Video: Determining the Type of Concavity of a Parametric Curve

Consider the parametric curve π‘₯ = cos πœƒ and 𝑦 = 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 cos πœƒ and 𝑦 is equal to sin πœƒ. Determine whether this curve is concave up, down, or neither at πœƒ is equal to πœ‹ 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 𝑦 with respect to π‘₯. We know that when d two 𝑦 by dπ‘₯ squared is greater than zero, our curve is concave down. And when it’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 𝑦 by dπ‘₯ squared. Now, we’ve been given our curve in terms of parametric equations. Therefore, we can use the following formula to find d two 𝑦 by dπ‘₯ squared. This formula tells us that d two 𝑦 by dπ‘₯ squared is equal to d by dπœƒ of d𝑦 by dπ‘₯ over dπ‘₯ by dπœƒ. In order to use this, we first need to find d𝑦 by dπ‘₯, which we also know a formula for. We have that d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ over dπ‘₯ by dπœƒ. Therefore, we can start by finding d𝑦 by dπœƒ and dπ‘₯ by dπœƒ.

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

At this stage, we found d two 𝑦 by dπ‘₯ squared. We just need to evaluate it at πœƒ is equal to πœ‹ by six. At πœƒ is equal to πœ‹ by six, we have that it is equal to negative csc cubed of πœ‹ 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œ‹ by six. We know that sin of πœ‹ 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 we’ve evaluated d two 𝑦 by dπ‘₯ squared at πœƒ is equal to πœ‹ by six to be equal to negative eight. Since negative eight is less than zero, this tells us that, at πœƒ is equal to πœ‹ by six, our curve is concave up.

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