I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-12 : Differentials

In this section we’re going to introduce a notation that we’ll be seeing quite a bit in the next chapter. We will also look at an application of this new notation.

Given a function \(y = f\left( x \right)\) we call \(dy\) and \(dx\) differentials and the relationship between them is given by,

\[dy = f'\left( x \right)dx\]Note that if we are just given \(f\left( x \right)\) then the differentials are \(df\) and \(dx\) and we compute them in the same manner.

\[df = f'\left( x \right)dx\]Let’s compute a couple of differentials.

- \(y = {t^3} - 4{t^2} + 7t\)
- \(w = {x^2}\sin \left( {2x} \right)\)
- \(f\left( z \right) = {{\bf{e}}^{3 - {z^4}}}\)

Before working any of these we should first discuss just what we’re being asked to find here. We defined two differentials earlier and here we’re being asked to compute a differential.

So, which differential are we being asked to compute? In this kind of problem we’re being asked to compute the differential of the function. In other words, \(dy\) for the first problem, \(dw\) for the second problem and \(df\) for the third problem.

Here are the solutions. Not much to do here other than take a derivative and don’t forget to add on the second differential to the derivative.

a \(dy = \left( {3{t^2} - 8t + 7} \right)dt\)

b \(dw = \left( {2x\sin \left( {2x} \right) + 2{x^2}\cos \left( {2x} \right)} \right)dx\)

c \(df = - 4{z^3}{{\bf{e}}^{3 - {z^4}}}dz\)

There is a nice application to differentials. If we think of \(\Delta x\)as the change in \(x\) then \(\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\) is the change in \(y\) corresponding to the change in \(x\). Now, if \(\Delta x\)is small we can assume that \(\Delta y \approx dy\). Let’s see an illustration of this idea.

First let’s compute actual the change in \(y\), \(\Delta y\).

\[\Delta y = \cos \left( {{{\left( {2.03} \right)}^2} + 1} \right) - 2.03 - \left( {\cos \left( {{2^2} + 1} \right) - 2} \right) = 0.083581127\]Now let’s get the formula for *dy*.

Next, the change in \(x\) from \(x = 2\) to \(x = 2.03\) is \(\Delta x = 0.03\) and so we then assume that \(dx \approx \Delta x = 0.03\). This gives an approximate change in \(y\) of,

\[dy = \left( { - 2\left( 2 \right)\sin \left( {{2^2} + 1} \right) - 1} \right)\left( {0.03} \right) = 0.085070913\]We can see that in fact we do have that \(\Delta y \approx dy\) provided we keep \(\Delta x\) small.

We can use the fact that \(\Delta y \approx dy\) in the following way.

First, recall the equation for the volume of a sphere.

\[V = \frac{4}{3}\pi {r^3}\]Now, if we start with \(r = 45\) and use \(dr \approx \Delta r = 0.01\) then \(\Delta V \approx dV\) should give us maximum error.

So, first get the formula for the differential.

\[dV = 4\pi {r^2}dr\]Now compute \(dV\).

\[\Delta V \approx dV = 4\pi {\left( {45} \right)^2}\left( {0.01} \right) = 254.47\,\,{\rm{in^{3}}}\]The maximum error in the volume is then approximately 254.47 in^{3}.

Be careful to not assume this is a large error. On the surface it looks large, however if we compute the actual volume for \(r = 45\) we get \(V = 381,703.51\,\,\rm{in^3}\). So, in comparison the error in the volume is,

\[\frac{{254.47}}{{381703.51}} \times 100 = 0.067\% \]

That’s not much possible error at all!