that it has a radius of length 5 mm with an increase of 0.01 mm, where V(r) = πr^3 (the volume
of a sphere of radius r) with dr = 0.01. We find the derivative of the volume: V′(r) = π(3r^2 ) =
4 πr^2 . Now we can plug into the formula: dV = 4πr^2 dr. If we plug in r = 5 and dr = 0.01, we
get dV = 4π(5)^2 (0.01) = π ≈ 3.142.
- (a) 1.963 m^3 ; (b) 15.708 m^3
(a) Recall the formula that we use when we want to approximate the error in a measurement:
dy = f′(x) dx. Here we want to approximate the error in the volume of a cylinder when we
know that it has a diameter of length 5 m (which means that its radius is 2.5 m) and its height is
20 m, with an error in the height of 0.1 m, where V = πr^2 h with dh = 0.1. Note that the radius is
exact, so when we take the derivative we will treat only the height as a variable. We find the
derivative of the volume: V′ = πr^2 . Now we can plug into the formula: dV = πr^2 dh. If we plug
in r = 2.5, h = 20, and dh = 0.1, we get dV = π(2.5)^2 (0.1) = 0.625π ≈ 1.963.
(b) Here we want to approximate the error in the volume of a cylinder when we know that it
has a diameter of length 5 m (which means that its radius is 2.5 m) and its height is 20 m, with
an error in the diameter of 0.1 m (which means that the error in the radius is 0.05 m), where V
= πr^2 h with dr = 0.05. Note that the height is exact, so when we take the derivative we will
treat only the radius as a variable. We find the derivative of the volume: V′ = 2πrh. Now we
can plug into the formula: dV = 2πrh dr. If we plug in r = 2.5, h = 20, and dr = 0.05, we get dV
= 2π(2.5)(20)(0.05) = 5π ≈ 15.708.
