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# Question Video: Determining the Type of Concavity of a Parametric Curve Mathematics • Higher Education

Consider the parametric curve π₯ = cos π and π¦ = 2 sin π. Determine whether this curve is concave up, down, or neither at π = π/6.

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### 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. Weβre 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, letβs 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 canβt do this directly because weβre given a parametric curve. Letβs 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, weβre given π₯ and π¦ in terms of π. So, we can do this. Letβs 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. Thatβs 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. Weβve 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, weβre 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. Weβre now ready to start finding our expression for our second derivative function.

Letβs start by finding the expression in our numerator. Thatβs 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.

Weβre 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, weβll 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 weβve 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.

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