(c) The altitude of a right circular cone is 15cm and is increasing at 0.2 cm/s. The
radius of the base is 10cm and is decreasing at 0.3cm/s. How fast is the volume
changing?
Answers
Given :-
- Altitude of a right circular cone = 15cm
- Altitude increasing speed = 0.2 cm/s .
- Radius = 10 cm.
- Decreasing speed of radius = 0.3 cm / s.
To Find :-
- How fast is the volume changing ?
Solution :-
we know that,
- Volume of cone = (1/3) * π * (radius)² * Height .
so, putting all values and differentiating we get,
→ V = (1/3) * π * (r)² * h
→ dV/dt = (π/3)[2rh*(dr/dt) + r²(dh/dt]
putting inc/dec speed we get,
→ dV/dt = (π/3)[2*10*15*(-0.3) + (10)² * 0.2]
→ dV/dt = (π/3)[300(-0.3) + 100 * 0.2]
→ dV/dt = (π/3)[-90 + 20]
→ dV/dt = (π/3) * (-70)
→ dV/dt = (22/7) * (1/3) * (-70)
→ dV/dt = {22 * (-10)}/3
→ dV/dt = (-220/3)
→ dV/dt ≈ (-73.34)
since, negative sign shows that, volume is decreasing.
Hence, Volume is decreasing at the rate of 73.34cm/s.
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