How will we solve this question - Maths - Application of Derivatives

Question

How will we solve this question??

Sneha Apar · Accepted Answer

Dear student,

let ABC be the conical vessel such that its radius OC = 5 cm and depth OA = 15 cm let V be the volume of water in the vessel at any time t let DE = r cm and AD = h cm therefore $V = \frac{1}{3} π r^{2} h$ (1) now triangles ADE and AOC are similar triangles therefore $\frac{A D}{D E} = \frac{A O}{O C} \frac{h}{r} = \frac{15}{5} = 3 h = 3 r o r r = \frac{h}{3} (2)$ given: $\frac{d V}{d t} = 0 1 c m^{3} / \sec$ (3) (i) from eq(1) and (2): $V = \frac{1}{3} π (\frac{h}{3})^{2} \cdot h V = \frac{π}{27} h^{3}$ now differentiating the equation wrt t : $\frac{d V}{d t} = \frac{π}{27} \cdot 3 h^{2} \cdot \frac{d h}{d t} \frac{d V}{d t} = π \cdot \frac{h^{2}}{9} \cdot \frac{d h}{d t} [f r o m (3)] 0 1 = π \cdot \frac{h^{2}}{9} \cdot \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{0 9}{{π h}^{2}}$ ${\frac{d h}{d t} |}_{a t h = 6} = \frac{0 9}{π \cdot 6 * 6} = \frac{0 9}{36 π} = \frac{1}{40 π} c m / \sec$ (4) (ii) let water surface area be A $A = π r^{2} \Rightarrow A = π (\frac{h}{3})^{2}$ $A = \frac{π h^{2}}{9}$ differentiating wrt t : $\frac{d A}{d t} = \frac{π}{9} \cdot 2 h \frac{d h}{d t} = \frac{2 π h}{9} \cdot \frac{d h}{d t} \frac{d A}{d t} |_{a t h = 6 c m} = \frac{2 π}{9} * 6 * \frac{1}{40 π} [f r o m e q (4)] = \frac{6}{9 * 20} = \frac{1}{30} c m^{2} / \sec$ (5) (iii) let the wetted surface of the vessel be S $S = π r l$ where $l = \sqrt{r^{2} + h^{2}}$ be the slant height of water cone $S = π r \sqrt{r^{2} + h^{2}} S = π \cdot \frac{h}{3} \cdot \sqrt{(\frac{h}{3})^{2} + h^{2}} S = \frac{π h}{3} \cdot \sqrt{\frac{h^{2}}{9} + h^{2}} S = \frac{π h}{3} \cdot h \sqrt{\frac{10}{9}} S = \frac{π h^{2} \sqrt{10}}{9}$ (6) differentiating wrt t : $\frac{d S}{d t} = \frac{\sqrt{10}}{9} \cdot π \cdot 2 h \frac{d h}{d t} \frac{d S}{d t} |_{a t h = 6 c m} = \frac{\sqrt{10}}{9} \cdot π \cdot 2 * 6 * \frac{1}{40 π} [u \sin g e q (4)] = \frac{\sqrt{10}}{30} c m^{2} / \sec$ Regards