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Step-by-Step Solution
Step 1: Understand the Problem
We have a conical water tank (with the vertex downward and axis vertical) being filled at a constant rate of 6 cubic meters per hour. The semi-vertical angle of the cone is given by
\theta = \tan^{-1}\left(\frac{3}{4}\right) .
We need to find the rate at which the wet curved surface area of the water is increasing at the instant when the water depth is 4 meters.
Step 2: Express the Radius–Height Relationship
From
\tan \theta = \frac{3}{4} = \frac{r}{h} \implies r = \frac{3}{4}h.
At the moment when the depth h=4 meters, substituting in the above relation gives
r = \frac{3}{4}\times 4 = 3 \text{ meters}.
Step 3: Volume of the Water
The volume V of water in a cone is given by:
V = \frac{1}{3}\,\pi\,r^2\,h.
Since r = \frac{3}{4}h, we can write:
r^2 = \left(\frac{3}{4}h\right)^2 = \frac{9}{16}h^2.
Hence,
V = \frac{1}{3}\,\pi\,\left(\frac{9}{16}h^2\right)\,h = \frac{3}{16}\,\pi\,h^3.
Step 4: Differentiate Volume with Respect to Time
Let \frac{dV}{dt} = 6 cubic meters per hour (given) and \frac{dh}{dt} be the rate of change of water depth. Differentiating
V = \frac{3}{16}\,\pi\,h^3
with respect to time t gives:
\frac{dV}{dt} = \frac{3}{16}\,\pi \times 3h^2 \,\frac{dh}{dt} = \frac{9}{16}\,\pi\,h^2\,\frac{dh}{dt}.
Substitute \frac{dV}{dt} = 6 :
6 = \frac{9}{16}\,\pi\,h^2 \,\frac{dh}{dt}.
When h=4 ,
6 = \frac{9}{16}\,\pi \times 16 \,\frac{dh}{dt} = 9\pi\,\frac{dh}{dt}.
Hence,
\frac{dh}{dt} = \frac{6}{9\pi} = \frac{2}{3\pi}.
Step 5: Relate \frac{dr}{dt} and \frac{dh}{dt}
Since
r = \frac{3}{4}h,
differentiate with respect to t :
\frac{dr}{dt} = \frac{3}{4}\,\frac{dh}{dt}
= \frac{3}{4} \times \frac{2}{3\pi}
= \frac{1}{2\pi}.
Step 6: Curved Surface Area of the Water
The wet curved surface area A of the water (which is in a conical shape) is
A = \pi r \sqrt{r^2 + h^2}.
Using r = \frac{3}{4}h, we can write:
r^2 = \frac{9}{16}h^2,\quad r^2 + h^2 = \frac{9}{16}h^2 + h^2 = \frac{25}{16}h^2,
thus
\sqrt{r^2 + h^2} = \sqrt{\frac{25}{16}h^2} = \frac{5}{4}h.
Therefore,
A = \pi \left(\frac{3}{4}h\right) \left(\frac{5}{4}h\right)
= \frac{15}{16}\,\pi\,h^2.
Step 7: Differentiate Curved Surface Area w.r.t. Time
Let \frac{dA}{dt} be the rate at which the curved surface area changes. Starting from
A = \frac{15}{16}\,\pi\,h^2,
we differentiate:
\frac{dA}{dt} = \frac{15}{16}\,\pi \times 2h\,\frac{dh}{dt}
= \frac{15}{8}\,\pi\,h \,\frac{dh}{dt}.
At h=4 and we already found
\frac{dh}{dt} = \frac{2}{3\pi},
so
\frac{dA}{dt} = \frac{15}{8}\,\pi \times 4 \times \frac{2}{3\pi}
= \frac{15}{8} \times 4 \times \frac{2}{3}
= \frac{15 \times 4 \times 2}{8 \times 3}.
Simplify:
\frac{15 \times 4 \times 2}{8 \times 3} = \frac{15 \times 8}{24} = \frac{120}{24} = 5.
Step 8: Final Answer
The rate (in square meter per hour) at which the wet curved surface area is increasing when the depth of water is 4 meters is
5.
