Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.63 of these notes.

## Question 1

Use differentials to estimate the amount of tin in a closed tin closed tin can with diameter 8 cm and height 12 cm if the can is 0.04 cm thick.

### Solution

Assuming that the measurements of 8 cm and 12 cm are taken from the outside of the can, then we could estimate the change in volume of a cylinder if the radius were increased by 0.04 cm to 4 cm and the height increased by 0.08 cm to 12 cm (convince yourself of this with a picture.). Now the tin in the can comprises the difference between a $(r,h)=(3.96,11.92)$ cylinder and a $(r,h)=(3,12)$ cylinder.

Now the volume of a cylinder is given by $V=\pi r^2h$.

We can use the differential of $V$, $dV$ (evaluated at $(r,h)=(3.96,11.92)$although the other way around would  also be a good estimate) to estimate the change in volume: $dv=\frac{\partial V}{\partial r}dr+\frac{\partial V}{\partial h}dh$,

where $dr=+0.04$ cm and $dh=+0.08$ cm. Now $\left.\frac{\partial V}{\partial r}=2\pi r h\right|_{(r,h)=(3.96,11.92)}=2\pi (3.96)(11.92)$, and $\left.\frac{\partial V}{\partial h}=\pi r^2\right|_{(r,h)=(3.96,11.92)}=\pi (3.96)^2$. $\Rightarrow dP=2\pi (3.96)(11.92)\times(+0.04)+\pi (3.96^2)\times (+0.8)\approx15.805$ cm ${}^3$.

## Question 2

Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm is diameter if the metal in the wall is 0.05 cm thick and the metal in the top and bottom is 0.1 cm thick.

### Solution

Using the same method, the differential is a good estimate: $dv=\frac{\partial V}{\partial r}dr+\frac{\partial V}{\partial h}dh$,

where in this case we take $(r,h)=(1.95,9.8)$, and $dh=+0.2$ and $dr=0.05$. Now $\left.\frac{\partial V}{\partial h}=\pi r^2\right|_{(r,h)=(1.95,9.8)}=\pi (1.95)^2$, $\left.\frac{\partial V}{\partial r}=2\pi r h\right|_{(r,h)=(1.95,9.8)}=2\pi(1.95)(9.8)$.

Hence $dP=\pi(1.95^2)\times(+0.2)+2\pi(1.95)(9.8)\times(0.05)\approx 8.393$ cm ${}^3$.

## Question 3

If $R$ is the total resistance of three resistors, connected in parallel, with the resistances $R_1,R_2$ and $R_3$, then $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.

If the resistances are measured as $R_1=25$ $\Omega$, $R_2=40$ $\Omega$ and $R_3=50$ $\Omega$, with possible errors of 5 % in each case, estimate the maximum error in the calculated value of $R$.

### Solution

Now first we want to get an expression for $R(R_1,R_2,R_3)$: $\frac{1}{R}=\frac{1}{R_1}\cdot\frac{R_2R_3}{R_2R_3}+\frac{1}{R_2}\cdot\frac{R_1R_3}{R_1R_3}+\frac{1}{R_3}\cdot\frac{R_1R_2}{R_1R_2}$, $\Rightarrow R(R_1,R_2,R_3)=\frac{R_1R_2R_3}{R_1R_2+R_1R_3+R_2R_3}.$

We can approximate the error in $R$, $\Delta R$ by: $\Delta R\approx dR=\left|\frac{\partial R}{\partial R_1}\right|dR_1+\left|\frac{\partial R}{\partial R_2}\right|dR_2+\left|\frac{\partial R}{\partial R_3}\right|dR_3$,

where $dR_i$ corresponds to the error in $R_i$ $\Delta R_i$.

Now the errors are 5 % hence: $dR_1=(0.05)(25)=1.25.$ $dR_2=(0.05)(40)=2.$ $dR_3=(0.05)(50)=2.5.$

Also, using the quotient rule: $\frac{\partial R}{\partial R_1}=\frac{(R_1R_2+R_1R_3+R_2R_3)(R_2R_3)-(R_1R_2R_3)(R_2+R_3)}{[R_1R_2+R_1R_3+R_2R_3]^2}.$ $=\frac{R_1R_2^2R_3+R_1R_2R_3^2+R_2^2R_3^2-R_1R_2^2R_3-R_1R_2R_3^2}{[R_1R_2+R_2R_3+R_1R_3]^2}.$

Similarly, $\frac{\partial R}{\partial R_2}=\frac{R_1^2R_3^2}{[R_1R_2+R_1R_3+R_2R_3]^2}$. $\frac{\partial R}{\partial R_3}=\frac{R_1^2R_2^2}{[R_1R_2+R_1R_3+R_2R_3]^2}$.

Note that these will be evaluated at $(R_1,R_2,R_3)=(25,40,50)$, using a calculator; $\frac{\partial R}{\partial R_1}=\frac{(40)^2(50)^2}{[(25)(40)+(25)(50)+(40)(50)]^2}=\frac{64}{289}.$

Similarly, $\frac{\partial R}{\partial R_2}=\frac{25}{289}.$ $\frac{\partial R}{\partial R_3}=\frac{16}{289}.$

Hence, $\Delta R\approx \frac{64}{289}\times(1.25)+\frac{25}{289}\times(2)+\frac{16}{289}\times(2.5)\approx 0.588\,\Omega$.

## Question 4

The moment of inertia of a body about an axis is given by $I=kbD^3$ where $k$ is a constant and $B$ and $D$ are the dimensions of the body. If $B$ and $D$ are measured as 2 m and 0.8 m respectively, and the measurement errors are 10 cm in $B$ and $8$ mm in $D$, determine the error in the calculated value of the moment of inertia using the measured values, in terms of $k$.

### Solution

In class we did an example where we estimated the change in $I$ due to changes in $B$ and $D$ (from the sample test). The only difference between that example and this one is that errors are always positive and we take absolute values; i.e: $\Delta I=\left|\frac{\partial I}{\partial B}\right|\Delta B+\left|\frac{\partial I}{\partial D}\right|\Delta D$.

## Question 5

The volume, $V$, of a liquid of viscosity coefficient $\eta$ delivered after a time $t$ when passed through a tube of length $l$ and diameter $d$ by a pressure $P$ is given by $V=\frac{Pd^4t}{128\eta l}.$

If the errors in $V$, $P$ and $l$ are 1 %, 2 % and 3 % respectively, determine the error in $\eta$. HINT: If the error in $A$ is $x$ % then the error is $x A_0/100$ when $A=A_0$.

### Solution

First we solve for a function $\eta(V,P,l,d,t)$: $\eta=\frac{Pd^4t}{128Vl}=\frac{d^4t}{128}PV^{-1}l^{-1}.$

Again we use the differential to estimate the error (as errors in $d$ and $t$ were not mentioned we will assume they don’t have errors): $\Delta \eta\approx \left|\frac{\partial \eta}{\partial V}\right|\Delta V+\left|\frac{\partial \eta}{\partial P}\right|\Delta P+\left|\frac{\partial \eta}{\partial l}\right|\Delta l$

Now we must look at the partial derivatives (verify the last steps yourself): $\left|\frac{\partial \eta}{\partial V}\right|=\left|-\frac{d^4t}{128}PV^{-2}l^{-1}\right|=\frac{\eta}{V}$. $\left|\frac{\partial \eta}{\partial P}\right|=\left|\frac{d^4t}{128}V^{-1}l^{-1}\right|=\frac{\eta}{P}$. $\left|\frac{\partial \eta}{\partial l}\right|=\left|\frac{d^4t}{128}PV^{-1}l^{-2}\right|=\frac{\eta}{l}.$

Now by the hint, the errors in $V,P$ and $l$: $\Delta V=\frac{V}{100}$, $\Delta P=\frac{2P}{100}$, and $\Delta l=\frac{3l}{100}$.

Hence, $\Delta \eta=\frac{\eta }{V}\times \frac{V}{100}+\frac{\eta }{P}\times \frac{2P}{100}+\frac{\eta}{l}\times \frac{3l}{100}$ $\Rightarrow\Delta \eta =\frac{6}{100}\eta$.

That is the error in $\eta$ is 6 %.