The area A of a circle is increasing at a constant rate of 1.5 cm\(^2s^{-1}\). Find, to 3 significant figures, the rate at which the radius r of the circle is increasing when the area of the circle is 2 cm\(^2\).
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Correct Answer: Option D
Explanation:
Area of a circle (A) = \(\pi r^2\)
Given
\(\frac{dA}{dt} = 1.5cm^2s^{-1}\)
\(\frac{dr}{dt}\) = ?
A = 2cm\(^2\)
Now
2 = \(\pi r^2\)
= r\(^2 = \frac {2}{\pi}\)
r = \(\sqrt \frac {2}{\pi}\) cm = 0.798cm
\(\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}\)
\(\frac {dA}{dr} = 2\pi r\) (differentiating A = \(\pi r^2)\)
\(\frac {dr}{dA} = \frac {1}{2\pi r}\)
\(\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199\)
\(\frac {dr}{dt} = 0.299cms^{-1}\) (to 3 s.f)
Area of a circle (A) = \(\pi r^2\)
Given
\(\frac{dA}{dt} = 1.5cm^2s^{-1}\)
\(\frac{dr}{dt}\) = ?
A = 2cm\(^2\)
Now
2 = \(\pi r^2\)
= r\(^2 = \frac {2}{\pi}\)
r = \(\sqrt \frac {2}{\pi}\) cm = 0.798cm
\(\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}\)
\(\frac {dA}{dr} = 2\pi r\) (differentiating A = \(\pi r^2)\)
\(\frac {dr}{dA} = \frac {1}{2\pi r}\)
\(\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199\)
\(\frac {dr}{dt} = 0.299cms^{-1}\) (to 3 s.f)