Differentiation of parametrized curves

If x and y are given as functions of t, as happens with a parametrized curve, then the formulas for derivatives of y with respect to x, in terms of the derivatives of x and y with respect to t, ought to fall directly out of the notation. Unfortunately, the usual notation for higher derivatives prevents this.

To see how this should work, consider the first derivative. There, the formula is dy/dx = (dy/dt)/(dx/dt). That is, simply divide both sides of the fraction by dt. Another even slicker way to do this would be to reinterpret the differentials as derivatives with respect to t; that is, writing a dot above a quantity to indicate differentiation with respect to t, write dy/dx = /.

If you try to do this with second derivatives, based on the usual notation for them, then you get a formula which is wrong: d2y/dx2 ≠ /2 = (d2y/dt2)/(dx/dt)2. (Here, I've written ‘≠’ to show that ‘=’ would have been wrong, but it's possible that these may happen to be equal in certain examples.) To get the correct formula, we simply need to differentiate / using the Quotient Rule: (d/dt)(/) = ( − )/2. Dividing by , the second derivative of y with respect to x is really ( − )/3 = /2 − / ⋅ /2; in other words, the naïve formula is only the first term of a two-term expression. This formula is a little long, but it will correctly give you the second derivative of y with respect to x using the first and second derivatives of x and y with respect to t.

There is a symbol for the second derivative using differentials that can serve as a mnemonic for this. To get it, we again differentiate dy/dx using the Quotient Rule, only now using the Quotient Rule for differentials rather than the Quotient Rule for derivatives: d(dy/dx) = (dx d(dy) − dy d(dx))/(dx)2 = (dx d2y − dy d2x)/dx2. Dividing by dx, the second derivative of y with respect to x is really (dx d2y − dy d2x)/dx3 = d2y/dx2 − dy/dx ⋅ d2x/dx2. As you can see, replacing d with d/dt gives the formula from the previous paragraph.

For this reason, I don't like to write d2y/dx2 for the second derivative of y with respect to x. Of course, nobody wants to write the formula from the previous paragraph when they just want a symbol for the second derivative; fortunately, you can write (d/dx)2y for that. This simply means that you apply the operation d/dx (find the differential and then divide by dx, or equivalently find the derivative with respect to x) twice to get the second derivative, which is certainly correct. You can even use this as a mnemonic for finding this second derivative: instead of interpeting d/dx as taking the differential and then dividing by dx, interpret it as taking the derivative with respect to t and then dividing by . This is essentially how the book tells you to take the second derivative.

Finally, whether you use either (dx d2y − dy d2x)/dx3 or (d/dx)2y, either way you can perform practical caclulations by interpreting the differentials literally as differentials. You simply have to write everything in terms of t, put dt and d2t in where they naturally appear, and find that the differentials of t cancel in the final answer. Alternatively, anticipating that the differentials of t will cancel, you can ignore them, which turns taking differentials into taking derivatives with respect to t again.

I'll do Example 10.2.2 on page 571 of the textbook to illustrate all of these approaches. Given x = t − t2, dx = dt − 2t dt, or  = dx/dt = 1 − 2t. Next, d2x = d2t − 2 dt2 − 2t d2t (applying the Product Rule in the second term), while  = −2. Similarly, given y = t − t3, dy = dt − 3t2 dt, or  = 1 − 3t2. Next, d2y = d2t − 6t dt2 − 3t2 d2t, while  = −6t.

To find (d/dx)y = dy/dx, either directly divide (dt − 3t2 dt)/(dt − 2t dt) and simplify this (by cancelling factors of dt) to (1 − 3t2)/(1 − 2t), or instead divide /, which again gives (1 − 3t2)/(1 − 2t). This is pretty much the same process, no matter how you go about it. Then to find (d/dx)2y, one way is to note that (d/dx)y = (1 − 3t2)/(1 − 2t) from the previous paragraph, so differentiate with respect to x again. Either take d((1 − 3t2)/(1 − 2t)) = (2 dt − 6t dt + 6t2 dt)/(1 − 2t)2 divided by dx = dt − 2t dt and simplify by cancelling factors of dt, or take (d/dt)((1 − 3t2)/(1 − 2t)) = (2 − 6t + 6t2)/(1 − 2t)2 divided by  = 1 − 2t; either way, you get (d/dx)2y = (2 − 6t + 6t2)/(1 − 2t)3. This is essentially how the textbook does this problem.

Alternatively, using (d/dx)2y = (dx d2y − dy d2x)/dx3, we immediately get ((dt − 2t dt)(d2t − 6t dt2 − 3t2 d2t) − (dt − 3t2 dt)(d2t − 2 dt2 − 2t d2t))/(dt − 2t dt)3, which simplifies drastically to (2 − 6t + 6t2)/(1 − 2t)3, the same answer as above. Notice that there is no need to work out dy/dx first. Or using (d/dx)2y = ( − )/3, you immediately get ((1 − 2t)(−6t) − (1 − 3t2)(−2))/(1 − 2t)3, which simplifies (somewhat less drastically) to (2 − 6t + 6t2)/(1 − 2t)3 again. I prefer this last method, which gets the answer is one step after the preliminary calculations and doesn't require quite as much algebra to simplify as the corresponding method using differentials.


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