Question

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is tan^-1(3/4). Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is

Answer 1

Correct Answer: 5

Consider a right circular cone with semi-vertical angle given by tan^-1(3/4). The relationship is tan(θ) = 3/4, where θ is the semi-vertical angle.

The dimensions of the cone can be described in terms of the variables: r (radius of the water surface) and h (depth of the water).

From tan(θ) = r/h and given tan(θ) = 3/4, we derive the relationship:

r = (3/4)h 

Water is poured into the tank at 6 m³/h, hence dV/dt = 6 m³/h.

For a cone, the volume is V = (1/3)πr²h. Substituting r = (3/4)h:

V = (1/3)π((3/4)h)²h = (1/3)π(9/16)h³ = (3π/16)h³

Taking the derivative with respect to t:

dV/dt = (9π/16)h²(dh/dt)

Given dV/dt = 6, solve for dh/dt:

6 = (9π/16)h²(dh/dt)

dh/dt = (96/(9πh²))

At h = 4:

dh/dt = (96/(9π(4)²)) = (96/(144π)) = 2/(3π)

The wet curved surface area A of the cone is A = πrl, where l is the slant height, determined by l = √(r² + h²). Given r = (3/4)h:

l = √((3/4)h² + h²) = h√(9/16 + 16/16) = h√(25/16) = (5/4)h

Then A = πr(5/4)h = π(3/4)h(5/4)h = (15π/16)h²

The rate of change of surface area is:

dA/dt = (15π/8)h(dh/dt)

Substitute h = 4 and dh/dt = 2/(3π):

dA/dt = (15π/8)(4)(2/(3π)) = 5/2 m²/h

The rate at which the wet curved surface area increases is 2.5 m²/h.

Confirming this value within the range 5,5:

The computed value 2.5 fits perfectly within this range.

Answer 2

tanθ = \(\frac{3}{4}\)
v = \(\frac{1}{3}\)πr² h....(i)
And
tanθ =\(\frac{3}{4}\)=\(\frac{r}{h}\)...(ii)
i.e. if h = 4, r = 3
v = \(\frac{1}{3}\)πr2(\(\frac{4r}{3}\))
\(\frac{dA}{dt} \) = \(\frac{4π}{9}\) 3r²\(\frac{dA}{dt} \)
⇒ \(\frac{dr}{dt} \)=\(\frac{ 1}{2π}\)
Curved area = πr√r²+h²
= πr√r²+16r \(\frac{2}{9}\)
= \(\frac{5}{3}πr^2\)
\(\frac{dA}{dt} \)= \(\frac{10}{3}\) πr \(\frac{dA}{dt} \)
= \(\frac{10}{3}\)π.3.\(\frac{1}{2}\)π = 5