How Does Differential Calculus Estimate Changes in Box Volume?

[p4nda]

Approximate, using differentials, the change in volume of a box if the length increases from 5 feet to 5.05 feet, the width decreases from 4 feet to 3.97 feet, and the depth increases from 3 feet to 3.02 feet. Then, find the actual change in volume and compare the results.



Any ideas on how to solve this problem or what I should use/start out with? (Don't worry, it's not a HW problem.)

 

[quasar987]

It doesn't matter if it's a HW problem per se or not. If it's a HW-type problem, it goes in the Homework Help forum.

The volume of a box is a function of 3 variables: l, w and h. And the derivative (and partial derivatives for that matter) give the rate at which the function changes. For instance, if V(l,w,h) is the volume function, then

[tex]\frac{\partial V}{\partial w}(l_0,w_0,h_0)[/itex] is a number with "units" of (volume)/(units of width). It gives the amount by which the volume would vary if you were to modify the width, <i>when</i> length is l_0, width is w_0 and height is h_0.[/tex]

 

[HallsofIvy]

Write out the equation for V (volume) as a function of x,y,z (height, width, length) [which is trivial]. Find dV as a function of x,y,z and dx,dy,dz. Plug in the values you are given.

 

[p4nda]

Sorry, but do I take the partial derivatives of the l, w, and h then plug it into the total differentiation formula? I am confused as to how I can take the partial derivative of (e.g. 5 feet to 5.05 feet)?

 

[nrqed]

Sorry, but do I take the partial derivatives of the l, w, and h then plug it into the total differentiation formula? I am confused as to how I can take the partial derivative of (e.g. 5 feet to 5.05 feet)?

One step at a time. Did you follow the instructions of HallsofIvy?
Once you have that, just plug in for dx, dy and dz the actual changes in those values.

 

[nrqed]

Sorry, but do I take the partial derivatives of the l, w, and h then plug it into the total differentiation formula? I am confused as to how I can take the partial derivative of (e.g. 5 feet to 5.05 feet)?

Another hint, in case you are trying to understand what Quasar said.

The volume is a function of x, y and z, right? So volume =V(x,y,z).

Now, if any of the variables changes, the volume will change by the amount

$$dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy<br /> + \frac{\partial V}{\partial z} dz$$

Just calculate this expression (in terms of variables, no numerical values yet) and then everywhere you see an x, plug in the initial value of the length along x, everywhere you see a y, plug in the value of the initial length along y, etc and for dx plug in the value of the change of length along x (put it negative if the length along x decreased) and so on for dy and dz.

Patrick