Class 12 Applications Of Derivatives · Case Study4 Markscase_based_application
Class 12 Maths · Applications Of Derivatives · Case Study #28
Water drains at 2 cm³/s from the bottom of an inverted conical tank whose semi-vertical angle is 45°. Initially its conical portion is full. At any later instant let r,h,l be the radius, height and slant height of the remaining water cone; r,h>0. All rates are signed unless a magnitude of decrease is requested.
Part (1) 1 Marks
Express the volume of remaining water in terms of r.
Answer: V=πr³/3
- tan45°=r/h=1, so h=r.
- V=πr²h/3=πr³/3.
Part (2) 1 Marks
Find dr/dt when r=2√2 cm.
Answer: −1/(4π) cm/s
- Differentiate V=πr³/3: dV/dt=πr²dr/dt.
- With dV/dt=−2 and r²=8, dr/dt=−2/(8π)=−1/(4π) cm/s.
Part (3) 2 Marks
Find the rate at which the wetted curved surface area decreases when r=2√2 cm.
Answer: 2 cm²/s (dS/dt=−2 cm²/s)
- l=√(r²+h²)=√2r, hence S=πrl=√2πr².
- dS/dt=2√2πr dr/dt=2√2π(2√2)(−1/(4π))=−2 cm²/s.
- The area decreases at 2 cm²/s.
Part (4) 2 Marks
Alternatively, find dh/dt when l=4 cm.
Answer: −1/(4π) cm/s
- h=r and l=√2h imply h=2√2 and V=πh³/3.
- Differentiate: −2=πh²dh/dt=8πdh/dt.
- Therefore dh/dt=−1/(4π) cm/s.