Problem Solving with or without calculus

Thursday 12th November 2020

As part of a recent Network Meeting we looked at the question below which could be used with GCSE students (without calculus), or A level students (with calculus).

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We looked at what assumptions students would need to make and what difficulties they may have, and briefly looked at the sourceOpens a new window of the question. Here are just two possible solutions:

Without calculus:

Let \(P\) be point \((x,y)\) and greatest distance from origin be \(D\).

Look for repeated root \(x^2+y^2=D^2\) and \(3x^2+y^2+15x=18\)

Solve simultaneously: \(2x^2+15x+D^2−18=0\)

Use \(b^2−4ac=0\)

\(15^2−(4)(2)(D^2−18)=0\)

\(225+144=8D^2\)

\(D^2=\frac{369}{8}\)

\(D=\frac{3\sqrt82}{4}\)

With calculus:

\(D^2=x^2+y^2\) and \(3x^2+y^2+15x=18\)

So \(D^2=18−15x−2x^2\)

Let \(D^2=L\)

\(\frac{dL}{dx}=−15−4x\) with stationary point when \(\frac{dL}{dx}=0, x=−\frac{15}{4}\)

\(\frac{d^2L}{dx^2}=-4\)\(<\)\(0\) so maximum at \(x=-\frac{15}{4}\)

When \(x=−\frac{15}{4}, y=\bar{+}\frac{3\sqrt57}{4}\)

\(D^2=(−\frac{15}{4})^2+(\frac{3\sqrt57}{4})^2\)

\(D=\frac{3\sqrt82}{4}\)

If you would be interested in attending a similar session, you can find out more information and apply for our next online Teacher Network meeting on the AMSP website.

By Jean Smith

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