Lesson Video: Second Derivatives of Parametric Equations | Nagwa Lesson Video: Second Derivatives of Parametric Equations | Nagwa

# Lesson Video: Second Derivatives of Parametric Equations Mathematics

In this video, we will learn how to find second derivatives and higher-order derivatives of parametric equations by applying the chain rule.

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### Video Transcript

Second Derivatives of Parametric Equations

In this video, we will learn how to find the second derivative of curves defined parametrically by applying the chain rule. To do this, letβs start with a pair of parametric equations: π₯ is equal to the function π of π‘ and π¦ is equal to the function π of π‘. We want to find an expression for the second derivative of π¦ with respect to π₯. We canβt do this directly because π¦ is not given as a function in π₯. If it was, we could just differentiate π¦ with respect to π₯ twice. So instead, because π¦ is given as a function in π‘ and π₯ is given us a function in π‘, we can apply the chain rule to find an expression for dπ¦ by dπ₯.

We recall if π is a differentiable function in π‘ and π is a differentiable function in π‘, then by applying the chain rule and the inverse function theorem, we can show the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘, provided dπ₯ by dπ‘ is nonzero. We want to use this to find an expression for the second derivative of π¦ with respect to π₯. Remember, we can find this by differentiating dπ¦ by dπ₯ with respect to π₯. And once again we have the same problem. We can note that dπ¦ by dπ₯ is a function in π‘, since dπ¦ by dπ‘ is a function in π‘ β particularly itβs π prime of π‘ β and dπ₯ by dπ‘ is also a function in π‘ β itβs π prime of π‘. So we canβt just directly differentiate this with respect to π₯. Weβre once again going to need to use the chain rule.

To help us apply the chain rule to this example, we can start by recalling that the chain rule tells us if capital πΉ is a function in π‘ and π‘ in turn is a function in π₯, then the derivative of capital πΉ with respect to π₯ is equal to the derivative of capital πΉ with respect to π‘ multiplied by the derivative of π‘ with respect to π₯. In this case, weβre going to set our function capital πΉ to be dπ¦ by dπ₯. And this is a function in π‘. And weβre differentiating this with respect to π₯. Applying the chain rule to this then gives us the derivative of dπ¦ by dπ₯ with respect to π‘ multiplied by dπ‘ by dπ₯.

And now we can see weβve simplified this expression slightly. dπ¦ by dπ₯ is a function in π‘, and now weβre differentiating this with respect to π‘. So we can attempt to do this by using our derivative results. However, we can notice weβre trying to differentiate π‘ with respect to π₯, but weβre not given π‘ as a function in π₯. Instead, weβre given π₯ as a function in π‘. But weβve seen how to get around this problem before when we tried to evaluate dπ¦ by dπ₯ by using the chain rule and the inverse function theorem. Applying the inverse function theorem to dπ‘ by dπ₯, we can show itβs equal to one divided by dπ₯ by dπ‘, provided the denominator is not equal to zero.

We can now rewrite dπ‘ by dπ₯ by using the inverse function theorem to find an expression for d two π¦ by dπ₯ squared. This gives us the second derivative of π¦ with respect to π₯ is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯ by dπ‘, provided dπ₯ by dπ‘ is nonzero. And this is a very useful result. If π¦ and π₯ are given parametrically, then we can find an expression for d two π¦ by dπ₯ squared by differentiating dπ¦ by dπ₯ with respect to π‘ and dividing this by the derivative of π₯ with respect to π‘. And then itβs worth noting to use this formula, we need to find an expression for dπ¦ by dπ₯. And to do this, we need to use our other formula. Letβs now see some examples of applying this formula to determine the second derivatives of parametric equations.

Given that π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus five π‘, find the second derivative of π¦ with respect to π₯.

In this question, weβre asked to find the second derivative of π¦ with respect to π₯. And we can notice something interesting. π¦ is not given as a function in π₯. Instead, weβre given a pair of parametric equations in terms of the variable π‘. This means weβre going to need to differentiate this by using parametric differentiation.

And before we recall our formula for finding the second derivative of parametric equations, we can note that we know π₯ is a differentiable function in π‘ and π¦ is a differentiable function in π‘ because theyβre both polynomials. And this helps justify that we can use this formula to determine the second derivative of π¦ with respect to π₯, which we recall is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯ by dπ‘. And this is, of course, provided that dπ₯ by dπ‘ is nonzero.

To use this to determine the second derivative of π¦ with respect to π₯, we first need to find an expression for dπ¦ by dπ₯. And we can recall if π₯ and π¦ are given as a pair of parametric equations in terms of a variable π‘, then the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π‘ divided by the derivative of π₯ with respect to π‘. And once again, this is provided that π¦ and π₯ are differentiable functions in π‘ and dπ₯ by dπ‘ is not equal to zero.

Weβre now ready to start finding our expression for d two π¦ by dπ₯ squared. We see we first need to find expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start by finding dπ¦ by dπ‘. This is equal to the derivative of three π‘ squared plus five π‘ with respect to π‘. Since this is a polynomial, we can do this term by term by using the power rule for differentiation. We multiply by the exponent of π‘ and then reduce this exponent by one. We get six π‘ plus five. We can follow the same process to determine dπ₯ by dπ‘. Thatβs the derivative of three π‘ squared plus one with respect to π‘. Once again, we apply the power rule for differentiation term by term. This time, we get six π‘.

We can now substitute these into our formula for dπ¦ by dπ₯. This then gives us that dπ¦ by dπ₯ is equal to six π‘ plus five divided by six π‘. And we could leave this expression like this. However, we know in our formula for d two π¦ by dπ₯ squared, weβre going to need to differentiate this with respect to π‘. And although we could evaluate this derivative by using the quotient rule, itβs easier to divide both terms in our numerator by six π‘. And since six π‘ divided by six π‘ is equal to one and five divided by six π‘ can be rewritten as five-sixths π‘ to the power of negative one, weβve shown dπ¦ by dπ₯ is equal to one plus five-sixths π‘ to the power of negative one.

And then we can just differentiate this by using the power rule for differentiation. And this is much easier to differentiate because we can just do this by using the power rule for differentiation. We could now start substituting our expressions into this formula. However, itβs easier to evaluate the numerator of the fraction on the right-hand side of the equation separately. We want to find the derivative of dπ¦ by dπ₯ with respect to π‘. Thatβs the derivative of five-sixths π‘ to the power of negative one with respect to π‘. And we can evaluate this term by term by using the power rule for differentiation. We get negative five-sixths π‘ to the power of negative two.

And we can simplify this slightly by using our laws of exponents. Multiplying by π‘ to the power of negative two is the same as dividing by π‘ squared. So we can rewrite this as negative five over six π‘ squared. Weβre now ready to substitute this and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared. This then gives us the second derivative of π¦ with respect to π₯ is equal to negative five over six π‘ squared all divided by six π‘. And we can then simplify this. Dividing by six π‘ is the same as multiplying by one over six π‘. And we can then evaluate this. Six π‘ squared times six π‘ is 36π‘ cubed. So this gives us negative five over 36π‘ cubed, which is our final answer.

Therefore, we were able to show if π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus five π‘, then the second derivative of π¦ with respect to π₯ is equal to negative five divided by 36π‘ cubed.

Letβs now see an example where we need to evaluate the second derivative of a pair of parametric equations at a given point.

If π¦ is equal to negative five π₯ cubed minus seven and π§ is equal to three π₯ squared plus 16, find the second derivative of π§ with respect to π¦ at π₯ is equal to one.

In this question, weβre asked to find the second derivative of π§ with respect to π¦. And weβre asked to do this when π₯ is equal to one because weβre not given π§ as a function in π¦. Instead, weβre given π§ and π¦ as a pair of parametric equations. So weβre going to want to do this by using parametric differentiation. And to do this, we should first note that π¦ and π§ are polynomials, so theyβre both differentiable functions in π§. We can then recall if π§ is a differentiable function in π₯ and π¦ is a differentiable function in π₯, then the second derivative of π§ with respect to π¦ is equal to the derivative of dπ§ by dπ¦ with respect to π₯ divided by dπ¦ by dπ₯.

However, we canβt just directly use this formula to answer this question, since we canβt differentiate π§ with respect to π¦ directly. Instead, weβre going to need to use the chain rule, which, if we use in conjunction with the inverse function theorem, tells us dπ§ by dπ¦ will be equal to dπ§ by dπ₯ divided by dπ¦ by dπ₯. And this will now allow us to answer this question. We need to find expressions for dπ¦ by dπ₯ and dπ§ by dπ₯. Letβs start by finding an expression for dπ¦ by dπ₯. Thatβs the derivative of negative five π₯ cubed minus seven with respect to π₯. This is a polynomial expression in π₯, so we can do this term by term by using the power rule for differentiation. We multiply by the exponent of π₯ and then reduce this exponent by one. This gives us dπ¦ by dπ₯ is negative 15π₯ squared.

We can follow the same process to find dπ§ by dπ₯. Itβs the derivative of three π₯ squared plus 16 with respect to π₯. And we apply the power rule for differentiation term by term to get six π₯. Now that we have expressions for dπ§ by dπ₯ and dπ¦ by dπ₯, we can substitute these into our formula to find dπ§ by dπ¦. Doing this, we get that dπ§ by dπ¦ is equal to six π₯ divided by negative 15π₯ squared, and we can simplify this. First, we can cancel a shared factor of π₯ in the numerator and denominator. Next, we can cancel a shared factor of three in the numerator and denominator. This then leaves us with negative two divided by five π₯.

Now we could substitute these expressions into our formula for d two π¦ by dπ₯ squared. However, itβs easier to calculate the expression in the numerator of the right-hand side of the equation separately. This then gives us the derivative of dπ§ by dπ¦ with respect to π₯ is equal to the derivative of negative two over five π₯ with respect to π₯. And we can evaluate this derivative by rewriting negative two over five π₯ as negative two-fifths π₯ to the power of negative one. We can then differentiate this by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. This gives us two-fifths π₯ to the power of negative two, which we can rewrite as two over five π₯ squared.

We can now substitute these expressions into our formula for d two π§ by dπ¦ squared. This then gives us the second derivative of π§ with respect to π¦ is equal to two divided by five π₯ squared all divided by negative 15π₯ squared. And we can simplify this. Dividing by negative 15π₯ squared is the same as multiplying by the reciprocal of negative 15π₯ squared. And since five π₯ squared multiplied by 15π₯ squared is 75π₯ to the fourth power, this simplifies to give us negative two divided by 75π₯ to the fourth power.

But weβre not done yet. Remember, the question wants us to determine the second derivative of π§ with respect to π¦ evaluated when π₯ is equal to one. And we can evaluate this by substituting π₯ is equal to one into our expression for d two π§ by dπ¦ squared. We get negative two divided by 75 times one to the fourth power, which we can calculate is equal to negative two divided by 75. Therefore, we were able to show if π¦ is equal to negative five π₯ cubed minus seven and π§ is equal to three π₯ squared plus 16, then the second derivative of π§ with respect to π¦ at π₯ is equal to negative one evaluates to give negative two over 75.

In our next example, weβll have to use parametric differentiation to determine the second derivative of π¦ with respect to π₯. However, one of our parametric equations will be defined implicitly.

If π₯ is equal to two times the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, find the second derivative of π¦ with respect to π₯.

In this question, weβre asked to find the second derivative of π¦ with respect to π₯. But we can see weβre not given π¦ as a function in π₯. Instead, weβre given π₯ as a function in π§. And weβre also given π¦ defined implicitly in terms of π§. This means to find the second derivative of π¦ with respect to π₯, weβre going to need to apply two results. Weβre going to need to use parametric differentiation to determine an expression for d two π¦ by dπ₯ squared.

And thereβs two different methods we could use to find an expression for dπ¦ by dπ§. Either we could square both sides of the equation and rearrange to find an equation for π¦ in terms of π§ or we could use implicit differentiation. Either method would work. We can start by recalling the following formula for the second derivative of π¦ with respect to π₯. Itβs equal to the derivative of dπ¦ by dπ₯ with respect to π§ divided by dπ₯ by dπ§, provided π₯ and π¦ are defined parametrically in terms of π§. And to use this result, we need to find an expression for dπ¦ by dπ₯ in terms of π§. And we can find this by using the chain rule and the inverse function theorem, which tells us that dπ¦ by dπ₯ would be equal to dπ¦ by dπ§ divided by dπ₯ by dπ§.

Therefore, we can find an expression for the second derivative of π¦ with respect to π₯ by finding an expression for dπ¦ by dπ§ and dπ₯ by dπ§. Letβs start by finding an expression for dπ₯ by dπ§. Thatβs the derivative of two sec of five π§ with respect to π§. We can do this by recalling the derivative of the sec of π times π for any real constant π is equal to π times the sec of ππ multiplied by the tan of ππ. So, dπ₯ by dπ§ is 10 sec of five π§ tan of five π§.

We now need to find an expression for dπ¦ by dπ§. And as stated before, thereβs two ways we could do this. And itβs personal preference which one we want to use. Weβll differentiate both sides of this equation with respect to π§. To differentiate root three π¦ with respect to π§, weβre going to need to use implicit differentiation. To make this easier, letβs start by rewriting root three π¦ as root three multiplied by π¦ to the power of one-half. We can now differentiate this with respect to π§ by using the chain rule.

First, π¦ is a function in π§, so we need to differentiate π¦ with respect to π§. Then we multiply this by the derivative of this expression with respect to π¦. We evaluate this by using the chain rule. We multiply by the exponent of π¦, which is one-half, and reduce this exponent by one. This gives us dπ¦ by dπ§ multiplied by root three over two times π¦ to the power of negative one-half. We then need to differentiate the right-hand side of this equation with respect to π§. Thatβs the derivative of the tan of five π§ with respect to π§. And we can evaluate this. Itβs five times the sec squared of five π§.

Remember, we want to find an expression for dπ¦ by dπ§. So weβre going to need to rearrange this equation to make dπ¦ by dπ§ the subject. We can multiply both sides of the equation through by root π¦. We can multiply both sides of the equation through by two and divide through by root three. This gives us dπ¦ by dπ§ is 10 over root three multiplied by root π¦ times the sec squared of five π§. But remember, weβre going to need to differentiate our expression for dπ¦ by dπ₯ with respect to π§. So we donβt want any π¦-terms in our expression. And we can get around this by noting we can find an expression for root π¦ in terms of π§. Weβre told that root three π¦ is equal to the tan of five π§. If we divide both sides of this equation through by root three, we get that root π¦ is equal to one over root three times the tan of five π§.

We can substitute this into our expression for dπ¦ by dπ§. This gives us 10 over root three multiplied by one over root three times the tan of five π§ multiplied by the sec squared of five π§. And we can simplify this slightly. Root three multiplied by root three is equal to three. So dπ¦ by dπ§ is ten-thirds the tan of five π§ multiplied by the sec squared of five π§. We can substitute this expression and our expression for dπ₯ by dπ§ to determine an expression for dπ¦ by dπ₯. This then gives us that the derivative of π¦ with respect to π₯ is equal to ten-thirds tan of five π§ times the sec squared of five π§ divided by 10 sec of five π§ tan of five π§. And we can simplify this. We can cancel a shared factor of sec of five π§ in the numerator and denominator. We can cancel a shared factor of tan of five π§ in the numerator and denominator. And we can also cancel a shared factor of 10 in the numerator and denominator. This just leaves us with one-third times the sec of five π§.

Weβre now almost ready to find an expression for second derivative of π¦ with respect to π₯. Letβs start by evaluating the numerator on the right-hand side of our formula. Letβs start by first clearing some space. We want to differentiate our expression for dπ¦ by dπ₯ with respect to π§. Thatβs the derivative of one-third sec of five π§ with respect to π§. And we can evaluate this. Itβs equal to five-thirds times the sec of five π§ multiplied by the tan of five π§.

We can now substitute this into our formula for d two π¦ by dπ₯ squared. This then gives us the second derivative of π¦ with respect to π₯ is equal to five-thirds times the sec of five π§ tan of five π§ all divided by 10 sec of five π§ multiplied by the tan of five π§. And we can evaluate this. We can cancel the shared factor of sec of five π§ times the tan of five π§ in the numerator and denominator. And we can also cancel a shared factor of five in the numerator and denominator. This gives us one-third divided by two, which we can just evaluate is equal to one-sixth, which is our final answer. Therefore, we were able to show if π₯ is two times the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, then the second derivative of π¦ with respect to π₯ is equal to one-sixth.

Letβs now go over the key points of this video. We were able to show if π¦ is a differentiable function π in π‘ and π₯ is a differentiable function π of π‘, then we can find an expression for the second derivative of π¦ with respect to π₯ by applying the chain rule and inverse function theorem. We showed that d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯ by dπ‘. And this was provided that dπ₯ by dπ‘ was nonzero and we can find dπ¦ by dπ₯ by dividing dπ¦ by dπ‘ by dπ₯ by dπ‘.