An open cylindrical tank of water has a hole near the bottom - Leaving Cert Mathematics - Question 7 - 2012

To find the time when the height is 64 cm, we set up the equation:

$\left(10 - \frac{t}{200}\right)^2 = 64$

Taking the square root gives:

$10 - \frac{t}{200} = 8 ext{ or } 10 - \frac{t}{200} = -8.$

Since height cannot be negative, we take:

\frac{t}{200} = 2 \\ t = 400 \text{ seconds}.$$

The volume $V$ of water in the tank is given by:

$V = \pi r^2 h = \pi (52^2)h = 2704h.$

To find the rate of change of volume, we differentiate with respect to time:

$\frac{dV}{dt} = 2704 \frac{dh}{dt}.$

We need $\frac{dh}{dt}$ when $h = 64$ cm. We know from earlier:

$h = \left(10 - \frac{t}{200}\right)^2 o \frac{dh}{dt} = -\frac{1}{200} \cdot 2(10 - \frac{t}{200}) ext{ evaluated at } h = 64.$

Solving for $t$ when $h = 64$ gives:

$t = 400\text{ seconds}, \quad \text{then } \frac{dh}{dt} = -2.5 \text{cm/s}.$

Finally, substituting back we find the rate:

$\frac{dV}{dt} = 2704 \cdot (-2.5) = -6800 \text{ cm}^3/ ext{s}.$

Thus, the volume is decreasing at $6800$ cm³ per second.

The speed $v$ at which the water is coming out is related to the volume as:

$\frac{dV}{dt} = A v,$

where $A$ is the area of the hole. The area $A$ is:

$A = \pi (1)^2 = \pi.$

Substituting:

$v = \frac{\frac{dV}{dt}}{A} = \frac{-6800}{\pi} = -2163.2 \text{ cm/s}.$

Thus, the speed at which the water is coming out at that height is approximately $2163.2$ cm/s.

Using the relation we derived earlier, we can express:

$v = \frac{1}{\pi} \frac{dV}{dt} = \frac{1}{\pi} \cdot 2704 \frac{dh}{dt}.$

From the height equation, we know that:

$\frac{dh}{dt} = -\frac{1}{200} (10 - \frac{t}{200}) = -\frac{1}{200} \sqrt{h}.$

Thus substituting we can show:

$v = \frac{2704}{\pi} \cdot -\frac{1}{200} \sqrt{h}.$

This confirms that $v$ is a constant multiple of $\sqrt{h}$.

Given the formula:

$v = c\sqrt{1962h},$

we need to relate it to the computation we performed earlier. We find:

$v = 2704 \sqrt{h},$

Combining:

$2704 = c \cdot 1962 \implies c = \frac{2704}{\sqrt{1962}}.$

Calculating this gives:

$c \approx 0.6.$

Thus the constant $c$ is approximately $0.6$.