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 %.