I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If and, then the tradition formula gives

, and

It is that last part, where you divide by , that bothers me. Where did the come from?

Then it occurred to me that dividing by is the same as multiplying by

It’s just implicit differentiation!

Since is a function of *t* you must begin by differentiating the first derivative with respect to *t*. Then treating this as a typical Chain Rule situation and *multiplying* by gives the second derivative. (There is a technical requirement here that given , then exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happening there as well:

The reason you to do it this way is that since x is given as a function of *t*, it may be difficult to solve for *t* so you can find *dt/dx* in terms of *x. *But you don’t have to; just divide by* dx/dt* which you already know.

Here is an example for both derivatives.

Suppose that and

Then , and and .

Then

And

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014

Revised: 2/8/2016

That is very interesting…I don’t like memorizing formulas as well….I’m just thinking, what about test taking…will there be a time to derive it?

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I don’t think there is any “formula” to derive. Given

dy/dxas a function oft, you differentiatedy/dxwith respect totand then multiply bydt/dxsince you want the second derivative with respect tox(in terms oft). The students have to pay attention to what variable they are using (as always).LikeLike

Thank you for this and for all your helpful posts. I can’t reason what happened to the 3; shouldn’t the answer to the second derivative be divided by 3?

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Whoops. Looks like a typo. I have corrected the last equation. Thanks for catching that.

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My pleasure. You have been a tremendous help for many years. Thank you.

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Thought provoking and immediately useful post (as always 🙂 )- Thanks! Does this imply that parametric equations can be thought of as compositions?

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In a way, I think they can. If

xis a function oft, thentis a function ofx, namely the inverse. When you find the second derivative with respect toxof the implicitly defineddy/dx, dividing bydx/dtis the the same as multiplying bydt/dx. At the very least, it is a good way to remember how to find the second derivative which in parametric situations is not just differentiating the first derivative.LikeLike