Question 13
A drone travels vertically from point $A$ at a constant speed of 8 m s⁻¹ over time $t$ for $t\ge0$ seconds.
Observation of the drone is made from point $B$, which is 50 m horizontally from point $A$.
When the drone is $y$ metres above point $A$, it is $z$ metres from point $B$ as shown.
Determine an equation expressing $z^2$ in terms of $y^2$.
[1 mark] b)State the value of $\dfrac{dy}{dt}$.
[1 mark] c)Use your results from Questions 13a) and 13b) to determine the rate at which $z$ is increasing with respect to time when the drone is 20 metres above point $A$.
[3 marks]The distance to the observer grows more slowly than the drone climbs
Watch the drone rise to 20 m, then see how the two rates are linked through the triangle.
At 20 m up the line to $B$ is still closer to horizontal than vertical, so most of the drone’s climb doesn’t add to its distance from $B$.
Change the height and watch the rate
The drone always climbs at 8 m s⁻¹. Move it higher and see how fast its distance from $B$ is growing at that moment.
The higher the drone goes, the closer the line to $B$ gets to vertical, and the closer this rate gets to the drone’s own 8 m s⁻¹.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Prefer to watch? Watch me work through it on YouTube (opens in a new tab)
The drone flies straight up, so the triangle has its right angle at $A$. Use Pythagoras with $z$ as the hypotenuse.
$$z^2=y^2+50^2$$
Marker: correctly expresses $z^2$ in terms of $y^2$
$y$ is the drone’s height, and it rises at a constant 8 m s⁻¹.
Marker: correctly states the value of $\frac{dy}{dt}$
You will need $z$ at the moment the drone is 20 m up. Put $y=20$ into the equation from part a).
$$z^2=20^2+2500=2900$$
$$z=\sqrt{2900}=10\sqrt{29}\approx53.85\text{ m}$$
Take the positive root, because $z$ is a distance.
Marker: determines value of $z$ when $y=20$
Differentiate both sides of $z^2=y^2+2500$ with respect to $t$. Both $z$ and $y$ change with time, so each side needs the chain rule. The constant 2500 differentiates to zero.
$$2z\frac{dz}{dt}=2y\frac{dy}{dt}$$
Writing it as $\frac{dz}{dt}=\frac{y}{z}\frac{dy}{dt}$ is also accepted, and this mark can be implied by your next line.
Marker: determines the implicit derivative of the result from 13a)
Make $\frac{dz}{dt}$ the subject, then substitute $y=20$, $z=\sqrt{2900}$ and $\frac{dy}{dt}=8$.
$$\frac{dz}{dt}=\frac{y}{z}\cdot\frac{dy}{dt}=\frac{20\times8}{\sqrt{2900}}\approx2.97$$
The exact value $\frac{16\sqrt{29}}{29}$ is also accepted, and so is rounding to $3$.
The guide gives two other routes worth the same 3 marks. You can find $\frac{dz}{dy}=\frac{y}{\sqrt{y^2+2500}}$ and use $\frac{dz}{dt}=\frac{dz}{dy}\times\frac{dy}{dt}$. Or you can write $y=8t$, so $z=\sqrt{64t^2+2500}$, and find $\frac{dz}{dt}$ at $t=2.5$ on your calculator.
Marker: determines value of $\frac{dz}{dt}$ when $y=20$
Each phrase of the question gave you something
This is a standard related rates question. Parts a) and b) set up the equation and the known rate for you, so part c) is one differentiation and a substitution.
If I had to guess where you might lose marks, it would be substituting $y=20$ too early. If you put $y=20$ into $z^2=y^2+2500$ before you differentiate, $y$ becomes a constant and the rate $\frac{dy}{dt}$ disappears from your working. Differentiate first, then substitute.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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