Video: Determining the Type of Concavity of a Parametric Curve

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

Here, we have been asked about the concavity of a curve. We know that, in order to determine the concavity of a curve, we need to consider the second derivative of π¦ with respect to π₯. We know that when d two π¦ by dπ₯ squared is greater than zero, our curve is concave down. And when itβs less than zero, our curve is concave up. From this, we can see that, in order to consider the concavity of our curve, we first need to find d two π¦ by dπ₯ squared. Now, weβve been given our curve in terms of parametric equations. Therefore, we can use the following formula to find d two π¦ by dπ₯ squared. This formula tells us that d two π¦ by dπ₯ squared is equal to d by dπ of dπ¦ by dπ₯ over dπ₯ by dπ. In order to use this, we first need to find dπ¦ by dπ₯, which we also know a formula for. We have that dπ¦ by dπ₯ is equal to dπ¦ by dπ over dπ₯ by dπ. Therefore, we can start by finding dπ¦ by dπ and dπ₯ by dπ.

Weβve been given that π₯ is equal to cos π. And π¦ is equal to sin π. Differentiating cos π with respect to π, we obtain that dπ₯ by dπ is equal to negative sin π. And differentiating sin π with respect to π, we obtain that dπ¦ by dπ is equal to cos π. And substituting these into our equation for dπ¦ by dπ₯, we obtain that dπ¦ by dπ₯ is equal to negative cos π over sin π or negative cot π. Next, we need to differentiate dπ¦ by dπ₯ with respect to π, which is equivalent to d by dπ of negative cot π. Now, we know that cot π differentiates to give negative csc squared π. And so we can see that negative cot π will differentiate to csc squared π. Now, we have found d by dπ of dπ¦ by dπ₯ and dπ₯ by dπ. And so we obtain that d two π¦ by dπ₯ squared is equal to csc squared π over negative sin π. Since csc π is equal to one over sin π, we have that d two π¦ by dπ₯ squared is equal to negative csc cubed π.

At this stage, we found d two π¦ by dπ₯ squared. We just need to evaluate it at π is equal to π by six. At π is equal to π by six, we have that it is equal to negative csc cubed of π by six. And we can use the fact that csc is equal to one over sin, giving us that this is equal to negative one over sin cubed of π by six. We know that sin of π by six is equal to one-half. So the denominator of our fraction is equal to one-half cubed. Now, one-half cubed is one-eighth. And one over one-eighth is simply eight. So weβve evaluated d two π¦ by dπ₯ squared at π is equal to π by six to be equal to negative eight. Since negative eight is less than zero, this tells us that, at π is equal to π by six, our curve is concave up.