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

Find the slope of the tangent line to the curve π = sin 2π at π = π/6.

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### Video Transcript

Find the slope of the tangent line to the curve π is equal to the sin of two π at π is equal to π by six.

Weβre given a curve defined by a polar equation. And we need to determine the slope of the tangent line to this polar curve at the point where π is equal to π by six. To answer this question, we first need to recall how we find the slope of the tangent line to a curve. This will be equal to dπ¦ by dπ₯ at this point, the rate of change of π¦ with respect to π₯. In this instance though, weβre not given π¦ as a function in π₯, so we canβt find this directly. Instead, weβre given a polar curve, π is equal to the sin of two π.

So to answer this question, weβre going to need to recall our formula for dπ¦ by dπ₯ for a polar curve. We recall by using the chain rule and the inverse function theorem on a polar curve, we can show that dπ¦ by dπ₯ will be equal to dπ¦ by dπ divided by dπ₯ by dπ. And this formula will work provided that dπ₯ by dπ is not equal to zero. So to find the slope of our tangent line when π is equal to π by six, weβre going to need to first find an expression for dπ¦ by dπ and dπ₯ by dπ. However, weβre not given π¦ or π₯ in terms of π. So, weβre going to need to find expressions for π¦ and π₯ in terms of π.

And to do this, weβll start by recalling the 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 this polar curve, π is equal to the sin of two π. So, we can substitute this into our expressions for π¦ and π₯. Substituting π is equal to the sin of two π into these two equations, we get π¦ is equal to the sin of two π times the sin of π and π₯ is equal to the sin of two π multiplied by the cos of π. We want to use these to find expressions for dπ¦ by dπ and dπ₯ by dπ.

Letβs start with dπ¦ by dπ. We know dπ¦ by dπ is the derivative of π¦ with respect to π. Thatβs the derivative of the sin of two π times the sin of π with respect to π. And this is the product of two differentiable functions, so we can evaluate this derivative by using the product rule. We recall the product rule tells us 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. We want to use this to find the derivative of the sin of two π multiplied by the sin of π with respect to π. So, weβll set π’ of π to be the sin of two π and π£ of π to be the sin of π.

To use the product rule, we need to find expressions for π’ prime of π and π£ prime π. Letβs start with π’ prime of π; thatβs the derivative of the sin of two π with respect to π. And we can evaluate this derivative by recalling one of our standard trigonometric derivative results. 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 ππ. In our case, the value of π is two. So, we get π’ prime of π is two times the cos of two π. We also need to find an expression for π£ prime of π; thatβs the derivative of the sin of π with respect to π. And we know this is just equal to the cos of π.

Weβre now ready to use the product rule to find an expression for the derivative of the sin of two π multiplied by the sin of π with respect to π. We just need to substitute in our expressions for π’ of π, π£ of π, π’ prime of π, and π£ prime of π. We get two times the cos of two π multiplied by the sin of π plus the cos of π multiplied by the sin of two π. This then allows us to evaluate the derivative of π¦ with respect to π. We just need to write this expression in. Doing this, we get dπ¦ by dπ is two cos of two π times the sin of π plus the cos of π multiplied by the sin of two π.

We now need to find an expression for dπ₯ by dπ; thatβs the derivative of the sin of two π times the cos of π with respect to π. And once again, we see this is the product of two differentiable functions. So, weβll evaluate this derivative by using the product rule. So, letβs clear some space and evaluate this by using the product rule. First, we notice that π’ of π will still be the sin of two π. So, weβve already found an expression for π’ prime of π. However, this time, weβll need to set π£ of π to be the cos of π. So, weβll need to find an expression for π£ prime of π. Thatβs the derivative of the cos of π with respect to π. And we know this is equal to negative the sin of π.

Weβre now ready to find an expression for the derivative of the sin of two π times the cos of π with respect to π by using the product rule. We just need to substitute in our expressions for π’ of π, π£ of π, π’ prime of π, and π£ prime of π into our formula from the product rule. This gives us two cos of two π times the cos of π plus negative sin of π multiplied by the sin of two π. And we can also simplify the coefficient in our second term. Now, just as we did before, we can use this to find an expression for dπ₯ by dπ.

Therefore, by using the product rule, weβve shown dπ₯ by dπ is equal to two cos of two π multiplied by the cos of π minus the sin of π times the sin of two π. Now that we found expressions for dπ¦ by dπ and dπ₯ by dπ, we can use these to find an expression for dπ¦ by dπ₯. Clearing some space and then substituting in our expressions for dπ¦ by dπ and dπ₯ by dπ, we get the following expression for dπ¦ by dπ₯.

But remember, the question is not just asking us for an expression for dπ¦ by dπ₯. We need to find the slope of our tangent line when π is equal to π by six. So, we need to substitute π is equal to π by six into this expression. So, weβve substituted π is equal to π by six into our expression for dπ¦ by dπ₯. This gives us the following expression for the slope of our tangent line to our polar curve when π is equal to π by six. And we could answer this by using what we know about trigonometric functions evaluated at standard angles or we could do this by using our calculator. Using either method, we can simplify our answer to get five root three divided by three. And this is our final answer.

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 the sin of two π at π is equal to π by six. We were able to show the slope of this tangent line would be five root three divided by three.