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Oil from a leaking container radiates outwards in the form of a circular film on the water surfaces.
If the area of the circle increases at rate of 900m²/s, how fast is the radius of the circle
increasing when the area is 1400 z m??
[6 marks)
[Ans:3.83m/s]

Sagot :

Trancescientidn

[tex] \frac{dr}{dt} = 6.79 \: \frac{m}{s} [/tex]

Step by Step Explanation:

The area a is defined as

[tex]a = \pi {r}^{2} [/tex]

Now take derivative of the above expression with respect to time:

[tex] \frac{da}{dt} = 2\pi r \frac{dr}{dt} [/tex]

Solving for dr/dt, we then get

[tex] \frac{dr}{dt} = \frac{1}{2\pi r} \frac{da}{dt} [/tex]

Note that da/dt = 900 m^2/s and when the area of the oil slick is 1400 m^2, the radius r is r = √(1400/π) = 21.1 m. Therefore,the rate at which the radius is increasing dr/dt is

[tex] \frac{dr}{dt} = \frac{1}{2\pi(21.1 \: m)}(900 \frac{ {m}^{2} }{s} ) = 6.79 \frac{m}{s} [/tex]

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