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.