The radius of a conical tank is 2.7 meters and the height of the tank is 4.3 meters. Water is flowing into the tank at a constant rate of 59.5 m3/minute. At the instant the the depth of the water is 0.6 meters, answer the following:
(A) At what rate is the depth of the water changing?
(B) At what rate is the radius of the waters surface changing?
(C) At what rate is the surface area of the exposed water changing?
To answer the questions, we need to make use of the related rates concept in calculus. We can relate the changing quantities using the formulas for the volume, slant height, and surface area of a cone.
(A) At what rate is the depth of the water changing?
We know that the volume of a cone is given by V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height. We differentiate this equation with respect to time to find how the volume is changing with respect to time.
dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt))
We have the values of r = 2.7 m and dh/dt = 0.6 m/min. Now we need to find dr/dt, the rate at which the radius is changing at the given instant.
(B) At what rate is the radius of the water's surface changing?
To find dr/dt, we can relate the radius to the height using similar triangles. The height of the tank is 4.3 m, and the depth of the water is 0.6 m. Using these values, we can write the equation:
r/h = (2.7 - r)/(4.3 - 0.6)
Simplifying this equation, we get:
r = (2.7h - 1.35)
Differentiating both sides with respect to time, we find:
dr/dt = 2.7(dh/dt)
We have the value of dh/dt from part (A). Now we can substitute those values to find dr/dt.
(C) At what rate is the surface area of the exposed water changing?
The surface area of the exposed water is given by A = πrl, where r is the radius and l is the slant height. We need to find dl/dt, the rate at which the slant height is changing at the given instant.
Using the Pythagorean theorem, we have:
l = sqrt(r^2 + h^2)
Differentiating both sides with respect to time, we get:
dl/dt = (1/l)(rr)(dr/dt) + (1/l)(hh)(dh/dt)
We have the values of r, h, dr/dt, and dh/dt. Now we can substitute those values to find dl/dt.
Thus, by solving these equations, we can find the rate at which the depth of the water is changing (A), the rate at which the radius of the water's surface is changing (B), and the rate at which the surface area of the exposed water is changing (C).
depth = h
dV/dh = area of surface (draw it :)
dV/dh = pi r^2 = surface area
r = (2.7/4.3)h
remember to use chain rule.
for example
dV/dt = 59.5 m^3/min
but
dV/dt = dV/dh dh/dt
so
dh/dt = 59.5/ (pi r^2)
etc