### Video Transcript

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

Weβre given a curve defined by a polar equation. And we need to determine the slope of the tangent line to this polar curve when π 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 dπ¦ by dπ₯, the rate of change of π¦ with respect to π₯. However, in this case, weβre not given π¦ as a function in π₯; instead this time, weβre given a polar curve. So, weβre going to need to use our formula for finding dπ¦ by dπ₯ for polar curves.

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

So by substituting π is equal to two minus the sin of three π, we get that π¦ is equal to two minus the sin of three π multiplied by the sin of π and π₯ is equal to two minus the sin of three π multiplied by the cos of π. So, now, we have π¦ and π₯ written in terms of π, so we can differentiate these to find expressions for dπ₯ by dπ and dπ¦ by dπ. Letβs start by finding an expression for dπ¦ by dπ.

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

The product rule tells us the derivative of the product of two differentiable functions π’ of π times π£ of π with respect to π is equal to π’ prime of π times π£ of π plus π£ prime of π times π’ of π. In our case, weβll set π’ of π to be the sin of three π and π£ of π to be the sin of π. Now, to use the product rule, weβre going to need to find expressions for π’ prime and π£ prime.

Letβs start with π’ prime. Thatβs the derivative of the sin of three π with respect to π. And to evaluate this, we need to recall for any real constant π, the derivative of the sin of π π with respect to π is equal to π times the cos of ππ. In our case, the value of π is equal to three. So, we get that π’ prime of π is equal to three times the cos of three π. We now need to find an expression for π£ prime of π. Thatβs the derivative of the sin of π with respect to π. This is a standard trigonometric derivative result we should commit to memory. Itβs the cos of π.

Weβre now ready to use the product rule to find an expression for the derivative of the sin of three π multiplied by the sin of π with respect to π. Substituting in our expressions for π’ of π, π£ of π, π’ prime of π, and π£ prime of π into our formula for the product rule, we get three times the cos of three π multiplied by the sin of π plus the cos of π multiplied by the sin of three π. And now that we found this, weβre ready to find an expression for dπ¦ by dπ.

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

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

So weβll 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 π’ of π will once again be the sin of three π. So, we already found π’ prime of π to be three times the cos of three π. This time, however, we need to set our function π£ of π to be the cos of π. We then need to find an expression for π£ prime of π. Thatβs the derivative of the cos of π with respect to π, and we know this is negative the sin of π.

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

First, the derivative of two times the cos of π with respect to π is equal to negative two sin of π. 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 dπ₯ by dπ. And at this point, we have two ways we could continue. We could substitute our expressions for dπ¦ by dπ and dπ₯ by dπ into our formula to find an expression for dπ¦ by dπ₯, and this would work. However, we donβt actually need to find an expression for dπ¦ by dπ₯; we just need to find the slope of our tangent line when π is equal to five π by four.

So, we just need to substitute π is equal to five π by four into our expression for dπ¦ by dπ₯. And to do that, we just need to substitute π is equal to five π by four into our expression for dπ¦ by dπ and π is equal to five π by four into our expression for dπ₯ by dπ. Then, we just need to take the quotient of these two values. So letβs clear some space and evaluate both of our derivatives that π is equal to five π by four.

Letβs start with dπ¦ by dπ. We just need to substitute in π 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 dπ₯ by dπ. We want to substitute in π 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 π is equal to five π by four. So, we get dπ¦ by dπ₯ at π 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, weβll want to take out a factor of negative one from our numerator. Then, all weβll 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 dπ¦ by dπ₯ for a polar curve to find the slope of the tangent line to the curve π is equal to two minus the sin of three π as π 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.