Functions of several variables

Total Approximation Differentiation

Question 1.

Assume that the approval rating of a Prime Minister is given by the function $A(d,e)$, where $d$ is defence spending (in billions) and $e$ is education spending (in billions). The output of the approval rating $A(d,e)$ itself is a percentage between $0$ and $100$.

It is desirable to predict how changes to defence and education spending impact upon the PM's approval. With current spending at $d_0$ and $e_0$, the rate that approval (in percentage) changes with respect to defence spending (in billions) is measured by Newspoll to be the partial derivative

$$\frac{\partial A}{\partial d}(d_0,e_0)=6.9,$$

so an increase in defence spending of $1$ billion dollars will translate to an increase in approval of $6.9\%$.
Similarly, the rate that approval changes with respect to education spending is measured to be the partial derivative

$$\frac{\partial A}{\partial e}(d_0,e_0)=\mathrm{5.8.}$$

Hence by the total differential approximation, for $[d,e]$ in the neighbourhood of $[d_0,e_0]$,

$$A(d,e)\approx A(d_0,e_0)+\frac{\partial A}{\partial d}(d_0,e_0)(d-d_0)+\frac{\partial A}{\partial e}(d_0,e_0)(e-e_0).$$

The current approval rating is

$$A(d_0,e_0)=51.$$

If defence spending is decreased by $0.8$ billion and education spending increased by $0.4$ billion, then the approval rating approximately changes to

$$\begin{array}{rl}A(d,e)& \approx 51+6.9(-0.8)+5.8(0.4)\\ & =51-5.52+2.32\\ & =\mathrm{47.8.}\end{array}$$

Therefore, the approval rating is approximately $47.8\%$.

Question 2.

The volume of toilet paper on the roll is thus given by the formula:

$$V(\ell ,r,R)=\pi \ell (R^2-r^2).$$

According to your measurements, the volume of toilet paper is (to the nearest cubic mm):

$$V(98,25,56)=\text{[ 773077 ]}{\text{ mm}}^3.$$

However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different.

The total differential approximation can be used to estimate how the measured volume $V(98,25,56)$ differs from the true value of $V(\ell ,r,R)$ as:

$$V(\ell ,r,R)\approx V(98,25,56)+\frac{\partial V}{\partial \ell }(98,25,56)(\ell -98)+\frac{\partial V}{\partial r}(98,25,56)(r-25)+\frac{\partial V}{\partial R}(98,25,56)(R-56).$$

We calculate the partial derivatives (to the nearest integer):

• $\frac{\partial V}{\partial \ell }(98,25,56)=\text{[ 7889 ]}$

• $\frac{\partial V}{\partial r}(98,25,56)=\text{[ -15394 ]}$

• $\frac{\partial V}{\partial R}(98,25,56)=\text{[ 34482 ]}$

Since we measured $\ell ,r$ and $R$ to the nearest mm,

$$|\ell -98|<0.5,|r-25|<0.5\text{and}|R-56|<\mathrm{0.5.}$$

Using the integer approximations above together with the triangle inequality,

$$|V(\ell ,r,R)-V(98,25,56)|\le \left|\frac{\partial V}{\partial \ell }(98,25,56)\right||\ell -98|+\left|\frac{\partial V}{\partial r}(98,25,56)\right||r-25|+\left|\frac{\partial V}{\partial R}(98,25,56)\right||R-56|.$$

And so our measured volume $V(98,25,56)$ differs from the true volume by

$$|V(\ell ,r,R)-V(98,25,56)|\le \text{[ 28882.5 ]}{\text{ mm}}^3.$$

To two decimal places, this represents an approximate error of no more than

$$\frac{|V(\ell ,r,R)-V(98,25,56)|}{V(98,25,56)}\le \text{3.74\%}.$$