Question

Use differentials to estimate the amount of material in a closed cylindrical can that is 60 cm high and 24 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.1 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses. The differential for the volume is

Answer 1

Answer:

$dV=542.9 cm^{3}$    

Step-by-step explanation:

Let's start with the equation of the volume of a cylinder:

$V=\pi r^{2}*h$

Where:

• r is the radius of the cylinder (r=D/2=24/2=12 cm)
• h is the high of the cylinder (h = 60 cm)

We can us partial derivatives to find the differential of this volume. So we will have:

$dV=\frac{\partial V}{\partial r}dr+\frac{\partial V}{\partial h}dh$ (1)

Now:

$\frac{\partial V}{\partial r}=2\pi r*h=2\pi 12*60=4523.9 cm^{2}$

$\frac{\partial V}{\partial h}=\pi 12^{2}=452.4 cm^{2}$

dr is a differential of the radius, so in our case it is 0.1 cm and dh, differential of the high, is 0.1*2 cm. We multiply by 2 because we need to consider the top and the bottom of the cylinder.

Now we just need to put all of this definitions in the equation (1).

$dV=4523.9*0.1+452.4*0.2=542.9 cm^{3}$    

I hope it helps you!