Give complete explanations of what you are doing, written in full sentences.
Solutions that have all the correct calculations and computations, but lack explanations,
will not get full marks!
Exercise3.8.1
Using the guidelines of Stewart section 4.4 [3.10.1], sketch the graph of the curve \(\tan ^{-1} \left(\frac{x-1}{x+1}\right)\).
Solution
First thing to note is we have a problem at \(x=-1\):
\begin{equation*}\at{\left(\frac{x-1}{x+1}\right)}{x=-1} = \frac{-1 -1}{-1 + 1} = \frac{-2}{0}\end{equation*}
The sage calculations below calculate finite limits at \(x=-1\). The graph does not
shoot of to \(\pm\infty\) at this value.
Next see how the gradient behaves:
\begin{align*}
f(x)&=\tan ^{-1} \left(\frac{x-1}{x+1}\right)\\
\text{Let }u&=\left(\frac{x-1}{x+1}\right)\\
\frac{df}{dx}&=\frac{df}{du}\frac{du}{dx}\\
\frac{d}{du}\tan ^{ - 1} u&=\frac{1}{1 + u^2 }\\
\frac{du}{dx}&=\frac{2}{(x+1)^2} \text{(Using quotient rule)}\\
\therefore \frac{df}{dx}&=\frac{1}{1 + u^2 }.\frac{2}{(x+1)^2}\\
&=\frac{1}{1+{\left(\frac{x-1}{x+1}\right)}^2}.\frac{2}{(x+1)^2}\\
&=\frac{(x+1)^2}{(x+1)^2+(x-1)^2}.\frac{2}{(x+1)^2}\\
&=\frac{1}{x^2+1}
\end{align*}
Strange! This function seems to have a continous and well defined gradient at \(x=-1\).
The sage commands below plot the graph and gardient, and also calculates the limits as \(x \to -1\)
and \(x \to \pm\infty\)
Out of interest lets plot some variants on this:
Exercise3.8.2
The manager of a 100-unit apartment complex knows from experience that all units will be occupied
if the rent is £800 per month. A market survey suggests that, on average, one additional unit will
remain vacant for each £10 increase in rent. What rent should the manager charge to maximise profit?
Answer\(900\) Solution
Let \(r\) be the rent per unit and \(R\) be the total rental income. We want to maximise \(R\).
\begin{equation*}
\begin{cases}
R = 100r & r\leq 800 \\
R = \left[100 - \frac{r-800}{10}\right]r & r > 800
\end{cases}
\end{equation*}
Rearranging the second part of this piecewise equation gives us the following equation to maximise:
\begin{equation*}R = 180r-\frac{r^2}{10}\end{equation*}Differentiate gives:
\begin{equation*}0=180-\frac{r}{5}\end{equation*}Solving gives us the stationary point (maxima):
\begin{equation*}r=900\end{equation*}Double check that the second derivative is negative for a maximum:
\begin{equation*}f''=-1/5\end{equation*}In this case this is easy to see because our equation is an upside down parabola.
Since this is a discrete problem and we solved using a continous derivative it is always
wise to check with some values:
\begin{align*}
\text{ Rent } = 880 \leadsto&R=880 \times 92 = 80960\\
\text{ Rent } = 890 \leadsto&R=890 \times 91 = 80990\\
\text{ Rent } = 900 \leadsto&R=900 \times 90 = 81000\\
\text{ Rent } = 910 \leadsto&R=910 \times 89 = 80990\\
\text{ Rent } = 920 \leadsto&R=920 \times 88 = 80960
\end{align*}
Exercise3.8.3
Two balls are thrown upward from the edge of a cliff \(140\) meters above the ground.
The first is thrown with a speed of \(15 m/s\), the other is thrown a second later with
a speed of \(8 m/s\). Do the balls ever pass each other?
Answer
The balls crossover below the cliff edge at the time \(4.33\) seconds and
at a distance of \(28.89\) metres below the cliff edge.
Solution
The acceleration of the ball is a downward (negative) force due to
the gravitational force on the earth's surface. This is commonly known as \(g\)
(units of \(m/s/s\) or \(ms^{-2}\). Normally distance is represented by \(s\).
The acceleration is the second derivative of distance. Speed is the first derivative.
\begin{align*}
\frac{d^2 s}{dt^2}&=a\\
\frac{ds}{dt}&=at+u_0\\
s&=\frac{at^2}{2}+u_0 t + s_0
\end{align*}
(where \(u_0\) is an integration constant representing the initial speed
and \(s_0\) is an integration constant representing the initial position.)
This is normally written as
\begin{equation*}s(t)=u_0t + \frac{1}{2}at^2\end{equation*}
For both balls \(s_0=0\) since we are free to set the zero of our
vertical axis to the height of the cliff edge.
For the first ball \(u_0=15\) and for the second ball \(u_0=8\).
Hence we have
\begin{equation*}s_1(t)=15t-\frac{1}{2}gt^2\end{equation*}
\begin{equation*}s_2(t)=8t-\frac{1}{2}gt^2\end{equation*}
\(g \approx 10\) hence
\begin{equation*}s_1(t)=15t-5t^2\end{equation*}
\begin{equation*}s_2(t)=8t-5t^2\end{equation*}
Since ball two is one second behind the first we can represent this as
\begin{equation*}s_1(t)=15t-5t^2\end{equation*}
\begin{equation*}s_2(t-1)=8(t-1)-5(t-1)^2\end{equation*}
These can be set equal to each other and solved
\begin{align*}
15t-5t^2&=8(t-1)-5(t-1)^2\\
15t-5t^2&=8t-8-5(t^2-2t+1)\\
15t-5t^2&=8t-8-5t^2+10t-5\\
15t&=18t-13\\
t&=\frac{13}{3}
\end{align*}
In other words there is a solution and the balls cross over.
See the sage plots below for a visual display of what is happening
Exercise3.8.4
At 2:00 PM a car’s speedometer reads \(50 km/h\). At 2:10 PM it reads
\(50 km/h\). Show that at some time between 2:00 and 2:10 the
acceleration is exactly \(90 km h^{-2}\).
Solution
All we can say about the acceleration in the interval is that its average has taken us from
\(50 km/h\) to \(65 km/h\) in a time of \(10\) minutes.
The only thing we have to be careful with is to keep the units consistent
\(\frac{km/hr}{hr} = km hr^{-2}\).
\begin{align*}
acc_{av} &=\frac{\text{change in speed}}{\text{change in time}}\\
acc_{av} &=\frac{65 - 50 \text{ km/hr}}{\frac{10}{60}\text{ hr}}\\
&=90 km h^{-2}\text{ (km per hour per hour)}
\end{align*}
But to show this mathematically you probably have to give a bigger explanation!
