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We had a number of correct answers to
this problem. Several solutions used a spreadsheet, for example the
one shown below is from an anonymous solver. We also received a
solution from someone who has done some calculus and was able to
solve the problem using this approach. I have included this
solution at the end for those of you who are interested.
My disappointment is that some of you are
still just sending in answers without explanation and I do want to
see how you arrived at your answers. Although the solution below
was found using a spreadsheet you might have used a calculator and
carried out a standard "trial and improvement" process to obtain
the same result.
Catherine, Ellean, Anna from The Mount School,
York add the following comment which means they were really
thinking about the numbers and the context...
"P.S she must be super woman to run and swim that fast."
How fast do you think you could run and swim in m/s?
Andisheh of Springfield School offered
the following explanation (well done):
First of all I tried using trial and error. I thought the more
she runs the quicker she would get thre but that did not work. I
used Pythagoras to work out the distances swum and run. I tried
drawing a table but that did not help a lot. Idrew a graph and trid
to find out the lowest point but I was a bit inaccurate. Then I
used Microsoft Excel and typed in the formula and tried it with
numbers from 0 to 100 and tried to find our the shortest time and
this worked.
Andisheh attached a spreadsheet and I
have used this and several other solutions employing a spreadhsheet
to create the extract below: Solvers were able to find that the
minimum time occurred with a distance between 76 and 77m and then
worked with one and then two decimal places to obtain greater
accuruacy.
"Swim Distance" was calculated using Pythagoras' theorem:
The right-angled triangle will have sides of 50m (the distance from
the beach) and $x$m, which is 100m (the distance along the beach)
minus the distance run.
$$\mbox{Swim}^2 = 50^2 + x^2 $$
$$\mbox{Distance swum} = \sqrt{50^2 + x^2}$$
The times were then found in seconds by dividing the distances by
the speeds and the total time just required the two times to be
added.
The final solution was sent in by an anonymous
solver who used calculus (a branch of mathematics normally met for
the first time at advanced level in the UK). It is a neat solution
but this high level mathematics was certainly not required!
Let $T$ be the total time.
Let $t_1$ be the time spent swimming.
Let $t_2$ be the time spent running.
By Pythagoras' theorem
\begin{eqnarray} PC^2 &=& 50^2 + x^2 \\ PC&=&
\sqrt{50^2 + x^2} \end{eqnarray}
So for the total time:
$$ T = \frac{\sqrt{x^2 + 50^2}}{3} + \frac{100-x}{7}$$
$$\frac{dT}{dx} = \frac{x(x^2 + 50^2)^\frac{-1}{2}}{3}-
\frac{1}{7}$$
$$\frac{dT}{dx} = 0 \mbox{ when } \frac{1}{7} = \frac{x(x^2 +
50^2)^\frac{-1}{2}}{3}$$ Important question to think about:
How would you check that this is a minimum value?
Solving the equation above for $x$
\begin{eqnarray} 3(x^2+50^2)^\frac{1}{2} &=& 7x\\
9(x^2+50^2)&=& 49x^2\\ x&=& 23.7 \mbox{ (1dp)}
\end{eqnarray}
Chris will need to land at the point on the beach 76.3m from the
family.