Question 16
A curve is defined by the parametric equations $x=2\tan(\theta)$ and $y=3\sin(2\theta)$, where $0\le\theta\lt\dfrac{\pi}{2}$.
Given that $\dfrac{dy}{dx}$ can be expressed in the form $a\cos^4(\theta)+b\cos^2(\theta)$, where $a,b\in R$, determine the values of $a$ and $b$.
One angle drives both coordinates
As the angle grows, the point traces the curve. The orange tangent rides along with it, and its gradient comes from two rates.
Each coordinate is given as a function of $\theta$, so the gradient is a ratio of two rates taken with respect to $\theta$. The triangle is drawn at three tenths of the true rates so it fits.
Slide along the curve, then choose the identity
Move the slider to change the angle. The readouts divide the two rates and check the answer against the required form at every point.
At the origin the curve leaves at its steepest. Both forms give a gradient of $3$, since $6-3=3$ when $\cos(0)=1$.
Here $2\theta$ is less than $\frac{\pi}{2}$, so $\cos(2\theta)$ is positive and the curve is rising. The last two lines agree because they are the same expression written two ways.
This is the top of the curve. $\cos\left(\frac{\pi}{2}\right)=0$ makes the rise rate zero, and the form agrees: $6\left(\frac14\right)-3\left(\frac12\right)=0$.
Past the top, $\cos(2\theta)$ is negative and the curve falls. It falls most steeply at $\theta=\frac{\pi}{3}$, where the gradient is $-\frac38$. After that $\frac{dx}{d\theta}=2\sec^2(\theta)$ grows without bound while $\frac{dy}{d\theta}$ stays between $-6$ and $0$, so the gradient climbs back towards zero as $\theta$ nears $\frac{\pi}{2}$.
Which double angle identity gets you to the form?
After dividing, you have $\frac{dy}{dx}=3\cos(2\theta)\cos^2(\theta)$. There are three versions of $\cos(2\theta)$. Pick one and see what it gives you.
Look at the required form before you choose. It has only cosines in it.
This version is true, but it leaves a sine term behind, and the form has no room for one. You would have to replace $\sin^2(\theta)$ with $1-\cos^2(\theta)$ and collect terms, which costs time and invites a slip.
This is the version the form asks for. Expanding gives cosines only, so you can read off $a=6$ and $b=-3$.
This version is also true, but it brings in a sine term. You would have to rewrite it with $\sin^2(\theta)=1-\cos^2(\theta)$ and collect terms before the expression matches the form.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA sample response.
Both coordinates are given in terms of $\theta$, so start by differentiating each one with respect to $\theta$. The derivative of $\tan(\theta)$ is $\sec^2(\theta)$.
$$\begin{aligned}x&=2\tan(\theta)\\\frac{dx}{d\theta}&=2\sec^2(\theta)\end{aligned}$$
Equivalent forms are accepted, such as $\frac{dx}{d\theta}=\frac{2}{\cos^2(\theta)}$. This mark can be implied by your later working.
Marker: correctly determines $\frac{dx}{d\theta}$
Use the chain rule on $\sin(2\theta)$. The inside function $2\theta$ has derivative $2$.
$$\begin{aligned}y&=3\sin(2\theta)\\\frac{dy}{d\theta}&=3\times2\cos(2\theta)\\&=6\cos(2\theta)\end{aligned}$$
This mark can be implied by your later working.
Marker: correctly determines $\frac{dy}{d\theta}$
Divide the rate for $y$ by the rate for $x$. This is the chain rule written as a quotient.
$$\frac{dy}{dx}=\frac{\ \frac{dy}{d\theta}\ }{\frac{dx}{d\theta}}\quad\text{or}\quad\frac{dy}{d\theta}\times\frac{d\theta}{dx}$$
$$\frac{dy}{dx}=\frac{6\cos(2\theta)}{2\sec^2(\theta)}$$
Equivalent forms are accepted, such as $\frac{dy}{dx}=\frac{6\cos(2\theta)\cos^2(\theta)}{2}$. If one of your derivatives was wrong, you can still earn this mark by follow-through.
Marker: determines an expression for $\frac{dy}{dx}$
Dividing by $\sec^2(\theta)$ is the same as multiplying by $\cos^2(\theta)$. The required form has only cosines, so use the version of the double angle identity written in cosines.
$$\cos(2\theta)=2\cos^2(\theta)-1$$
$$\begin{aligned}\frac{dy}{dx}&=3\cos(2\theta)\cos^2(\theta)\\&=3\left(2\cos^2(\theta)-1\right)\cos^2(\theta)\\&=6\cos^4(\theta)-3\cos^2(\theta)\end{aligned}$$
Marker: determines an expression for $\frac{dy}{dx}$ in the required form
Match your expression with $a\cos^4(\theta)+b\cos^2(\theta)$. The coefficient of $\cos^4(\theta)$ is $a$ and the coefficient of $\cos^2(\theta)$ is $b$.
Write the values out at the end, even though they are visible in your last line. This mark is for stating them.
Marker: states the values of $a$ and $b$
Each phrase of the question gave you something
The question gives you the destination and nothing about the route. You have to decide to differentiate parametrically, then see that dividing by $\sec^2(\theta)$ turns into a factor of $\cos^2(\theta)$, then pick the one version of the double angle identity that lands on the form. Each of those is a skill you have used before, but here you choose and join them yourself. It stays familiar because the wording still points at a practised method. Parametric equations plus $\frac{dy}{dx}$ point straight at dividing one rate by the other, and the target form is printed for you to check against.
If I had to guess where you might lose marks, it would be the fourth mark. The marking guide gives it only for an expression in the required form, so stopping at $3\cos(2\theta)\cos^2(\theta)$ leaves it behind. Reaching for $\cos^2(\theta)-\sin^2(\theta)$ out of habit leaves a sine term you then have to remove. The first two marks can be implied by your later working, and the third allows follow-through, but a lost factor of $2$ from the chain rule on $\sin(2\theta)$ will change your values of $a$ and $b$.
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 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it