5 Engineering Example 1

5.1 Plastic bottle design

Introduction

Manufacturing containers is a large and varied industry and optimum packaging can save companies millions of pounds. Although determining the capacity of a container and amount of material needed can be done by physical experiment, mathematical modelling provides a cost-effective and efficient means for the designer to experiment.

Problem in words

A manufacturer is designing a new plastic bottle to contain 900 ml of fabric softener. The bottle is circular in cross section, with a varying radius given by

$r=4+0.5z-0.07z^2+0.002z^3$

where z is the height above the base in cm.

1. Find an expression for the volume of the bottle and hence show that the fill level needs to be approximately 18 cm.
2. If the wall thickness of the plastic is 1 mm, show that this is always small compared to the bottle radius.
3. Hence, find the volume of plastic required to manufacture a bottle which is 20 cm tall (include the plastic in the base and side walls), using a numerical method.

A graph the radius against z is shown below:

Figure 10

Mathematical statement of problem

Calculate all lengths in centimetres.

1. The formula for the volume of a solid of revolution, revolved round the z axis between $z=0$ and $z=d$ is ${\int \; <!--nolimits\; -->}_0^d\pi r^{2}dz$ . We have to evaluate this integral.
2. To show that the thickness is small relative to the radius we need to find the minimum radius.
3. Given that the thickness is small compared with the radius, the volume can be taken to be the surface area times the thickness. Now the surface area of the base is easy to calculate being $\pi \times 4^2$ , but we also need to calculate the surface area for the sides, which is much harder.

   For an element of height $dz$ this is $2\pi z\times$ (the slant height) of the surface between $z$ and $z+dz$ . The slant height is, analytically $\left(\sqrt{1+{\left(\frac{dr}{dz}\right)}^2}\right)\times dz$ , or equivalently the distance between $\left(r\left(z\right),z\right)$ and $\left(r\left(z+dz\right),z+dz\right)$ , which is easier to use numerically.

Analytically the surface area to height $20$ is ${\int \; <!--nolimits\; -->}_0^{20}2\pi r\sqrt{1+{\left(\frac{dr}{dz}\right)}^2}dz$ ; we shall approximate this numerically. This will give the area of the side surface.

Mathematical analysis

1. We could calculate this integral exactly, as the volume is ${\int \; <!--nolimits\; -->}_0^d\pi {\left(4+0.5z-0.07z^2+0.002z^3\right)}^{2}dz$ but here we do this numerically (which can often be a simpler approach and possibly is so here). To do that we need to keep an eye on the likely error, and for this problem we shall ensure the error in the integrals is less than 1 ml. The formula for the error with the trapezium rule, with step $h$ and integrated from 0 to 20 (assuming from the problem that we shall not integrate over a larger range) is $\frac{20}{12}{h}^2\text{max}\left|f^{″}\right|$ . Doing this crudely with $f=\pi g^2$ where $g\left(z\right)=4+0.5z-0.07z^2+0.002z^3$ we see that

   $\left|g\left(z\right)\right|\le 4+10+28+16=58$ (using only positive signs and   $\left|z\right|\le 20$ )

   and $\left|g^{\prime }\left(z\right)\right|\le 0.5+0.14z+0.006z^2\le 0.5+2.8+2.4=5.7<6$ ,

   and $\left|g^{″}\left(z\right)\right|\le 0.14+0.012z\le 0.38$ .

   Therefore

   $f^{″}=2\pi \left(g{g}^{″}+{\left(g^{\prime }\right)}^2\right)\le 2\left(58\times 0.38+6^2\right)\pi <117\pi$ , so $\frac{20}{12}{h}^2$ max $\left|f^{″}\right|\le 613h^2$ .

   We need $h^2<1∕613$ , or $h<0.0403$ . We will use $h=0.02$ , and the error will be at most $0.25$ .

   The approximation to the integral from $0$ to $18$ is

   $\frac{1}{2}\pi g^2\left(0\right)0.02+{\sum }_{i=1}^{899}\pi g^2\left(0.02i\right)0.02+\frac{1}{2}\pi g^2\left(18\right)0.02$

   (recalling the multiplying factor is a half for the first and last entries in the trapezium rule). This yields a value of $899.72$ , which is certainly within $1$ ml of $900$ .

2. From the graph the minimum radius looks to be about $2$ at about $z=18$ . Looking more exactly (found by solving the quadratic to find the places where the derivative is zero, or by plotting the values and by inspection), the minimum is at $z=18.93$ , when $r=1.948$ cm. So the thickness is indeed small (always less than $0.06$ of the radius at all places.)
3. For the area of the side surface we shall calculate ${\int \; <!--nolimits\; -->}_0^{20}2\pi r\sqrt{1+{\left(\frac{dr}{dz}\right)}^2}dz$ numerically, using the trapezium rule with step $0.02$ as before. $\sqrt{1+{\left(\frac{dr}{dz}\right)}^2}dz=\sqrt{{\left(dz\right)}^2+{\left(dr\right)}^2}$ , which we shall approximate at point $z_n$ by $\sqrt{{\left(z_{n+1}-z_n\right)}^2+{\left(r_{n+1}-r_n\right)}^2}$ , so evaluating $r\left(z\right)$ at intervals of $0.02$ gives the approximation

   $\pi r\left(0\right)\sqrt{{\left(0.02\right)}^2+{\left(r\left(0.02\right)-r\left(0\right)\right)}^2}+{\sum }_{i=1}^{999}2\pi r\left(0.02i\right)\sqrt{{\left(0.02\right)}^2+{\left(r\left(0.02\left(i+1\right)\right)-r\left(0.02i\right)\right)}^2}$

   $+\pi r\left(20\right)\sqrt{{\left(0.02\right)}^2+{\left(r\left(20\right)-r\left(19.98\right)\right)}^2}$ .

   Calculating this gives $473{\text{cm}}^2$ . Approximating the analytical expression by a direct numerical calculation gives $474{\text{cm}}^2$ . (The answer is between $473.5$ and $473.6{\text{cm}}^2$ , so this variation is understandable and does not indicate an error.) The bottom surface area is $16\pi =50.3{\text{cm}}^2$ , so the total surface area we may take to be $474+50=524{\text{cm}}^2$ , and hence the volume of plastic is $524\times 0.1=52.4{\text{cm}}^3$ .

Mathematical comment

An alternative to using the trapezium rule is Simpson’s rule which will require many fewer steps.

When using a computer program such as Microsoft Excel having an efficient method may not be important for a small problem but could be significant when many calculations are needed or computational power is limited (such as if using a programmable calculator).

The reader is invited to repeat the calculations for (1) and (3) using Simpson’s rule.

The analytical answer to (1) is given by

${\int \; <!--nolimits\; -->}_0^{18}\pi \left(16+4z-0.31z^2-0.054z^3+0.0069z^4-0.00028z^5+0.000004z^6\right)dz$

which gives $899.7223$ to 4 d.p.

Exercises

1. Using 4 subintervals in the composite trapezium rule approximate

   ${\int \; <!--nolimits\; -->}_1^5\sqrt{x}dx.$

2. The function $f$ is known to have a second derivative with the property that

   $\left|f^{″}\left(x\right)\right|<12$

   for $x$ between 2 and 3. Using the error bound given earlier in this Section determine how many subintervals are required so that the composite trapezium rule used to approximate

   ${\int \; <!--nolimits\; -->}_2^{3}f\left(x\right)dx$

   can be guaranteed to have an error in it that is less than 0.001.

3. Using 4 subintervals in the composite Simpson rule approximate

   ${\int \; <!--nolimits\; -->}_1^5\sqrt{x}dx.$

4. The function $f$ is known to have a fourth derivative with the property that

   $\left|f^{\left(iv\right)}\left(x\right)\right|<6$

   for $x$ between $-1$ and 5 . Determine how many subintervals are required so that the composite Simpson’s rule used to approximate

   ${\int \; <!--nolimits\; -->}_{-1}^{5}f\left(x\right)dx$

   incurs an error that is less than 0.001.

5. Determine the minimum number of steps needed to guarantee an error not exceeding $\pm 0.000001$ when numerically evaluating

   ${\int \; <!--nolimits\; -->}_2^{4}ln\left(x\right)dx$

   using Simpson’s rule.

Answer