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Chapter 12: Problem 40

A circular conical vessel is being filled with ink at a rate of \(10 \mathrm{~cm}^{3} / \mathrm{s} .\) How fast is the level rising after \(20 \mathrm{~cm}^{3}\) have been poured in? The cone has height \(50 \mathrm{~cm}\) and radius \(20 \mathrm{~cm}\) at its brim. (The volume of a cone of height \(h\) and cross-sectional radius \(r\) at its brim is given by \(V=\frac{1}{3} \pi r^{2} h .\).)

Short Answer

Expert verified
After 20 cm鲁 of ink have been poured into the circular conical vessel, the level of ink is rising at a rate of \(\frac{15}{8}\) cm/s.

Step by step solution

01

Write down the volume equation of a cone

First, let's write down the formula for the volume of a cone, which is given by: \(V = \frac{1}{3} \pi r^2 h\).
02

Determine a relationship between r and h

Since the cone is filled with ink, the proportions remain constant throughout. We can find a relationship between radius r and height h by using similar triangles. Let R and H be the dimensions at the brim (20 cm and 50 cm). We have: \(\frac{r}{h} = \frac{R}{H}\) Solve for r: \(r = \frac{hR}{H}\).
03

Replace r in the volume equation

Replace r in the volume expression with the relationship we just derived: \(V = \frac{1}{3}\pi \left(\frac{hR}{H}\right)^2 h\)
04

Implicit differentiation

Differentiate both sides of the equation with respect to time to obtain a related rates equation. Remember we are given the rate of how fast ink is being poured (dV/dt = 10 cm鲁/s). \(\frac{dV}{dt} = \frac{1}{3}\pi \left(\frac{R^2}{H^2}\right)(2hh'\))
05

Solve for the rate of height change, dh/dt

Now we know dV/dt (10 cm鲁/s) and the dimensions R and H. Plug in the given values and solve for the rate of height change dh/dt after 20 cm鲁 (V) have been poured in. \(10 = \frac{1}{3}\pi \left(\frac{20^2}{50^2}\right)(2(20)h')\) Solve for \(h'\) (the rate of height change): \(h' = \frac{15}{8}\) Therefore, after 20 cm鲁 of ink have been poured into the circular conical vessel, the level of ink is rising at a rate of \(\frac{15}{8}\) cm/s.

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique used in calculus to find the derivative of an equation that defines one variable implicitly in terms of another variable. In other words, rather than having a function stated as y = f(x), you may have an equation like g(x, y) = 0 that involves both x and y. When you perform implicit differentiation, you take the derivative of both sides of the equation with respect to x, and then solve for dy/dx. This process is particularly useful in related rates problems where two or more quantities are changing over time.

In related rates problems, you often deal with two variables changing with respect to time. By treating time as an independent variable and implicitly differentiating, you can find the relationship between the rates of change of the dependent variables. It's important to apply the chain rule when differentiating each term with respect to time. For instance, if h was a function of time t, then its derivative would be represented as dh/dt, the rate at which h changes over time. Applying implicit differentiation in this manner allows us to solve complex dynamic scenarios, such as the changing water level in a cone.
Cone Volume Formula
The volume of a cone can be calculated using the formula V = 1/3 * 蟺 * r^2 * h, where V is the volume, r is the radius of the base, and h is the height of the cone. This formula is derived from the fact that a cone can be seen as a pyramid with a circular base. Just as the volume of a pyramid is one-third the product of its base area and height, so too is the volume of the cone.

The cone's volume formula is essential when solving related rates problems involving cones, such as when a liquid fills or empties from the cone at a certain rate. It's important to note that if a cone's dimensions are increasing or decreasing at a given rate, the radius and height are not independent; they are related through the geometric proportions of the cone, as shown by similar triangles in the solution above. Understanding this relationship helps in setting up equations correctly to determine the rate at which the volume changes.
Rate of Change
The rate of change is a concept that measures how much a quantity changes over time. In calculus, it is often represented as a derivative, which describes the slope of a function at any given point. Rates of change can be constant, like the speed of a car cruising at a steady 60 mph, or variable, like a rocket accelerating as it leaves the atmosphere.

In our conical vessel example, two rates of change are significant: the rate at which the vessel is being filled (dV/dt) and the rate at which the ink level rises (dh/dt). By understanding rates of change and how they relate to each function, you can determine one rate by knowing the other. This concept is fundamental in all sorts of applications, from physics to economics to biology, and in particular, for solving related rates problems, where we seek to find out how the change in one variable affects the change in another over time.

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Most popular questions from this chapter

The volume of paint in a right cylindrical can is given by \(V=4 t^{2}-t\) where \(t\) is time in seconds and \(V\) is the volume in \(\mathrm{cm}^{3} .\) How fast is the level rising when the height is \(2 \mathrm{~cm}\) ? The can has a height of \(4 \mathrm{~cm}\) and a radius of \(2 \mathrm{~cm}\). HINT [To get \(h\) as a function of \(t\), first solve the volume \(V=\pi r^{2} h\) for \(h\).]

My aunt and I were approaching the same intersection, she from the south and I from the west. She was traveling at a steady speed of 10 miles/hour, while I was approaching the intersection at 60 miles/hour. At a certain instant in time, I was one-tenth of a mile from the intersection, while she was one- twentieth of a mile from it. How fast were we approaching each other at that instant?

Rewrite the statements and questions in mathematical notation. The population \(P\) is currently 10,000 and growing at a rate of 1,000 per year.

A right circular conical vessel is being filled with green industrial waste at a rate of 100 cubic meters per second. How fast is the level rising after \(200 \pi\) cubic meters have been poured in? The cone has a height of \(50 \mathrm{~m}\) and a radius of \(30 \mathrm{~m}\) at its brim. (The volume of a cone of height \(h\) and crosssectional radius \(r\) at its brim is given by \(V=\frac{1}{3} \pi r^{2} h .\).)

As the new owner of a supermarket, you have inherited a large inventory of unsold imported Limburger cheese, and you would like to set the price so that your revenue from selling it is as large as possible. Previous sales figures of the cheese are shown in the following table: $$\begin{array}{|r|c|c|c|}\hline \text { Price per Pound, } p & \$ 3.00 & \$ 4.00 & \$ 5.00 \\ \hline \text { Monthly Sales in Pounds, } q & 407 & 287 & 223 \\\\\hline\end{array}$$ a. Use the sales figures for the prices \(\$ 3\) and \(\$ 5\) per pound to construct a demand function of the form \(q=A e^{-b p}\), where \(A\) and \(b\) are constants you must determine. (Round \(A\) and \(b\) to two significant digits.) b. Use your demand function to find the price elasticity of demand at each of the prices listed. c. At what price should you sell the cheese in order to maximize monthly revenue? d. If your total inventory of cheese amounts to only 200 pounds, and it will spoil one month from now, how should you price it in order to receive the greatest revenue? Is this the same answer you got in part (c)? If not, give a brief explanation.
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