 𝜃. That’s the derivative of the sin of 𝜃 with respect to 𝜃. This is a standard trigonometric derivative result we should commit to memory. It’s the cos of 𝜃.

We’re now ready to use the product rule to find an expression for the derivative of the sin of three 𝜃 multiplied by the sin of 𝜃 with respect to 𝜃. Substituting in our expressions for 𝑢 of 𝜃, 𝑣 of 𝜃, 𝑢 prime of 𝜃, and 𝑣 prime of 𝜃 into our formula for the product rule, we get three times the cos of three 𝜃 multiplied by the sin of 𝜃 plus the cos of 𝜃 multiplied by the sin of three 𝜃. And now that we found this, we’re ready to find an expression for d𝑦 by d𝜃.

We’ll differentiate our expression for 𝑦 term by term with respect to 𝜃. First, the derivative of two times the sin of 𝜃 with respect to 𝜃 is equal to two times the cos of 𝜃. Next, we’ll subtract the derivative of the sin of three 𝜃 multiplied by the sin of 𝜃. And by clearing some space and using the fact that we found this earlier, we could just write this into our expression for d𝑦 by d𝜃. And at this point, there are a few options.

We could start simplifying our expression for d𝑦 by d𝜃. However, remember, we’re eventually just going to substitute 𝜃 is equal to five 𝜋 by four into this expression. So often, it’s easier not to simplify, since we’re just going to substitute our value of 𝜃 into this expression. So, in this case, we’ll leave this as it is. We now need to find an expression for d𝑥 by d𝜃. And this time, we’ll distribute the cos of 𝜃 over our parentheses. We get two cos of 𝜃 minus the sin of three 𝜃 times the cos of 𝜃. And once again, to differentiate this term by term, we see we’re going to need to use the product rule to evaluate the derivative of the sin of three 𝜃 multiplied by the cos of 𝜃 with respect to 𝜃.

So we’ll rewrite out the product rule and clear some space to help us evaluate this derivative. To apply the product rule on this product, we see our function 𝑢 of 𝜃 will once again be the sin of three 𝜃. So, we already found 𝑢 prime of 𝜃 to be three times the cos of three 𝜃. This time, however, we need to set our function 𝑣 of 𝜃 to be the cos of 𝜃. We then need to find an expression for 𝑣 prime of 𝜃. That’s the derivative of the cos of 𝜃 with respect to 𝜃, and we know this is negative the sin of 𝜃.

Now, to evaluate the derivative of this product, we just need to substitute our expressions for 𝑢, 𝑣, 𝑢 prime, and 𝑣 prime into our expression for the product rule. We get three times the cos of three 𝜃 multiplied by the cos of 𝜃 plus negative the sin of 𝜃 multiplied by the sin of three 𝜃. And we can then simplify our second term to give us the following expression. We’re now ready to use this to find an expression for d𝑥 by d𝜃. We want to differentiate our expression for 𝑥 term by term with respect to 𝜃.

First, the derivative of two times the cos of 𝜃 with respect to 𝜃 is equal to negative two sin of 𝜃. Next, we need to subtract the derivative of our product. And we already found this by using the product rule. So, we can substitute this in to find our expression for d𝑥 by d𝜃. And at this point, we have two ways we could continue. We could substitute our expressions for d𝑦 by d𝜃 and d𝑥 by d𝜃 into our formula to find an expression for d𝑦 by d𝑥, and this would work. However, we don’t actually need to find an expression for d𝑦 by d𝑥; we just need to find the slope of our tangent line when 𝜃 is equal to five 𝜋 by four.

So, we just need to substitute 𝜃 is equal to five 𝜋 by four into our expression for d𝑦 by d𝑥. And to do that, we just need to substitute 𝜃 is equal to five 𝜋 by four into our expression for d𝑦 by d𝜃 and 𝜃 is equal to five 𝜋 by four into our expression for d𝑥 by d𝜃. Then, we just need to take the quotient of these two values. So let’s clear some space and evaluate both of our derivatives that 𝜃 is equal to five 𝜋 by four.

Let’s start with d𝑦 by d𝜃. We just need to substitute in 𝜃 is equal to five 𝜋 by four. This gives us the following expression. And we can evaluate this expression either by using what we know about trigonometric functions evaluated at standard angles or by using our calculator. Either way, when we evaluate this expression, we get one minus the square root of two. We can then do exactly the same thing for d𝑥 by d𝜃. We want to substitute in 𝜃 is equal to five 𝜋 by four, and this gives us the following expression. And once again, we can evaluate this by using what we know about trigonometric functions evaluated at standard angles or by using a calculator. Either method gives us the square root of two plus two.

Now, by using our formula, the quotient of these two values will give us the slope of our tangent line when 𝜃 is equal to five 𝜋 by four. So, we get d𝑦 by d𝑥 at 𝜃 is equal to five 𝜋 by four is equal to one minus the square root of two all divided by root two plus two. And we could leave our answer like this. However, remember, this represents the slope of a line, and we know that the square root of two is bigger than one. So this slope is negative. So, to highlight this fact, we’ll want to take out a factor of negative one from our numerator. Then, all we’ll do is reorder the two terms in our numerator. And this gives us our final answer of negative one times the square root of two minus one all divided by the square root of two plus two.

In this question, we were able to use our formula for finding d𝑦 by d𝑥 for a polar curve to find the slope of the tangent line to the curve 𝑟 is equal to two minus the sin of three 𝜃 as 𝜃 is equal to five 𝜋 by four. We got this slope was equal to negative one times the square root of two minus one all divided by the square root of two plus two.