Question Video: Finding the Second Derivative of Trigonometric Parametric Equations Mathematics • Higher Education

If π‘₯ = cot 𝑑 and 𝑦 = csc 𝑑, find d²𝑦/dπ‘₯Β².

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Video Transcript

If π‘₯ is equal to the cot of 𝑑 and 𝑦 is equal to the csc of 𝑑, find the second derivative of 𝑦 with respect to π‘₯.

In this question, we’re asked to determine the second derivative of 𝑦 with respect to π‘₯. However, we’re not given 𝑦 as a function in π‘₯. Instead, we’re given a pair of parametric equations in terms of the variable 𝑑. And since 𝑦 is not given as a function in π‘₯, we can’t just differentiate our expression for 𝑦 with respect to π‘₯ twice to determine d𝑦 by dπ‘₯ squared. Instead, we’re going to need to use parametric differentiation.

We can then recall that d two 𝑦 by dπ‘₯ squared will be equal to the derivative of d𝑦 by dπ‘₯ with respect to 𝑑 divided by the derivative of π‘₯ with respect to 𝑑. And there is one small thing worth noting about this formula. We need to use d𝑦 by dπ‘₯. However, 𝑦 is not giving as a function of π‘₯, so we can’t just differentiate 𝑦 with respect to π‘₯. And we can find this by once again using parametric differentiation. Or alternatively, it’s an application of the chain rule. d𝑦 by dπ‘₯ will be equal to d𝑦 by d𝑑 divided by dπ‘₯ by d𝑑.

Therefore, to find the second derivative of 𝑦 with respect to π‘₯, we’re going to first need to find expressions for d𝑦 by d𝑑 and dπ‘₯ by d𝑑. Let’s start by finding the derivative of 𝑦 with respect to 𝑑. That’s the derivative of the csc of 𝑑 with respect to 𝑑. And we can do this by recalling the derivative of the csc of πœƒ with respect to πœƒ is equal to negative the cot of πœƒ multiplied by the csc of πœƒ. Therefore, the derivative of 𝑦 with respect to 𝑑 is equal to negative the cot of 𝑑 multiplied by the csc of 𝑑.

We can follow a very similar process to determine the derivative of π‘₯ with respect to 𝑑. That’s the derivative of the cot of 𝑑 with respect to 𝑑. In this case, we recall the derivative of the cot of πœƒ with respect to πœƒ is equal to negative the csc squared of πœƒ. Therefore, dπ‘₯ by d𝑑 is equal to negative the csc squared of 𝑑. We can now substitute our expressions for d𝑦 by d𝑑 and dπ‘₯ by d𝑑 into our formula to determine d𝑦 by dπ‘₯. Doing this, we get that d𝑦 by dπ‘₯ is equal to negative the cot of 𝑑 times the csc of 𝑑 divided by negative the csc squared of 𝑑.

And we can simplify this. We can start by canceling the shared factor of negative one in the numerator and denominator. And we can also cancel one shared factor of csc of 𝑑 in the numerator and denominator. This leaves us with the cot of 𝑑 divided by the csc of 𝑑. Remember, to find d two 𝑦 by dπ‘₯ squared, we’re going to need to differentiate d𝑦 by dπ‘₯ with respect to 𝑑. So it’s a good idea to try and make this expression as simple as possible. So while we could leave this as cot of 𝑑 divided by the csc of 𝑑 and differentiate this by using the quotient rule, let’s simplify this by using our trigonometric identities.

We’re going to use the fact that the csc of 𝑑 is equal to the cos of 𝑑 divided by the sin of 𝑑. And one over the csc of 𝑑 is equal to the sin of 𝑑. Therefore, we can rewrite our numerator as the cos of 𝑑 divided by the sin of 𝑑. And we’re multiplying this by one over the csc of 𝑑. We can instead multiply it by the sin of 𝑑. This gives us the cos of 𝑑 divided by the sin of 𝑑 multiplied by the sin of 𝑑. And we can simplify this by canceling the shared factor of sin of 𝑑 in the numerator and denominator.

Therefore, d𝑦 by dπ‘₯ is equal to the cos of 𝑑. We’re now ready to use our formula to find an expression for the second derivative of 𝑦 with respect to π‘₯. Let’s start by clearing some space. Substituting in our expression for d𝑦 by dπ‘₯ and our expression for dπ‘₯ by d𝑑 into our formula for d two 𝑦 by dπ‘₯ squared, we get the second derivative of 𝑦 with respect to π‘₯ is equal to the derivative of the cos of 𝑑 with respect to 𝑑 divided by negative the csc squared of 𝑑.

Let’s start by evaluating the numerator. We recall the derivative of the cos of πœƒ with respect to πœƒ is negative the sin of πœƒ. So the derivative of the cos of 𝑑 with respect to 𝑑 is negative the sin of 𝑑. We can also recall that dividing by the csc of 𝑑 is the same as multiplying by the sin of 𝑑. So dividing by the csc squared of 𝑑 is the same as multiplying by the sin squared of 𝑑. We can also move the negative one from our denominator into our numerator. However, this will just cancel with our other factor of negative one. This just leaves us with the sin of 𝑑 times the sin squared of 𝑑, which is of course equal to the sin cubed of 𝑑, which is our final answer.

Therefore, we were able to show if π‘₯ is the cot of 𝑑 and 𝑦 is the csc of 𝑑, then the second derivative of 𝑦 with respect to π‘₯ is the sin cubed of 𝑑.

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