1) Water is running into a conical vessel, 15cm deep and 5cm in radius at the rate of 0 1cm3/s - Maths - Application of Derivatives

Question

1) Water is running into a conical vessel, 15cm deep and 5cm in
radius at the rate of 0 1cm3/s When the water is 6cm
deep find at what rate
(i) water level is raising (ii) the water surface area is is
increasing (iii) the wetted surface of the vessel is increasing
(iv) slant height is increasing
2) A man 1 5m tall moves away from a pole 9 5m tall at a rate of
3m/s Find the rate at which elevation of top of the pole is
changing when the man is 6m away from the pole?

Ajanta Trivedi · Accepted Answer

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 : $\frac{d S}{d t} = \frac{\sqrt{10}}{9} \cdot π \cdot 2 h \frac{d h}{d t} \frac{d S}{d t} |_{a t h = 6 c m} = \frac{\sqrt{10}}{9} \cdot π \cdot 2 * 6 * \frac{1}{40 π} [u \sin g e q (4)] = \frac{\sqrt{10}}{30} c m^{2} / \sec$ (iv) let the slant height be $l = \sqrt{r^{2} + h^{2}}$ $l = \sqrt{(\frac{h}{3})^{2} + h^{2}} = \sqrt{\frac{h^{2}}{9} + h^{2}} l = h \cdot \frac{\sqrt{10}}{3} (7)$ differentiating wrt t: $\frac{d l}{d t} = \frac{\sqrt{10}}{3} \cdot \frac{d h}{d t} \frac{d l}{d t} |_{a t h = 6 c m} = \frac{\sqrt{10}}{3} \cdot \frac{1}{40 π} [u \sin g e q (4)] = \frac{\sqrt{10}}{120 π} c m / \sec$ hope this helps you cheers!!