### Video Transcript

Consider the parametric curve π₯ is equal to the cos of π and π¦ is equal to two times the sin of π. Determine whether this curve is concave up, down, or neither at π is equal to π by six.

The question gives us a curve defined by a pair of parametric equations. Weβre given π₯ is some function of π and π¦ is some function of π. We need to determine the concavity of this curve at the point where π is equal to π by six. To start, letβs recall what we mean by the concavity of a curve.

The concavity of a curve tells us whether the tangent lines lie above or below the curve. And one way of checking this is to check the sin of the second derivative of π¦ with respect to π₯. If d two π¦ by dπ₯ squared is positive at a point, then our curve is concave upwards at this point. And similarly, if d two π¦ by dπ₯ squared is negative at a point, then our curve is concave downwards at this point. So, we want to find an expression for d two π¦ by dπ₯ squared. However, we canβt do this directly because weβre given a parametric curve. Letβs recall some of our rules for differentiating parametric curves.

If π¦ is some function of π and π₯ is some function of π, we recall by using the chain rule, we can see that dπ¦ by dπ₯ will be equal to dπ¦ by dπ divided by dπ₯ by dπ. And this will be true everywhere except where our denominator dπ₯ by dπ is equal to zero. And we can arrive at another formula by using a very similar method. We know d two π¦ by dπ₯ squared is the second derivative of π¦ with respect to π₯. So to find d two π¦ by dπ₯ squared, we need to differentiate dπ¦ by dπ₯ with respect to π₯. Once again, we can do this by using the chain rule. Doing this, we get d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with respect to π divided by dπ₯ by dπ. And again, this will be true everywhere except where our denominator dπ₯ by dπ is equal to zero.

So to find the concavity of our curve, we want to find an expression for d two π¦ by dπ₯ squared. But to do this, we need to find an expression for dπ¦ by dπ₯. And to find an expression for this, we need to find dπ₯ by dπ and dπ¦ by dπ. And, of course, weβre given π₯ and π¦ in terms of π. So, we can do this. Letβs start by finding an expression for dπ₯ by dπ.

We know that π₯ is equal to the cos of π. So, this is just the derivative of the cos of π with respect to π. And this is a standard trigonometric derivative result which we should commit to memory. The derivative of the cos of π with respect to π is negative the sin of π. We can do the same to find an expression for dπ¦ by dπ. Thatβs the derivative of two times the sin of π. And if we evaluate this, we get two times the cos of π.

Now that we found expressions for dπ¦ by dπ and dπ₯ by dπ, we can find an expression for our gradient function dπ¦ by dπ₯. We have dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. Weβve already shown dπ¦ by dπ is two cos of π and dπ₯ by dπ is negative the sin of π. So, we get two cos of π divided by negative the sin of π. We could leave our answer like this; however, remember, weβre going to need to differentiate this with respect to π. So, instead of differentiating this by using the quotient rule, we could simplify this.

To do this, we recall that the cot of π is the cos of π divided by the sin of π. And we know trigonometric derivative results that help us differentiate the cot of π. So, using this to simplify our expression, we get negative two times the cot of π. And we can see this is a lot easier to differentiate with respect to π. Weβre now ready to start finding our expression for our second derivative function.

Letβs start by finding the expression in our numerator. Thatβs the derivative of dπ¦ by dπ₯ with respect to π. We already found that dπ¦ by dπ₯ is negative two cot of π. So, we need to differentiate negative two cot of π with respect to π. And once again, we can use a standard trigonometric derivative result. The derivative of the cot of π with respect to π is negative the csc squared of π. So by applying this result, we get negative two times negative the csc squared of π. We can cancel the two factors of negative one, which just leaves us with two times the csc squared of π.

Weβre now ready to use our formula to find an expression for d two π¦ by dπ₯ squared. First, in our numerator, we get the derivative of dπ¦ by dπ₯ with respect to π. We found this was equal to two csc squared of π. And we need to divide this by dπ₯ by dπ. We already found this was equal to negative the sin of π. And we can simplify this expression. Remember, the csc of π is equivalent to one divided by the sin of π. So, multiplying by the csc squared of π is the same as dividing by the sin squared of π. This means we can rearrange our equation to see that d two π¦ by dπ₯ squared is equal to negative two divided by the sin cubed of π.

However, remember, the question wants us to find the concavity of the curve at the point where π is equal to π by six. To do this, weβll substitute π is π by six into our equation for d two π¦ by dπ₯ squared. Substituting π is equal to π by six, we get negative two divided by the sin cubed of π by six. And if we evaluate this expression, we get negative 16. So, what weβve shown is the second derivative of π¦ with respect to π₯ when π is equal to π by six is negative. Therefore, our curve must be concave downwards when π is equal to π by six.

Therefore, we were able to show the parametric curve π₯ is equal to the cos of π and π¦ is equal to two times the sin of π is concave downward at the point where π is equal to π by six.