Two cyclists are at the same road intersection. One cyclist travels north at \(20\,{\text{km}}\,{{\text{h}}^{ - 1}}\). The other cyclist travels west at \(15\,{\text{km}}\,{{\text{h}}^{ - 1}}\).

Use calculus to show that the rate at which the distance between the two cyclists changes is independent of time.

METHOD 1

attempt to set up (diagram, vectors)     (M1)

correct distances \(x = 15t,{\text{ }}y = 20t\)     (A1) (A1)

the distance between the two cyclists at time \(t\) is \(s = \sqrt {{{(15t)}^2} + {{(20t)}^2}}  = 25t{\text{ (km)}}\)     A1

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = 25{\text{ (km}}\,{{\text{h}}^{ - 1}})\)     A1

hence the rate is independent of time     AG

METHOD 2

attempting to differentiate \({x^2} + {y^2} = {s^2}\) implicitly     (M1)

\(2x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 2y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 2s\frac{{{\text{d}}s}}{{{\text{d}}t}}\)     (A1)

the distance between the two cyclists at time \(t\) is \(\sqrt {{{(15t)}^2} + {{(20t)}^2}}  = 25t{\text{ (km)}}\)     (A1)

\(2(15t)(15 + 2(20t)(20) = 2(25t)\frac{{{\text{d}}s}}{{{\text{d}}t}}\)     M1

Note:     Award M1 for substitution of correct values into their equation involving \(\frac{{{\text{d}}s}}{{{\text{d}}t}}\).

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = 25{\text{ (km}}\,{{\text{h}}^{ - 1}})\)     A1

hence the rate is independent of time     AG

METHOD 3

\(s = \sqrt {{x^2} + {y^2}} \)     (A1)

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = \frac{{x\frac{{{\text{d}}x}}{{{\text{d}}t}} + y\frac{{{\text{d}}y}}{{{\text{d}}t}}}}{{\sqrt {{x^2} + {y^2}} }}\)     (M1)(A1)

Note:     Award M1 for attempting to differentiate the expression for \(s\).

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = \frac{{(15t)(15) + (20t)(20)}}{{\sqrt {{{(15t)}^2} + {{(20t)}^2}} }}\)     M1

Note:     Award M1 for substitution of correct values into their \(\frac{{{\text{d}}s}}{{{\text{d}}t}}\).

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = 25{\text{ (km}}\,{{\text{h}}^{ - 1}})\)     A1

hence the rate is independent of time     AG

[5 marks]

Reasonably well done. Most successful candidates determined that \(s = 25t \Rightarrow \frac{{{\text{d}}s}}{{{\text{d}}t}} = 25\) from \(x = 15t\) and \(y = 20t\). A number of candidates did not use calculus while a few candidates correctly used implicit differentiation.