Question Video: Determining the Type of Concavity of a Parametric Curve Involving Using the Chain Rule

Consider the parameric curve π‘₯ = cosΒ³ πœƒ and 𝑦 = sinΒ³ πœƒ. Determine whether this curve is concave up, down, or neither at πœƒ = πœ‹/6.

07:59

Video Transcript

Consider the parametric curve π‘₯ is equal to the cos cubed of πœƒ and 𝑦 is equal to the sin cubed of πœƒ. Determine whether this curve is concave up, down, or neither at πœƒ is equal to πœ‹ by six.

The question gives us a curve which is defined by a pair of parametric equations. We’re given π‘₯ in terms of πœƒ and 𝑦 in terms of πœƒ. We need to determine the concavity of this curve at the point where πœƒ is equal to πœ‹ by six.

To start, we need to recall what is meant by the concavity of a curve. The concavity tells us whether the tangent lines to the curve lie above or below our curve. And in particular, we can measure this by using our second derivative. We know if d two 𝑦 by dπ‘₯ squared is greater than zero at a point, then our curve is concave upward at this point. And we also know if d two 𝑦 by dπ‘₯ squared is less than zero at a point, then our curve is concave downwards at this point.

So, we can check the concavity of our curve by looking at d two 𝑦 by dπ‘₯ squared. However, we can’t differentiate 𝑦 with respect to π‘₯ directly in this case because we’re given our curve as a parametric curve. We’re not given 𝑦 in terms of π‘₯. We’re given 𝑦 in terms of πœƒ and π‘₯ in terms of πœƒ. So, we’re going to need to use our rules for differentiating parametric curves.

The first thing we’ll recall is an application of the chain rule. If 𝑦 is a function of πœƒ and π‘₯ is a function of πœƒ, then we can find by using the chain rule d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And this is, of course, only valid when our denominator dπ‘₯ by dπœƒ is not equal to zero.

Of course, we need to find an expression for the second derivative of 𝑦 with respect to π‘₯. That’s the derivative of d𝑦 by dπ‘₯ with respect to π‘₯. And once again, we can do this by using the chain rule. We get d two 𝑦 by dπ‘₯ squared is equal to the derivative of d𝑦 by dπ‘₯ with respect to πœƒ divided by dπ‘₯ by dπœƒ. And just as before, this won’t be valid when our denominator dπ‘₯ by dπœƒ is equal to zero. So, to find our expression for d two 𝑦 by dπ‘₯ squared, we need to find an expression for d𝑦 by dπ‘₯. But to find an expression for d𝑦 by dπ‘₯, we need to differentiate 𝑦 with respect to πœƒ and π‘₯ with respect to πœƒ. And we’re given 𝑦 and π‘₯ as functions of πœƒ. So, we can do this.

Let’s start by finding an expression for dπ‘₯ by dπœƒ. We know π‘₯ is the cos cubed of πœƒ, so we need to differentiate the cos cubed of πœƒ with respect to πœƒ. Now, there’s a couple of different ways we can evaluate this derivative. For example, we could use the chain rule. However, in this case, we can use the general power rule. Recall, the general power rule tells us for any real number 𝑛 and differentiable function 𝑓 of πœƒ, the derivative of 𝑓 of πœƒ to the 𝑛th power with respect to πœƒ is equal to 𝑛 times 𝑓 prime of πœƒ multiplied by 𝑓 of πœƒ to the power of 𝑛 minus one.

In our case, we want to differentiate the cos cubed of πœƒ. So, we’ll set 𝑓 of πœƒ to be the cos of πœƒ and 𝑛 to be equal to three. Then, to use the general power rule, we need to find an expression for 𝑓 prime of πœƒ. Of course, this is a standard trigonometric derivative result. The derivative of the cos of πœƒ with respect to πœƒ is negative the sin of πœƒ. Now, we can apply the general power rule. We get dπ‘₯ by dπœƒ is equal to three times negative the sin of πœƒ multiplied by the cos of πœƒ raised to the power of three minus one.

And, of course, we can simplify this to give us negative three sin of πœƒ times the cos squared of πœƒ. Now, we’ll clear our working and do something very similar to find an expression for d𝑦 by dπœƒ. This time, we need to differentiate the sin cubed of πœƒ with respect to πœƒ. Again, we could do this by using the chain rule, but we’ll use the general power rule.

This time, we’re differentiating the sin cubed of πœƒ, so we’ll set 𝑓 of πœƒ to be the sin of πœƒ and our value of 𝑛 equal to three. This time, we need to use the fact that the derivative of the sin of πœƒ with respect to πœƒ is equal to the cos of πœƒ. So, we find 𝑓 prime of πœƒ is the cos of πœƒ. Now, by applying the general power rule, we get d𝑦 by dπœƒ is equal to three cos of πœƒ multiplied by the sin of πœƒ raised to the power of three minus one. And, of course, we can simplify this to get three cos of πœƒ sin squared of πœƒ.

Now that we found expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ, let’s clear our working and use these to find an expression for d𝑦 by dπ‘₯. So, by using our formula, we get d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. Then, we substitute in our expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ. We get d𝑦 by dπ‘₯ is equal to three cos of πœƒ sin squared of πœƒ all divided by negative three sin of πœƒ cos squared of πœƒ.

And we can simplify this expression. First, we’ll cancel the shared factor of three in our numerator and our denominator. Next, we’ll cancel the shared factor of the sin of πœƒ in our numerator and our denominator. Finally, we’ll cancel the shared factor of the cos of πœƒ in our numerator and our denominator. This leaves us with the sin of πœƒ divided by negative the cos of πœƒ. And we can rewrite this as negative the tan of πœƒ.

We’re now ready to start finding our expression for our second derivative of 𝑦 with respect to π‘₯. Let’s start by finding the numerator of our formula. The numerator of this expression is the derivative of d𝑦 by dπ‘₯ with respect to πœƒ. But we already showed d𝑦 by dπ‘₯ is equal to negative the tan of πœƒ. So, we need to differentiate negative the tan of πœƒ with respect to πœƒ. And this is one of our standard trigonometric derivative results. The derivative of the tan to πœƒ with respect to πœƒ is equal to the sec squared of πœƒ.

So, by using this, we’ve shown the derivative of d𝑦 by dπ‘₯ with respect to πœƒ is equal to negative the sec squared of πœƒ. Now, all we need to do to find an expression for our second derivative of 𝑦 with respect to π‘₯ is divide this expression by dπ‘₯ by dπœƒ. But we already found dπ‘₯ by dπœƒ. It’s negative three sin of πœƒ times the cos squared of πœƒ. So, by using our formula and substituting in the expressions we found for the numerator and the denominator, we get d two 𝑦 by dπ‘₯ squared is equal to negative the sec squared of πœƒ divided by negative three sin of πœƒ multiplied by the cos squared of πœƒ.

And we can simplify this slightly since negative one divided by negative one is just equal to one. Now, we could start simplifying this expression even more. But remember, we’re only interested in the concavity of this curve at a point. And to find this, we’re only interested in if our output when πœƒ is equal to πœ‹ by six is positive or negative. Well, we know the sec squared of πœƒ is a square; it will always be greater than or equal to zero. Similarly, the cos squared of πœƒ is also a square. It will be greater than or equal to zero for any input value of πœƒ.

So, in actual fact, the only part of this expression which can be negative is the sin of πœƒ. So, we can just evaluate this. The sin of πœ‹ by six, we know, is equal to one-half. And remember, this formula is only valid if dπ‘₯ by dπœƒ is not equal to zero. So, we should check that the cos of πœ‹ by six is also not equal to zero. Of course, we know the cos of πœ‹ by six is the square root of three divided by two. So, this is not equal to zero.

So, what we’ve shown is when πœƒ is equal to πœ‹ by six, the numerator of d two 𝑦 by dπ‘₯ squared is positive. It’s a nonzero number squared. Similarly, the denominator is also positive. It’s the product of three nonzero positive numbers. So, when πœƒ is equal to πœ‹ by six, d two 𝑦 by dπ‘₯ squared is the quotient of two positive numbers. Therefore, it’s positive.

And remember, if d two 𝑦 by dπ‘₯ squared is positive at a point, then we know that our curve is concave upward at this point. Therefore, given the parametric curve π‘₯ is equal to the cos cubed of πœƒ and 𝑦 is equal to the sin cubed of πœƒ. We were able to show that the curve is concave upward at the point where πœƒ is equal to πœ‹ by six.

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