Q.Find an angle $\theta$, $0 < \theta < \dfrac{\pi}{2}$, which increases twice as fast as its sine.

The key idea is the Increasing Function Test: if two quantities are changing with respect to the same variable, their rates of change are related by derivatives.

We are told that $\theta$ increases twice as fast as $\sin \theta$. This means the rate of change of $\theta$ with respect to time is double the rate of change of $\sin \theta$:

Differentiate $\sin \theta$ using the chain rule:

$\frac{d\theta}{dt} = 2 \left( \cos \theta \cdot \frac{d\theta}{dt} \right)$ …

The key idea here is that "increases twice as fast" is a statement about rates of change with respect to time. When we say one quantity increases twice as fast as another, we mean their derivatives with respect to time are in the ratio 2:1.

Let’s unpack that. If $\theta$ and $\sin\theta$ are both changing as time passes, then:

• The rate at which $\theta$ increases is $\frac{d\theta}{dt}$.
• The rate at which $\sin\theta$ increases is $\frac{d}{dt}(\sin\theta) = \cos\theta \cdot \frac{d\theta}{dt}$ (by the chain rule).

The condition “$\theta$ increases twice as fast as its sine” means:

Now substitute the derivative of $\sin\theta$:

Assuming $\frac{d\theta}{dt} \neq 0$ (the angle is actually changing), we can divide both sides by $\frac{d\theta}{dt}$:

So:

Within the interval $0 < \theta < \frac{\pi}{2}$, the angle whose cosine is $\frac{1}{2}$ is: …

Method: Comparing the Rate of a Variable to the Rate of a Function of It

Some related-rates problems never mention time explicitly — instead they compare how fast a quantity itself changes to how fast some function of that quantity changes (e.g. "$\theta$ increases twice as fast as $\sin\theta$"). The trick is to translate the words into an equation of time-derivatives and let $t$ cancel out.

Steps

Step 1: Translate the comparison into an equation of rates.

"$P$ increases $n$ times as fast as $Q=g(P)$" becomes

Step 2: Differentiate $Q=g(P)$ using the chain rule.

Step 3: Substitute and cancel $\frac{dP}{dt}$.

$\frac{dP}{dt}=n\,g'(P)\,\frac{dP}{dt}.$ …

Common Mistakes

Mistake 1: Dropping the chain-rule factor when differentiating $\sin\theta$ with respect to time

Why it's wrong: writing $\frac{d}{dt}(\sin\theta)=\cos\theta$ (missing the $\frac{d\theta}{dt}$ factor) treats $\theta$ as the independent variable of differentiation instead of a function of $t$, which breaks the whole "twice as fast" comparison of rates. Correct approach: $\frac{d}{dt}(\sin\theta)=\cos\theta\cdot\frac{d\theta}{dt}$, always.

Mistake 2: Dividing both sides by $\frac{d\theta}{dt}$ without justifying it's nonzero

Why it's wrong: the cancellation $1=2\cos\theta$ is only valid because $\theta$ is actually increasing ($\frac{d\theta}{dt}\neq0$); skipping this check (even though it's true here) is a rigor gap examiners look for in a "prove/find" style question. Correct approach: explicitly state the angle is increasing, so $\frac{d\theta}{dt}\ne0$, before dividing it out. …