Question Video: Determining the Type of Concavity of a Parametric Curve

Consider the parametric curve π‘₯ = cos πœƒ and 𝑦 = 2 sin πœƒ. Determine whether this curve is concave up, down, or neither at πœƒ = πœ‹/6.

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Video Transcript

Consider the parametric curve π‘₯ is equal to the cos of πœƒ and 𝑦 is equal to two times the sin 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 defined by a pair of parametric equations. We’re given π‘₯ is some function of πœƒ and 𝑦 is some function 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, let’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 𝑦 with respect to π‘₯. If d two 𝑦 by dπ‘₯ squared is positive at a point, then our curve is concave upwards at this point. And similarly, if d two 𝑦 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 𝑦 by dπ‘₯ squared. However, we can’t do this directly because we’re given a parametric curve. Let’s recall some of our rules for differentiating parametric curves.

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

So to find the concavity of our curve, we want to find an expression for d two 𝑦 by dπ‘₯ squared. But to do this, we need to find an expression for d𝑦 by dπ‘₯. And to find an expression for this, we need to find dπ‘₯ by dπœƒ and d𝑦 by dπœƒ. And, of course, we’re given π‘₯ and 𝑦 in terms of πœƒ. So, we can do this. Let’s start by finding an expression for dπ‘₯ by dπœƒ.

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

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

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

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

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

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

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

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