Question

Water is dripping out from a conical funnel of semi-vertical angle π/4at the uniform rate of 2 cm²/sec in
its surface area through a tiny hole at the vertex in the bottom. When the slant height of the water is 4
cm, find the rate of decrease of the slant height of the water.​

Answer 1

Hi,

Answer:

Let’s consider at any time t the slant height of the cone be “l”, radius be “r” and height of the cone be “h”.

Now, the water is dripping out of the funnel at a uniform rate of 2 cm^2/sec. So, we can write the equation for the rate of decreasing surface area as

d(πr²) / dt = - 2 cm²/s

or, 2πr(dr/dt) = -2cm²/s …… (i)

From the figure below we can see that at any instance  

tanθ = r/h

or, h tanθ = r

on differentiating the equation on both sides

or, (dh/dt) tanθ = dr/dt ….. (ii)

Also, at the instance when the water is decreasing, we can say

sinθ = r / l

or, 1/√2 = r/l ……...... [since the semi vertical angle is given as π/4 and sinπ/4 = 1/√2]

or, r = 2√2 cm ….. (iii) [at this instance the slant height “l” is given as 4 cm]

Similarly, as θ = π/4 when water is decreasing, we can also write  

h tan(π/4) = r and (dh/dt) tanθ = dr/dt

as, h = r = 2√2 cm and dh / dt = dr / dt …… (iv) [∵tan(π/4) = 1]  

Putting the value of r from (iii) in (i), we get

2πr(dr/dt) = -2

Or, 2π * 2√2 * (dr/dt) = -2

Or, dr/dt = -1 / (π * 2√2) cm/sec = dh/dt….. (v)

Considering ∆OAB from the figure and by using Pythagoras theorem we get,

l² = h² + r²

on differentiating the above equation

or, 2l(dl/dt) = 2h(dh/dt) + 2r(dr/dt)

substituting the values from (iv) & (v),

or,  4 * (dl/dt) = [2√2 * {-1 / (π * 2√2)}] + [2√2 * {-1 / (π * 2√2)}]

or, (dl/dt) = - 2 / 4π = -1/2π cm/s

Hence, the rate of decrease of slant height of the water is (1/2π) cm/s.

Hope this helps!!!!!