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

# 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 = cos (π/3) at π = π/2.

06:16

### Video Transcript

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

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 π is equal to π by two. To do this, we need to start by recalling the slope of the tangent line to a curve is given by dπ¦ by dπ₯, the rate of change of π¦ with respect to π₯. And normally to find this, we would differentiate π¦ with respect to π₯. However, this time, weβre not given π¦ in terms of π₯. Instead, weβre given a polar equation. So instead, weβre going to need to recall our formula for finding dπ¦ by dπ₯ for a polar curve.

We recall, by using the chain rule and the 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 an expression for dπ¦ by dπ₯, we first need to find expressions 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 these expressions. And to help us find these, we need to recall our standard polar equations, π¦ is equal to π sin π and π₯ is equal to π cos π. And for our polar curve, we know that π is equal to the cos of π divided by three, so we can substitute this into our expressions.

Substituting in π is equal to the cos of π divided by three, we get π¦ is equal to the cos of π over three times the sin of π and π₯ is equal to the cos of π over three multiplied by the cos of π. And we want to find expressions for dπ¦ by dπ and dπ₯ by dπ. And we can see both of these are the product of two differentiable functions, so we can find both of these by using the product rule. We recall 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.

We can use this to find expressions for both dπ¦ by dπ and dπ₯ by dπ. Weβll start with dπ¦ by dπ, so we need to set π’ of π as the cos of π over three and π£ of π to be the sin of π. Now, to apply the product rule, weβre going to need to find expressions for π’ prime of π and π£ prime of π. Weβll start with π’ prime of π. Thatβs the derivative of the cos of π over three with respect to π. And to help us differentiate this, it might be worth recalling that π over three can be rewritten as one-third times π.

Then, we can evaluate this derivative by recalling one of our standard trigonometric derivative results. For any real constant π, the derivative of the cos of ππ with respect to π is equal to negative π times the sin of ππ. So we can apply this with our value of π equal to one-third. This gives us negative one-third sin of π over three. We then want 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 the cos of π. We can now use the product rule to help us find dπ¦ by dπ. Itβs equal to π’ prime of π times π£ of π plus π£ prime of π times π’ of π.

Substituting in our expressions for π’, π£, π’ prime, and π£ prime, we get dπ¦ by dπ is equal to negative one-third sin of π over three times the sin of π plus the cos of π multiplied by the cos of π divided by three. So weβve now found an expression for dπ¦ by dπ. We now need to do the same to find an expression for dπ₯ by dπ. We want to do this by using the product rule again. We need to set π’ of π to be the cos of π divided by three, which is exactly what we had it before. However, this time, weβll need to set π£ of π to be the cos of π.

Since π’ of π is exactly how we had it before, weβve already found an expression for π’ prime of π. We just need to find π£ prime of π, which is the derivative of the cos of π with respect to π. And of course, we know this is negative the sin of π. Now we just need to substitute our expressions for π’, π£, π’ prime, and π£ prime into our formula for the product rule to find an expression for dπ₯ by dπ. We get negative one-third times the sin of π over three multiplied by the cos of π plus negative the sin of π times the cos of π over three. And of course, we can simplify the coefficient of our second term to be negative one.

And now that we found expressions for dπ¦ by dπ and dπ₯ by dπ, we can find an expression for dπ¦ by dπ₯. So letβs clear some space and substitute in our expressions for dπ¦ by dπ and dπ₯ by dπ into our formula for dπ¦ by dπ₯. We get the following expression for dπ¦ by dπ₯. And now that we found an expression for dπ¦ by dπ₯, the only thing left to do to find the slope of our tangent line is substitute in π is equal to π by two.

However, before we do this, we can notice something interesting. We know the cos of π by two is equal to zero. So both of these terms the cos of zero will be equal to zero when we substitute in π is equal to π by two. So we donβt need to include these parts of the expression since they will have a factor of zero. Substituting in π is equal to π by two and removing the terms which are equal to zero, we get dπ¦ by dπ₯ at π is equal to π by two is equal to negative one-third sin of π by two over three times the sin of π by two all divided by negative the sin of π by two multiplied by the cos of π by two over three.

And we can start simplifying. First, we know the sin of π by two is equal to one, so we can remove these two terms. Next, both our numerator and our dominator share a factor of negative one. We also see we have one-third written in our numerator. We can write this is as a three in our denominator. Next, our last two remaining arguments are π by two over three. We can simplify these to give us π by six. So this entire expression simplifies to give us the sin of π by six divided by three times the cos of π by six.

And at this point, thereβs several different ways we could evaluate this expression. Weβll write the sin of π by six over the cos of π by six as the tan of π by six. So we have one-third times the tan of π by six, and we know the tan of π by six is root three over three. And root three over three multiplied by one-third is equal to root three over nine, which is our final answer.

Therefore, we were able to find the slope of the tangent line to the curve π is equal to the cos of π over three at π is equal to π by two by using our formula. We got the slope was equal to the square root of three divided by nine.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy