Video: Second Derivatives of Parametric Equations

Video Transcript

Second Derivatives of Parametric Equations

In this video, we’ll learn how to find the second derivatives of parametric equations with respect to 𝑥 and how to find higher-order derivatives of parametric equations by using the chain rule. And we’ll also cover a variety of examples and uses of the second derivatives of parametric equations. So let’s start with a pair of parametric equations. Let’s say that 𝑥 is equal to some function 𝑓 of 𝑡 and 𝑦 is equal to some function 𝑔 of 𝑡. And we already know how to find an expression for d𝑦 by d𝑥 assuming 𝑓 of 𝑡 and 𝑔 of 𝑡 are differentiable and d𝑥 by d𝑡 is not equal to zero.

We saw by applying the chain rule and the inverse function theorem to our parametric equations 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡, we get d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡 provided d𝑥 by d𝑡 is not equal to zero. But we want to find an expression for the second derivative of 𝑦 with respect to 𝑥. Of course, the second derivative of 𝑦 with respect to 𝑥 is the derivative of d𝑦 by d𝑥 with respect to 𝑥. Normally. we would just differentiate this directly. However, in this case, there’s a little problem. We see from our formula for d𝑦 by d𝑥 that the numerator d𝑦 by d𝑡 is a function in 𝑡 and the denominator d𝑥 by d𝑡 is also a function in 𝑡. So in this case, d𝑦 by d𝑥 will be a function in 𝑡.

So when we try to evaluate d two 𝑦 by d𝑥 squared by differentiating d𝑦 by d𝑥 with respect to 𝑥, we’ll have a problem. We’re trying to differentiate a function in 𝑡 with respect to 𝑥. However, we can get around this problem by applying the chain rule. We need to recall the chain rule tells us if capital 𝐹 is a function in 𝑡 and 𝑡 is a function in 𝑥, then we can calculate d𝐹 by d𝑥 by calculating d𝐹 by d𝑡 and multiplying this by d𝑡 by d𝑥. In our case, capital 𝐹 will be the differential d𝑦 by d𝑥. So by applying the chain rule, we now have that d two 𝑦 by d𝑥 squared is equal to the derivative of d𝑦 by d𝑥 with respect to 𝑡 multiplied by d𝑡 by d𝑥.

And now we know that d𝑦 by d𝑥 is a function in 𝑡. So we can differentiate this with respect to 𝑡 by using one of our derivative rules. However, we don’t have 𝑡 as a function in 𝑥. Instead, we’re given 𝑥 as a function in 𝑡. But in actual fact, we’ve seen how to get around this problem before when we found d𝑦 by d𝑥 by using the inverse function theorem and the chain rule. When we were finding this expression, instead of finding d𝑡 by d𝑥 directly, we used the inverse function theorem to find d𝑥 by d𝑡. We can do this in this case as well. We set 𝑡 to be the inverse function 𝑓 of 𝑥. Then we can find an expression for d𝑡 by d𝑥 by using the inverse function theorem. We get that d𝑡 by d𝑥 is equal to one divided by d𝑥 by d𝑡 provided d𝑥 by d𝑡 is not equal to zero.

Now, all we need to do is substitute this into our equation. And this gives us a formula for finding d two 𝑦 by d𝑥 squared for parametric equations in this form. We get that d two 𝑦 by d𝑥 squared is equal to the derivative of d𝑦 by d𝑥 with respect to 𝑡 divided by d𝑥 by d𝑡 provided that d𝑥 by d𝑡 is not equal to zero. We’re now ready to start using this formula to find the second derivative of 𝑦 with respect to 𝑥 for parametric equations. However, it’s worth reiterating we are still going to need to use our formula for finding the first derivative of 𝑦 with respect to 𝑥 in parametric equations, since this is directly used in our formula for the second derivative.

Given that 𝑥 is equal to three 𝑡 squared plus one and 𝑦 is equal to three 𝑡 squared plus five 𝑡, find the second derivative of 𝑦 with respect to 𝑥.

In this question, we’re given a pair of parametric equations. And we need to use these to find the second derivative of 𝑦 with respect to 𝑥. And we know we have a formula for finding the second derivative of 𝑦 with respect to 𝑥 for parametric equations. But to use this formula, we do need to check that our function for 𝑥 in terms of 𝑡 and our function for 𝑦 in terms of 𝑡 are both differentiable. In this case, they’re both polynomial, so we do in fact know they’re differentiable. So let’s now recall this formula. We have d two 𝑦 by d𝑥 squared is equal to the derivative of d𝑦 by d𝑥 with respect to 𝑡 divided by d𝑥 by d𝑡. And this is only valid when d𝑥 by d𝑡 is not equal to zero.

But to use this formula, we see we first need to find an expression for d𝑦 by d𝑥. But of course, we’re given 𝑥 in terms of 𝑡 and 𝑦 in terms of 𝑡. So to find an expression for d𝑦 by d𝑥, we’re going to need to use another formula. Once again, for a pair of parametric equations, 𝑥 given in terms of 𝑡 and 𝑦 given in terms of 𝑡, we know d𝑦 by d𝑥 will be equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡 provided that d𝑥 by d𝑡 is not equal to zero. So to find d𝑦 by d𝑥, we’re going to need to first find d𝑦 by d𝑡 and d𝑥 by d𝑡.

Let’s start by finding d𝑥 by d𝑡. First, we know that 𝑥 is equal to three 𝑡 squared plus one. So d𝑥 by d𝑡 will be the derivative of three 𝑡 squared plus one with respect to 𝑡. Of course, since this is a polynomial, we can do this term by term by using the power rule for differentiation. We want to multiply by our exponent of 𝑡 and then reduce this exponent by one. This gives us d𝑥 by d𝑡 is equal to six 𝑡. And of course, we know that the constant one doesn’t change as 𝑡 changes. So its rate of change with respect to 𝑡 is equal to zero. So we have d𝑥 by d𝑡 is equal to six 𝑡.

We now need to find an expression for d𝑦 by d𝑡. We know 𝑦 is equal to three 𝑡 squared plus five 𝑡. This means d𝑦 by d𝑡 will be the derivative of three 𝑡 squared plus five 𝑡 with respect to 𝑡. Once again, we can do this term by term by using the power rule for differentiation. And of course, when we do this, we get six 𝑡 plus five. So we’ve shown d𝑦 by d𝑡 is equal to six 𝑡 plus five. We now have expressions for both d𝑥 by d𝑡 and d𝑦 by d𝑡. So we can use these in our formula to find an expression for d𝑦 by d𝑥. Substituting in our expressions for d𝑦 by d𝑡 and d𝑥 by d𝑡, we get d𝑦 by d𝑥 is equal to six 𝑡 plus five divided by six 𝑡. And of course, we know this is not valid when d𝑥 by d𝑡 is equal to zero.

So this won’t be valid when our denominator six 𝑡 is equal to zero. And of course, six 𝑡 being equal to zero means that 𝑡 is equal to zero. In other words, we know that our formula for d𝑦 by d𝑥 is valid for all values of 𝑡 except when 𝑡 is equal to zero. But in this question, we don’t actually need to find the values of 𝑡 where our second derivative will be valid. So we don’t need to include this. However, it’s worth thinking about. Remember, we found that expression for d𝑦 by d𝑥 to help us find an expression for d two 𝑦 by d𝑥 squared.

We can see in this expression, we need to find the derivative of d𝑦 by d𝑥 with respect to 𝑡. So we’re going to need to differentiate our expression for d𝑦 by d𝑥 with respect to 𝑡. Using this, we get the derivative of d𝑦 by d𝑥 with respect to 𝑡 is equal to the derivative of six 𝑡 plus five all divided by six 𝑡 with respect to 𝑡.

Since this is the quotient of two differentiable functions, we might be tempted to use the quotient rule here. However, remember, our value of 𝑡 is not allowed to be equal to zero. So in fact, we could just divide both terms in our numerator by six 𝑡. So by dividing each term in our numerator separately by the denominator, we now have the derivative of six 𝑡 divided by six 𝑡 plus five divided by six 𝑡 with respect to 𝑡. Once again, we know that 𝑡 is not equal to zero. So we can simplify our first fraction to just give us one. And since we need to differentiate this, it might be easier to rewrite 𝑡 from the denominator in our numerator by using our laws of exponents. Instead of dividing by 𝑡, we’ll multiply by 𝑡 to the power of negative one.

Now we can evaluate this derivative term by term by using the power rule for differentiation. First, the derivative of the constant one with respect to 𝑡 is equal to zero. Next, the derivative of five times 𝑡 to the power of negative one divided by six is equal to negative five 𝑡 to the power of negative two divided by six. And once again, we’ll use our laws of exponents to simplify this expression. Instead of multiplying by 𝑡 to the power of negative two, we’ll instead divide by 𝑡 squared. So we’ve now found the derivative of d𝑦 by d𝑥 with respect to 𝑡. We got negative five divided by six 𝑡 squared.

We’re now ready to use this to find an expression for d two 𝑦 by d𝑥 squared. We’ve already found an expression for both the numerator and the denominator of this formula. The numerator, the derivative of d𝑦 by d𝑥 with respect to 𝑡, is equal to negative five divided by six 𝑡 squared. And the denominator d𝑥 by d𝑡 is equal to six 𝑡. Now, by substituting these two expressions into our formula, we get d two 𝑦 by d𝑥 squared is equal to negative five divided by six 𝑡 squared all divided by six 𝑡. And once again, we know this won’t be valid when 𝑡 is equal to zero. And we can simplify this expression. Instead of dividing by six 𝑡, we can instead multiply by the reciprocal of six 𝑡. So by using this, we get d two 𝑦 by d𝑥 squared is equal to negative five divided by six 𝑡 squared all multiplied by one over six 𝑡.

And now we can just simplify this expression. First, in our denominator, six multiplied by six simplifies to give us 36. And by using our laws of exponents, 𝑡 squared multiplied by 𝑡 simplifies to give us 𝑡 cubed. And this gives us our final answer. If 𝑥 is equal to three 𝑡 squared plus one and 𝑦 is equal to three 𝑡 squared plus five 𝑡, we were able to show that d two 𝑦 by d𝑥 squared is equal to negative five divided by six 𝑡 cubed. This formula can be useful in helping us find the concavity of curves defined by parametric equations as we’ll see in our next example.

Consider the parametric curve 𝑥 is equal to the cos of 𝜃 and 𝑦 is equal to the sin of 𝜃. Determine whether this curve is concave up, down, or neither at 𝜃 is equal to 𝜋 by six.

Here, we’ve been asked about the concavity of a curve. And we know to check the concavity of a curve, we’re going to want to find an expression for the second derivative of 𝑦 with respect to 𝑥. We know when the second derivative of 𝑦 with respect to 𝑥 is greater than zero, then the slope of our tangent lines are increasing. This means that these tangent lines will be below our curve, and so our curve will be concave upward on this interval. And we also know when the second derivative of 𝑦 with respect to 𝑥 is less than zero, the slope of our tangent lines will be decreasing. This means that our tangent lines will all lie above our curve, so our curve will be concave downward.

So one way to determine the concavity of our curve is to find an expression for d two 𝑦 by d𝑥 squared. And we can see that we’re given 𝑥 and 𝑦 as a pair of parametric equations. So to find d two 𝑦 by d𝑥 squared, we’re going to need to recall our formula for finding these parametric equations. We have d two 𝑦 by d𝑥 squared is equal to the derivative of d𝑦 by d𝑥 with respect to 𝜃 divided by d𝑥 by d𝜃. And we know this formula won’t be valid when d𝑥 by d𝜃 is equal to zero.

And we can see to use this formula, we’re going to first need to find an expression for d𝑦 by d𝑥. And once again, we can do this by using our formulas involving parametric curves. We get d𝑦 by d𝑥 is equal to d𝑦 by d𝜃 divided by d𝑥 by d𝜃. And once again, this formula will only be valid provided d𝑥 by d𝜃 is not equal to zero. And now we can see that we’re given 𝑦 in terms of 𝜃 and 𝑥 in terms of 𝜃. So we can find expressions for d𝑦 by d𝜃 and d𝑥 by d𝜃.

Let’s start by finding d𝑥 by d𝜃. First, we’re told that 𝑥 is equal to the cos of 𝜃. So d𝑥 by d𝜃 will be the derivative of the cos of 𝜃 with respect to 𝜃. And this is a standard trigonometric derivative result. We know this is equal to negative the sin of 𝜃. So we’ve shown d𝑥 by d𝜃 is negative the sin of 𝜃. We can do the same to find d𝑦 by d𝜃. We’re told 𝑦 is the sin of 𝜃, so d𝑦 by d𝜃 will be the derivative of the sin of 𝜃 with respect to 𝜃. And we know that this is equal to the cos of 𝜃. Now, we could substitute both of these into our expression for d𝑦 by d𝑥. And this gives us that d𝑦 by d𝑥 is equal to the cos of 𝜃 divided by negative the sin of 𝜃. And we can simplify this using our trigonometric identities to give us negative the cot of 𝜃.

And remember, we found this to find an expression for d two 𝑦 by d𝑥 squared. In this numerator, we now need to differentiate d𝑦 by d𝑥 with respect to 𝜃. So we need to differentiate negative the cot of 𝜃 with respect to 𝜃. And once again, this is a standard trigonometric derivative result we should commit to memory. The derivative of the cot of 𝜃 with respect to 𝜃 is equal to negative the cosec squared of 𝜃.

So this tells us that the negative cot of 𝜃 differentiates to give us the cosec squared of 𝜃. And now we’re ready to find an expression for d two 𝑦 by d𝑥 squared. We’ve already found an expression for the derivative of d𝑦 by d𝑥 with respect to 𝜃, and we’ve already found an expression for the derivative of 𝑥 with respect to 𝜃. So by substituting these expressions in, we get d two 𝑦 by d𝑥 squared is equal to the cosec squared of 𝜃 divided by negative the sin of 𝜃. And remember, by using our trigonometric identities, dividing by the sin of 𝜃 is the same as multiplying by the cosec of 𝜃. So we can rewrite this as negative the cosec cubed of 𝜃.

Now we found an expression for d two 𝑦 by d𝑥 squared. We just need to evaluate this at 𝜃 is equal to 𝜋 by six. Substituting 𝜃 is equal to 𝜋 by six, we get d two 𝑦 by d𝑥 squared at 𝜋 by six is equal to negative the cosec cubed of 𝜋 by six. And we can use the fact that the cosec of 𝜃 is one divided by the sin of 𝜃 to make this easier. Doing this, we get negative one divided by the sin cubed of 𝜋 by six. And in fact, we know the sin of 𝜋 by six is just equal to one-half. So the denominator of our fraction simplifies to give us one-half cubed. And, of course, one-half cubed is equal to one-eighth, and negative one divided by one-eighth is equal to negative eight.

So we’ve shown d two 𝑦 by d𝑥 squared at 𝜃 is equal to 𝜋 by six is equal to negative eight. Since this is negative, we can conclude our curve must be concave downward at this value of 𝜃. So we were able to show when 𝜃 is equal to 𝜋 by six, our curve is concave downward.

Now let’s consider what happens when we try to find higher-order derivatives of parametric equations. Let’s start with d three 𝑦 by d𝑥 cubed. We can say this will be equal to the derivative of d two 𝑦 by d𝑥 squared with respect to 𝑥. And of course, we already found an expression for d two 𝑦 by d𝑥 squared for a pair of parametric equations. And now we have the same problem we had before. Our numerator is a function in 𝑡 and our denominator is a function in 𝑡. So d two 𝑦 by d𝑥 squared is a function in 𝑡, which means to evaluate this derivative, we’re going to need to use the chain rule. This gives us d three 𝑦 by d𝑥 cubed is equal to the derivative of d two 𝑦 by d𝑥 squared with respect to 𝑡 multiplied by d𝑡 by d𝑥. And just as we did before, we can apply the inverse function theorem to rewrite d𝑡 by d𝑥 as one divided by d𝑥 by d𝑡.

And it’s worth pointing out since we used our formula for d two 𝑦 by d𝑥 squared, we already know d𝑥 by d𝑡 is not equal to zero. Now, by applying this, we have d three 𝑦 by d𝑥 cubed is equal to the derivative of d two 𝑦 by d𝑥 squared with respect to 𝑡 divided by d𝑥 by d𝑡. And now we’re starting to see a pattern emerging. In our expression for d two 𝑦 by d𝑥 squared, in the numerator, we have the derivative of d𝑦 by d𝑥 with respect to 𝑡. And in the denominator, we have d𝑥 by d𝑡. And in our expression for d three 𝑦 by d𝑥 cubed, we still have d𝑥 by d𝑡 in our denominator. However, in our numerator, instead of the first derivative, we have the second derivative. And in fact, this pattern continues for higher-order derivatives. We get d 𝑛 𝑦 by d𝑥 𝑛 is equal to the derivative of d 𝑛 minus one 𝑦 by d𝑥 𝑛 minus one with respect to 𝑡 divided by d𝑥 by d𝑡.

And just like before, we need to be careful when d𝑥 by d𝑡 is equal to zero. We can then use this formula to find higher-order derivatives of parametric equations. We’ll see how to do this in our final example.

If 𝑥 is equal to four 𝑡 plus three and 𝑦 is equal to two 𝑒 to the power of 𝑡 minus 𝑡 cubed, find d three 𝑦 by d𝑥 cubed.

We’ve been given a pair of parametric equations and asked to find the third derivative of 𝑦 with respect to 𝑥. We first need to recall our formula for finding higher-order derivatives of pairs of parametric equations. We recall the following formula, which is valid so long as d𝑥 by d𝑡 is not equal to zero. We want to find the third derivative of 𝑦 with respect to 𝑥. So we need our value of 𝑛 equal to three. So we can just substitute 𝑛 is equal to three into this expression.

To use this, we need to find an expression for d two 𝑦 by d𝑥 squared. And we also have a formula to find d two 𝑦 by d𝑥 squared. However, to use this formula, we also need an expression for d𝑦 by d𝑥. Luckily, we also have a formula to find this for parametric equations which are differentiable. So to find our third derivative, we first need to find expressions for d𝑦 by d𝑡 and d𝑥 by d𝑡. Let’s start with d𝑥 by d𝑡. That’s the derivative of four 𝑡 plus three with respect to 𝑡. This is a linear function, so its derivative will be the coefficient of 𝑡 which is four. We also need to find d𝑦 by d𝑡. That’s the derivative of two 𝑒 to the power of 𝑡 minus 𝑡 cubed with respect to 𝑡. And we can do this term by term.

First, the derivative of the exponential function is equal to itself. So our first term differentiates to give us two 𝑒 to the power of 𝑡. Next, we can differentiate negative 𝑡 cubed by using the power rule for differentiation. We get negative three 𝑡 squared. We can substitute these expressions into our formula for d𝑦 by d𝑥. We get two 𝑒 to the power of 𝑡 minus three 𝑡 squared all divided by four. We can then differentiate this with respect to 𝑡 to get d by d𝑡 of d𝑦 by d𝑥 is equal to two 𝑒 to the power of 𝑡 minus six 𝑡 all divided by four. And we can cancel a shared factor of two.

To find d two 𝑦 by d𝑥 squared, we need to divide this by d𝑥 by d𝑡, which we found was four. This gives us d two 𝑦 by d𝑥 squared is equal to 𝑒 to the power of 𝑡 minus three 𝑡 all divided by eight. To find d three 𝑦 by d𝑥 cubed, we need to differentiate this with respect to 𝑡. And if we do this, we get 𝑒 to the power of 𝑡 minus three all divided by eight. Finally, in our formula for d three 𝑦 by d𝑥 cubed, we need to divide this by d𝑥 by d𝑡. So we divide through by four, giving us 𝑒 to the power of 𝑡 minus three all divided by 32.

Let’s now go through the key points of this video. First, we were able to find an expression for d two 𝑦 by d𝑥 squared for a curve defined by a pair of parametric equations. And we saw this was useful for helping us to determine the concavity of a curve defined by a pair of parametric equations. Finally, we also saw how we could extend this definition to higher-order derivatives in our pairs of parametric equations.