Suppose a cylindrical can is manufactured with radius r. The manufacturing process makes the length r imprecise; r can be off by ±0.03 cm. The manufacturing process makes the height of the can precisely equal to three times the radius r (including any error in r, but not introducing any new error).
a) Use ∆r= ±0.03 to approximate the corresponding error in the volume of the can by using differentials (or the linear approximation)
b) The fraction ∆r/r is called the relative error of percentage error in the radius. If the relative error in the radius is z, approximate the corresponding relative error in the volume.
V = pi r^2 h = pi r^2 * 3r = 3 pi r^3
(a) If r becomes r + ∆r ,
V + ∆V = 3 pi (r + ∆r)^3
=3 pi [r^3+ 3r^2 ∆r + 3r(∆r)^2 + (∆r)^3]
V [1 +3(∆r)/r +3(∆r)^2/r^2 +(∆r)^3/r^3]
Since ∆r/r <<1, this linearizes to
V(1 + 3 ∆r/r)
So ∆V = 3V *(∆r/r)
(b) ∆V/V = 3z
To approximate the corresponding error in the volume of the can, we can use differentials or the linear approximation. Let's use differentials to find the error in the volume.
a) Differentials allow us to find approximate changes in a function based on small changes in its variables. In this case, we want to find the error in the volume (∆V) given the error in the radius (∆r).
The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height.
First, let's find the derivative of the volume function with respect to r to determine how the volume changes with a small change in r:
dV/dr = 2πrh
Next, we can use differentials to relate the changes in volume (∆V) and radius (∆r):
dV = (2πrh)dr
Now, we can substitute the given information that the height is precisely three times the radius, so h = 3r:
dV = (2πr * 3r)dr
= 6πr^2dr
To approximate the corresponding error in the volume (∆V), we multiply the differential equation by the magnitude of the radius error (∆r):
∆V = 6πr^2∆r
Now we can substitute the value of ∆r = ±0.03 cm:
∆V = 6πr^2 * 0.03
b) To find the corresponding relative error in the volume, we need to approximate the value using the relative error in the radius (∆r/r). Let's assume the relative error in the radius is z, so ∆r = zr.
∆V = 6πr^2 * zr
Now, let's find the relative error in the volume (∆V/V):
(∆V/V) = (6πr^2 * zr) / (πr^2 * 3r) (substituting h = 3r)
(∆V/V) = (6z) / 3
Simplifying, we get:
(∆V/V) = 2z
Therefore, the corresponding relative error in the volume is twice the relative error in the radius (z).