Tutorial : Determining Approximate Values Problem 1

Tutorial — By Christopher on February 19, 2009 at 12:00 AM

A cylinder has a fixed height of 12 cm and a radius of 5cm. If the radius shrinks by 0.05cm, find
(a) the approximate change in the volume of the cylinder.
(b) the final volume of the cylinder.

Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. pg 214 Practice 21 Question 3

Recall that y_new = y_initial + δy = y_initial + (dy/dx)δx.
Note that δy/δx = dy/dx.
Hence, δy = (dy/dx)δx.

To solve part the problem, we have to recall the formula to determine the volume of a cylinder.
The volume of a cylinder is given by πr²h, where r is the radius and h is the height.

(a) the approximate change in the volume of the cylinder

We know that V (Volume of cylinder) is defined by πr²h.
Find dV/dr.
Recall that dy/dx = anx^n-1.

dV/dr = 2πr^{2 – 1}h, whereby a = πh and n = 2 (Note that h is a constant!)
Simplify and we get dV/dr = 2πrh.

Given r = 5cm and h= 12cm,
dV/dr = 2π(5)(12)
= 120π cm² (Note the units!)

To find the approximate change, we use the formula δy/δx = dy/dx.
We can rewrite the equation to represent the problem.
δV/δr = dV/dr.

The problem requires us to find the approximate change in volume of the cylinder.
Therefore, we shift δr to the right hand side.

δV = (dV/dr) δr

We are given that δr is -0.05 cm (because the radius shrinks by 0.05 cm).
We have calculated earlier that dV/dr = 120π cm².

Replacing the values into δV = (dV/dr) δr,
δV = (120π cm²) (-0.05 cm)
δV = -6π cm³

Therefore, the approximate change in volume of the cylinder is -6π cm³.

(b) the final volume of the cylinder

Recall that y_new = y_initial + (dy/dx)δx.
We can rewrite it to represent the problem.
V_new = V_initial + (dV/dr)δr.