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At time t seconds the radius of a sphere is r cm, its volume is V cm³ and its surface area is S cm² - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 8
Question 7
At time t seconds the radius of a sphere is r cm, its volume is V cm³ and its surface area is S cm². [You are given that V = \( \frac{4}{3} \pi r^3 \) and that S = ... show full transcript
Worked Solution & Example Answer:At time t seconds the radius of a sphere is r cm, its volume is V cm³ and its surface area is S cm² - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 8
Step 1
Find \( \frac{dr}{dt} \) when the radius of the sphere is 4 cm
Answer
To find ( \frac{dr}{dt} ), we start by differentiating the volume formula with respect to time t:
[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) = 4\pi r^2 \frac{dr}{dt} ]
We know that ( \frac{dV}{dt} = 3 ) cm³ s⁻¹.
Setting up the equation:
[ 3 = 4\pi r^2 \frac{dr}{dt} ]
Substituting ( r = 4 ) cm:
[ 3 = 4\pi (4)^2 \frac{dr}{dt} ]
Simplifying gives:
[ 3 = 64\pi \frac{dr}{dt} ]
Solving for ( \frac{dr}{dt} ):
[ \frac{dr}{dt} = \frac{3}{64\pi} ]
Calculating this:
[ \frac{dr}{dt} \approx 0.01492\text{ cm s}^{-1} ]
Rounding to 3 significant figures yields:
[ \frac{dr}{dt} \approx 0.0149 \text{ cm s}^{-1} ]
Step 2
Find the rate at which the surface area of the sphere is increasing when the radius is 4 cm
Answer
To find the rate of change of the surface area S with respect to time, we differentiate:
[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt} ]
Substituting ( r = 4 ) cm and the value of ( \frac{dr}{dt} ):
[ \frac{dS}{dt} = 8\pi (4) \left(\frac{3}{64\pi}\right) ]
Canceling out ( \pi ):
[ \frac{dS}{dt} = 8 \times 4 \times \frac{3}{64} = 1.5 \text{ cm}^2 s^{-1} ]
Hence, the rate at which the surface area is increasing when the radius is 4 cm is ( 1.5 \text{ cm}^2 s^{-1} ).