Calculus I - Differentials

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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.

Example 1 Compute the differential for each of the following.

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

Show Solution

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.

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

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

$df = - 4{z^3}{{\bf{e}}^{3 - {z^4}}}dz$

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.

Example 2 Compute

$dy$ and

$\Delta y$ if

$y = \cos \left( {{x^2} + 1} \right) - x$ as

$x$ changes from

$x = 2$ to

$x = 2.03$ .

Show Solution

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.

$dy = \left( { - 2x\sin \left( {{x^2} + 1} \right) - 1} \right)dx$

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.

Example 3 A sphere was measured and its radius was found to be 45 inches with a possible error of no more that 0.01 inches. What is the maximum possible error in the volume if we use this value of the radius?

Show Solution

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 in³.

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!