The radius of a 12 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.2 inch. What is the resulting possible error in the volume of the cylinder? Include units in your answer.
V = πr²h = 16πr²
dV/dr = 32πr
dV = 16πr dr
Let dV and dr represent the change (or error) in volume and radius. In this case, dr = ±0.1 inch, r = 4 inches, and we are asked for dV, the resulting possible error in the cylinder's volume.
dV = 16πr dr
dV = 16π(4) ±0.2
dV = ±40.2 inches³ ---> MY ANSWER
Hmmm. h is constant at 12, so
V = πr^2h = 12πr^2
dV = 24πr dr
...
is this the correct method though?
my answer would then be 30.16
To calculate the resulting possible error in the volume of the cylinder, we first need to find the derivative of the volume formula with respect to the radius. This will give us the rate at which the volume changes with respect to changes in the radius.
The volume formula for a right circular cylinder is V = πr²h, where r is the radius and h is the height.
Taking the derivative of V with respect to r, we get dV/dr = 2πr².
Next, we can multiply both sides of the equation by dr to find the change in volume (dV) for a given change in radius (dr):
dV = 2πr² dr.
Now let's substitute in the values given in the problem. The radius, r, is 4 inches, and the possible error in the radius, dr, is ±0.2 inches. Note that we divide the error by 2 because it accounts for both the possible increase and decrease in the radius.
dV = 2π(4)² ±0.2/2
= 2π(16) ±0.1
= 32π ±0.1.
So, the resulting possible error in the volume of the cylinder is ±32π ±0.1 inches³.