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NCERT Exemplar · Q5

Q.Find an angle θ\theta, 0<θ<π20 < \theta < \dfrac{\pi}{2}, which increases twice as fast as its sine.

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The problem asks for an angle θ\theta in (0,π/2)(0,\pi/2) where the rate of increase of θ\theta is double the rate of increase of sin⁡θ\sin\theta. Using the derivative interpretation, this means dθdt=2ddt(sin⁡θ)\frac{d\theta}{dt} = 2 \frac{d}{dt}(\sin\theta), which simplifies to 1=2cos⁡θ1 = 2\cos\theta, giving θ=π3\theta = \frac{\pi}{3}.

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 θ\theta and sin⁡θ\sin\theta are both changing as time passes, then:

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

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

dθdt=2⋅ddt(sin⁡θ)\frac{d\theta}{dt} = 2 \cdot \frac{d}{dt}(\sin\theta)

Now substitute the derivative of sin⁡θ\sin\theta:

dθdt=2(cos⁡θ⋅dθdt)\frac{d\theta}{dt} = 2 \left( \cos\theta \cdot \frac{d\theta}{dt} \right)

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

1=2cos⁡θ1 = 2 \cos\theta

So:

cos⁡θ=12\cos\theta = \frac{1}{2}

Within the interval 0<θ<π20 < \theta < \frac{\pi}{2}, the angle whose cosine is 12\frac{1}{2} is: …

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