Question

Draining Hemispherical Reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius \(13 \mathrm{m},\) shown here in profile. Answer the following questions given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) units deep. (a) At what rate is the water level changing when the water is 8 m deep? (b) What is the radius \(r\) of the water's surface when the water is \(y\) m deep? (c) At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

Short answer

a) The rate at which the water level is changing when the water is 8 m deep is approx -0.054 m/min. b) The radius of the water's surface when the water is y m deep is \( r = \sqrt{R^{2}-(R-y)^{2}} \) . c) The rate at which the radius of the water surface is changing when the water is 8 m deep is approx 0.078 m/min.

Step-by-step solution

Calculate rate of change of water level

The equation for the volume V of water in the spherical tank as a function of the water depth y is: \(V=(\pi / 3) y^{2}(3 R-y)\). The problem statement states that the volume V is decreasing at a rate of 6 \(m^3/min\) (\(dV/dt = -6m^3/min\)). To find how fast the water level height y is decreasing when y = 8m, we take the derivative of V with respect to time t, and solve for \(dy/dt\) when y = 8m and R= 13m.

Calculating the radius of the water surface

The radius \(r\) of the water's surface when the water is \(y m\) deep can be found by visualizing the cross-section of the water in the tank as a circle. The radius of this circle will be a right triangle's leg, with the other leg being \(R-y\) and the hypotenuse being R. Thus we can use the Pythagorean theorem \(a^2 + b^2 = c^2\) to derive \(r^2 = R^2 - (R - y)^2\).

Compute rate of change of radius

To compute the rate at which the radius \(r\) of the water surface is changing when the water is 8m deep, we need to compute \(dr/dt\) when y = 8m. Compute the derivative of r (used in the previous step) with respect to t, and use Pythagorean theorem and Chain Rule. Replace \(dy/dt\) with its calculated value found in step 1, set y = 8m and R = 13m, and solve for \(dr/dt\)

Key concepts

Rate of Change

The rate of change in calculus is a fundamental concept that measures how a quantity changes over time. In fluid dynamics, we often want to know how the volume or level of fluid changes as time progresses. In our hemispherical reservoir example, the rate at which water is draining from the tank is given as 6 cubic meters per minute.

This rate, \( dV/dt \), is the derivative of the volume V with respect to time t, and it tells us the speed at which the volume decreases. When the problem asks about the rate at which the water level is changing at a specific depth, it's asking for the rate of change of the depth of water (\textbf{height}) with respect to time, denoted as \(dy/dt\). Calculating \(dy/dt\) requires us to know the relationship between volume and depth, which leads to the concept of related rates.

Volume of a Hemisphere

The volume of a hemisphere is the amount of space inside the shape. For a hemisphere with radius R and a water depth y, the volume can be calculated with the formula provided: \(V=(\pi / 3) y^{2}(3 R-y)\).

This formula helps us to establish a link between the volume and the vertical height (depth) of water in the container. In the drainage problem, understanding this relationship is key to solving both the rate at which the water level drops and the rate at which the radius of the water's surface changes. It's crucial to comprehend how volume adjusts as the shape of the water changes because only the upper portion of the hemisphere that is filled with water contributes to the actual volume.

Related Rates

In calculus, related rates problems involve finding the rate at which one quantity changes with respect to time when multiple quantities are related and changing over time. This kind of problem requires taking the derivative of an equation that relates different variables with respect to time, known as the time derivative.

For our hemisphere tank, we are concerned with how the depth of the water (y) and the radius of the water’s surface (r) change over time. The rate of change of the water’s volume directly affects these variables. As the water drains, not only does the depth decrease, but the radius of the water surface also decreases. To solve related rates problems, we use the chain rule to differentiate implicitly with respect to time, linking the rates of change of each related variable.

Derivatives

A derivative represents the rate at which one quantity changes with respect to another. In our context, we're concerned with derivatives in time, specifically \(dV/dt\) (rate of volume change), \(dy/dt\) (rate of water level change), and \(dr/dt\) (rate of radius change of the water surface).

The derivative of the volume V with respect to the depth y, \(dV/dy\), gives us insight into how sensitive the volume is to small changes in depth. When we calculate the derivative of V with respect to t and use the given rate of \(dV/dt\), it allows us to find \(dy/dt\) and \(dr/dt\) — the rates at which the depth and radius are changing over time. Using the chain rule, we link these rates to solve for the unknown rates we're trying to find.