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Chapter 3: Problem 32

Sand is poured into a conical pile with the height of the pile equalling the diameter of the pile. If the sand is poured at a constant rate of \(5 \mathrm{m}^{3} / \mathrm{s}\), at what rate is the height of the pile increasing when the height is 2 meters?

Short Answer

Expert verified
The height of the pile is increasing at a rate of \(5/( \pi) m/s\) when the height is 2 meters.

Step by step solution

01

Formulating the volume of the cone

First, let's write the formula for the volume of a cone, which is \(V = 1/3 * \pi * r^{2} * h\). Since the height equals the diameter, and the diameter is twice the radius, we replace the radius \(r\) with \(h/2\). Thus, we get \(V = 1/12 * \pi * h^{3}\).
02

Differentiating both sides

We next differentiate both sides of the equation with respect to time \(t\). This gives us \(dV/dt = 1/4 * \pi * h^{2} * dh/dt\). Here, \(dV/dt\) is the rate of change of volume and \(dh/dt\) is what we want to find - the rate of change of height.
03

Inserting values

We know that the sand is being poured at a rate of \(5 m^{3}/s\), so \(dV/dt = 5\). We substitute this and the height \(h = 2\) into the differentiated equation which gives us \(5 = 1/4 * \pi * 2^{2} * dh/dt\).
04

Solving for dh/dt

Solving the above equation for \(dh/dt\) (the change in height with time), we get \(dh/dt = 5 / (1/4 * \pi * 2^{2})\). After simplifying, the final value of \(dh/dt\) becomes \(5/( \pi)\), the rate at which the height of the pile of sand is increasing when the height reaches 2 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics that helps us understand changes. It's the study of how things change over time or space. When dealing with problems like sand pouring into a conical pile, calculus gives us a way to calculate these changes. Two main tools within calculus are derivatives and integrals.

Derivatives show the rate of change of a quantity. In our question, we want to find out how fast the height of the cone changes as sand is added. This involves using a derivative, specifically differentiating the equation that describes the cone's volume. Differentiation is the process of finding a derivative, which helps us figure out rates of changes.

This exercise involves related rates, a common application of derivatives. It relates different quantities and their rates, like how the volume of the sand affects the height of the pile. It's helpful for solving real-world problems, giving us insights into time-dependent scenarios like this one.
Volume of a Cone
The volume of a cone is how much space is inside the cone. We calculate it using the formula: \[ V = \frac{1}{3} \pi r^2 h \] where
  • \( V \) is the volume,
  • \( r \) is the radius of the base, and
  • \( h \) is the height of the cone.
In our scenario, the pile of sand forms a cone where the height equals the diameter. This makes understanding and calculation easier.

To accommodate this, we replace the radius in the standard formula with half of the height, as the diameter is twice the radius. This adjustment helps simplify our problem. The new formula becomes: \[ V = \frac{1}{12} \pi h^3 \]This adjusted equation allows us to directly relate the volume of sand being added to the changing height of the pile.
Differentiation
Differentiation is how we calculate the rate of change of a quantity. It is fundamental for solving related rates problems. In the context of our problem, we need differentiation to find out how quickly the height of the sand pile increases.

By differentiating the volume equation with respect to time, we relate the rate at which sand is poured into the cone (\(dV/dt\)) to the rate at which the cone's height increases (\(dh/dt\)). This process involves taking the derivative of the volume formula concerning time.

For the given equation \[ V = \frac{1}{12} \pi h^3 \]we differentiate both sides with respect to time \(t\), resulting in:\[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

Differentiation reveals these relationships, enabling us to solve for \(dh/dt\), giving the rate at which the height of our sand pile changes.

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Most popular questions from this chapter

The rate \(R\) of an enzymatic reaction as a function of the substrate concentration \([S]\) is given by \(R=\frac{[S] R_{m}}{K_{m}+[S]},\) where \(R_{m}\) and \(K_{m}\) are constants. \(K_{m}\) is called the Michaelis constant and \(R_{m}\) is referred to as the maximum reaction rate. Show that \(R_{m}\) is not a proper maximum in that the reaction rate can never be equal to \(R_{m}\)

A 10 -foot ladder leans against the side of a building as in example \(8.2 .\) If the bottom of the ladder is pulled away from the wall at the rate of \(3 \mathrm{ft} / \mathrm{s}\) and the ladder remains in contact with the wall, find the rate at which the top of the ladder is dropping when the bottom is 6 feet from the wall.

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In this exercise, we will explore the family of functions \(f(x)=x^{3}+c x+1,\) where \(c\) is constant. How many and what types of local extrema are there? (Your answer will depend on the value of \(c .\) ) Assuming that this family is indicative of all cubic functions, list all types of cubic functions.

The spectral radiancy \(S\) of an ideal radiator at constant temperature can be thought of as a function \(S(f)\) of the radiant frequency \(f .\) The function \(S(f)\) attains its maximum when \(3 e^{-c f}+c f-3=0\) for the constant \(c=10^{-13} .\) Use Newton's method to approximate the solution.
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