Question

Filling a Bowl (a) Volume A hemispherical bowl of radius \(a\) contains water to a depth \(h\) . Find the volume of water in the bowl. (b) Related Rates Water runs into a sunken concrete hemispherical bowl of radius 5 \(\mathrm{m}\) at a rate of 0.2 \(\mathrm{m}^{3} / \mathrm{sec} .\) How fast is the water level in the bowl rising when the water is 4 \(\mathrm{m}\) deep?

Short answer

The volume of water in the bowl when filled to a depth of 4m is approximately 310m\(^3\). The rate at which the water level in the bowl is rising when the water is 4m deep is approximately 0.04m/sec.

Step-by-step solution

Establish the formula for volume of a spherical cap

If you cut a sphere by a plane, the part of the sphere that is above the plane is referred to as a spherical cap. The volume \(V\) of a spherical cap of radius \(a\) and height \(h\) can be expressed as \( \frac{1}{6}\pi h(3a^2+h^2)\). The volume of water in the bowl will be equal to the volume of the spherical cap.

Substitute the values of \(a\) and \(h\) into the formula

From the given problem, the radius \(a\) of the hemisphere is 5 m and the depth \(h\) of the water is 4 m. Substitute these values into the volume formula from step 1 to compute the volume of water in the bowl, which is \( \frac{1}{6}\pi(4)(3(5)^2+(4)^2)\).

Solve the Related Rates

Use the formula for the volume of a spherical cap \(V = \frac{1}{6}\pi h(3a^2 + h^2) \). Given \(\frac{dV}{dt} = 0.2\ m^3/s\), we need to find \(\frac{dh}{dt}\) when \(h = 4\ m\) which can be done by using the Chain Rule / Implicit Differentiation. Applying the chain rule to the volume formula gives \(\frac{dV}{dt} = \frac{1}{6}\pi(3a^2 + 4h)\cdot\frac{dh}{dt}\). Solve this equation for \(\frac{dh}{dt}\) to find the rate the water level is rising.

Key concepts

Volume of a Spherical Cap

Understanding the volume of a spherical cap can initially seem daunting, but it's quite an interesting topic in calculus. Imagine slicing through a sphere with a flat plane. The portion of the sphere that lies above the plane is the 'spherical cap'. To find the volume of this cap, you use the formula: \[ V = \frac{1}{6}\pi h(3a^2 + h^2) \] where \( V \) is the volume, \( h \) is the height of the cap from the base to the top, and \( a \) is the radius of the entire sphere (not just the cap). This formula is an essential tool in solving problems where spherical caps are involved, such as finding the volume of fluid in a tank or measuring certain types of sediment layers. Imagine you're pouring water into a hemisphere, the resulting water level would form a spherical cap, which gives a practical use of this concept.

When applying this to homework problems or real-world scenarios, it's all about identifying the relevant dimensions and substituting them correctly into this formula. Doing so will give you the volume contained within the spherical cap - a concept as useful in engineering as it is in pure mathematics.

Hemispherical Bowl Volume

A hemispherical bowl is like half of a perfectly round ball with a flat surface at the bottom — a dome, to visualize it easily. To find the volume of a hemispherical bowl, you could use the volume formula for a full sphere and then divide by two. However, in the context of calculus problems involving related rates, what we're often interested in is the volume of liquid within the bowl up to a certain height.

Since the bowl is essentially a spherical cap that extends all the way down to the flat diameter, the volume formula for a spherical cap comes in handy yet again. This means that for a hemispherical bowl of radius \(a\), the full volume is \( \frac{2}{3}\pi a^3 \), but when partially filled to a height \(h\), the volume of the liquid is given by the same formula as a cap's volume. Mastery of these volume concepts is crucial for students in fields like fluid dynamics, architecture, and any discipline where managing volumes of materials is required.

Implicit Differentiation

Implicit differentiation is a powerful technique used when dealing with equations that are not easily solved for one variable in terms of another. In other words, you use implicit differentiation when you have variables intermingled with each other, and you can't (or it's not convenient to) isolate them.

This approach is often used in related rates problems, where you have a rate of change (derivatives) in terms of another variable that is not the direct variable of differentiation. In our example with the spherical cap, we're dealing with an equation that relates the volume \(V\) of the water to the height \(h\) of the water level in the bowl. Instead of isolating \(h\) and differentiating directly, we differentiate both sides of the volume equation with respect to time \(t\), bringing into play rates of changes—like the rate at which height changes over time, \(dh/dt\), and the rate at which volume changes over time, \(dV/dt\). This makes it essential to understand implicit differentiation fully to grasp how variables relate to each other in a changing context.

Chain Rule in Calculus

The chain rule stands as one of the most fundamental principles in calculus, particularly when we delve into problems involving rates. It's the go-to rule when we need to find the derivative of composite functions - in simpler terms, functions nested within other functions. Imagine a formula nested like Russian dolls; the chain rule helps us differentiate from the outermost to the innermost layer.

In the context of related rates, which is the essence of our exercise, the chain rule allows us to relate the rate of change of the volume \(dV/dt\) with the rate of change of the height \(dh/dt\). By applying this rule, we 'chain' these two rates together. In our problem, we differentiated the volume of the spherical cap with respect to time to find how fast the water level rises in terms of height per time. This shows how, with the chain rule, one can observe the ripple effect of one changing quantity upon another—providing essential insights into understanding dynamic systems in calculus and beyond.