/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 At what rate is soda being sucke... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 30

At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has a radius of 2 in? The depth of the soda decreases at a constant rate of 0.25 in / s.

Short Answer

Expert verified
Answer: The rate at which the soda is being sucked out is approximately -3.14 cubic inches per second.

Step by step solution

01

Write down the given information

The height of the glass, h = 6 in, the radius of the glass, r = 2 in, and the rate of the depth decrease, dh/dt = -0.25 in/s. We need to find dV/dt, the rate at which the volume of soda is being sucked out.
02

Write the formula for the volume of a cylinder

The volume of a cylinder can be calculated using the formula: V = 蟺r^2h, where V is the volume, r is the radius, and h is the height. Since the radius remains constant throughout this problem, we can consider it as a constant.
03

Differentiate the volume formula with respect to time

We want to find the rate at which the volume is changing, so we need to differentiate the volume formula with respect to time (t): dV/dt = 蟺r^2(dh/dt)
04

Substitute the values into the differentiated equation

Now we can substitute the values for r and dh/dt into the equation: dV/dt = 蟺(2)^2(-0.25)
05

Solve for the rate of volume decrease

Simplify the equation to find the rate of volume decrease: dV/dt = -蟺(4)(0.25) = -蟺 So, the rate at which the soda is being sucked out is approximately -3.14 cubic inches per second.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cylindrical Volume
Understanding how the volume of a cylinder changes is crucial for solving related rates problems. A cylinder is a three-dimensional shape with two parallel bases, often circular. To calculate the volume of a cylinder, we use the formula:\[ V = \pi r^2 h \]In this formula, \( V \) represents the volume, \( r \) is the radius of the base, and \( h \) is the height of the cylinder. For a cylinder, the volume depends on both radius and height.
If either of these dimensions changes, the volume will change as well. In real-world applications, understanding these changes can be very useful in situations involving fluid dynamics, manufacturing, and more.
Differentiation with respect to Time
Differentiation is a mathematical tool that allows us to compute the rate of change with respect to certain variables. In related rates problems, we use differentiation concerning time to understand how quantities change over time. This is crucial in scenarios where different quantities are interdependent.
When solving these problems, we typically have a function that describes a physical situation, and we differentiate this function with respect to time to find how one quantity is changing as another changes. For the given problem, we looked at how the volume of the soda in the glass changes as the depth of the soda changes. We apply the chain rule here to differentiate the formula for volume concerning time. This gives us a new function:\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]This equation allows us to solve for the given rate by substituting known values.
Volume of a Cylinder
The volume of a cylinder is fundamental when solving problems involving cylindrical objects. Cylinders are frequently encountered in everyday life 鈥� from soda cans to oil barrels. The reason we pay attention to their volume is that it directly relates to the storage capacity or how much fluid they can hold.
For our problem, we use the volume formula \( V = \pi r^2 h \), recognizing that while the radius \( r \) remains constant, the height \( h \) can change with respect to time. This change affects the overall volume of the cylinder over time. The constant radius means our problem simplifies by only having to deal with changes in height when applying the related rates technique. By understanding these core principles, we can effectively work out the rate at which the volume decreases as the soda is being consumed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general, $$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\)

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

A 500-liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min} .\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank (in liters) is given by \(V(t)=500-0.5 t\). a. Graph the mass function and verify that \(M(0)=0\). b. Graph the volume function and verify that the tank is empty when \(t=1000\) min. c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) and \(\lim _{\theta \rightarrow 000^{-}} C(t) ?\) \(t \rightarrow 1\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\). e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\). f. For what times is the concentration of the solution increasing? Decreasing?

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.