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A stone drops into a pond, creating a circular ripple - HSC - SSCE Mathematics Extension 1 - Question 8 - 2017 - Paper 1
Question 8
A stone drops into a pond, creating a circular ripple. The radius of the ripple increases from 0 cm, at a constant rate of 5 cm s⁻¹. At what rate is the area enclos... show full transcript
Worked Solution & Example Answer:A stone drops into a pond, creating a circular ripple - HSC - SSCE Mathematics Extension 1 - Question 8 - 2017 - Paper 1
Step 1
At what rate is the area enclosed within the ripple increasing when the radius is 15 cm?
Answer
To find the rate at which the area is increasing, we start with the formula for the area of a circle:
ho^2$$ where \(A\) is the area and \(r\) is the radius. To determine how the area changes with respect to time, we'll apply the chain rule: $$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$$ 1. First, find \(\frac{dA}{dr}\): Since \(A = \pi r^2\), we have: $$\frac{dA}{dr} = 2\pi r$$ 2. Next, substitute \(r = 15 \text{ cm}\) into the formula: $$\frac{dA}{dr} = 2\pi \times 15 = 30\pi$$ 3. Now, we know the radius increases at a rate of \(\frac{dr}{dt} = 5 \text{ cm s}^{-1}\). We can now compute \(\frac{dA}{dt}\): $$\frac{dA}{dt} = 30\pi \cdot 5 = 150\pi \, \text{cm}^2 \, \text{s}^{-1}$$ Thus, the area enclosed within the ripple is increasing at a rate of \(150\pi \, \text{cm}^2 \, \text{s}^{-1}\) when the radius is 15 cm.