Question 37

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Jambmaths question: 

$\begin{align}  & \text{The radius of a circle is increasing at the rate of 0}\text{.02cm}{{\text{s}}^{-1}}.\text{ Find the rate at which the area is } \\ & \text{increasing when the radius of the circle is 7cm} \\ & \text{(A)  0}\text{.88c}{{\text{m}}^{2}}{{s}^{-1}}\text{  (B) }0.75c{{m}^{2}}{{s}^{-1}}\text{ (C) }0.53c{{m}^{2}}{{s}^{-1}}\text{ (D) }0.35c{{m}^{2}}{{s}^{-1}} \\\end{align}$

Jamb Maths Solution: 

$\begin{align}  & \frac{dr}{dt}=0.02cm{{s}^{-1}} \\ & A=\pi {{r}^{2}} \\ & \frac{dA}{dr}=2\pi r \\ & \frac{dA}{dt}=\frac{dA}{dr}\times \frac{dr}{dt}=2\pi r\times 0.02=0.04\pi r \\ & \frac{dA}{dt}=0.04\times \frac{22}{7}\times \frac{7}{1}=0.88c{{m}^{2}}{{s}^{-1}} \\\end{align}$

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