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Calculus I (Notes) / Applications of Derivatives / Differentials   [Notes] [Practice Problems] [Assignment Problems]

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October 27, 2016

Calculus I - Notes
Derivatives Previous Chapter   Next Chapter Integrals
Linear Approximations Previous Section   Next Section Newton's Method


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  we call dy and dx differentials and the relationship between them is given by,




Note that if we are just given  then the differentials are df and dx and we compute them in the same manner.




Let’s compute a couple of differentials.


Example 1  Compute the differential for each of the following.






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.









There is a nice application to differentials.  If we think of  as the change in x then  is the change in y corresponding to the change in x.  Now, if  is small we can assume that .  Let’s see an illustration of this idea.


Example 2  Compute dy and  if  as x changes from  to .



First let’s compute actual the change in y, .




Now let’s get the formula for dy.



Next, the change in x from  to  is  and so we then assume that .  This gives an approximate change in y of,




We can see that in fact we do have that  provided we keep  small.


We can use the fact that  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?



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


Now, if we start with  and use  then  should give us maximum error.


So, first get the formula for the differential.



Now compute dV.



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  we get .  So, in comparison the error in the volume is,



That’s not much possible error at all!

Linear Approximations Previous Section   Next Section Newton's Method
Derivatives Previous Chapter   Next Chapter Integrals

Calculus I (Notes) / Applications of Derivatives / Differentials    [Notes] [Practice Problems] [Assignment Problems]

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