Differentiation - Connected Rates of Change
Differentiation - Connected Rates of Change
IB AA SL: core and optional content
Core: geometric area formula. Optional: rates with respect to time.
The figure shows a circle with radius of \(10\) cm.
- Show that the area, \(A\) cm\(^{2}\), of the region bounded by the chord \(DB\), the diameter \(DC\) and the arc length \(BC\) is \(A = 50(\theta + \sin \theta)\).
- If the length of the arc \(BC\) is \(k\) cm and \(k\) is increased at a constant rate of \(0.5\) cm/s, find
- the rate of change of \(\theta\) in radians per second,
- the rate of change of the area \(A\) when \(\theta = 0.5\) radians.
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Answer:(a) \(A=50(\theta+\sin\theta)\) (b)(i) \(\frac{\mathrm{d}\theta}{\mathrm{d}t}=\frac{1}{20}\text{ rad s}^{-1}\) (ii) \(\frac{\mathrm{d}A}{\mathrm{d}t}=4.69\text{ cm}^2\text{ s}^{-1}\) (3 s.f.)
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