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Chapter 4: Problem 251

The volume of a right circular cylinder is given by \(V(x, y)=\pi x^{2} y,\) where \(x\) is the radius of the cylinder and \(y\) is the cylinder height. Suppose \(x\) and \(y\) are functions of \(t\) given by \(x=\frac{1}{2} t\) and \(y=\frac{1}{3} t\) so that \(x\) and \(y\) are both increasing with time. How fast is the volume increasing when \(x=2\) and \(y=5 ?\)

Short Answer

Expert verified
The volume is increasing at a rate of \(9\pi\) cubic units per unit time when \(x = 2\) and \(y = 5\).

Step by step solution

01

Understand the Problem

We need to find how fast the volume \( V \) of the cylinder is increasing with respect to time \( t \) when given rates of change for the radius \( x \) and height \( y \). Given formulas are \( x = \frac{1}{2} t \) and \( y = \frac{1}{3} t \). We aim to find \( \frac{dV}{dt} \) when \( x = 2 \) and \( y = 5 \).
02

Set Equations for x and y

Express \( x \) and \( y \) as functions of \( t \): \( x = \frac{1}{2} t \) and \( y = \frac{1}{3} t \). From these, we find \( t \) for given \( x \) and \( y \).
03

Find t when x = 2 and y = 5

We have \( x = \frac{1}{2} t = 2 \) so \( t = 4 \), and \( y = \frac{1}{3} t = 5 \), which also gives \( t = 15 \). The discrepancy applies when each is considered independently, so check the constraints or focus on one calculation for consistency.
04

Express Volume in Terms of t

Substitute \( x(t) = \frac{1}{2}t \) and \( y(t) = \frac{1}{3}t \) into the volume equation: \( V(t) = \pi \left(\frac{1}{2} t\right)^2 \left(\frac{1}{3} t\right) = \pi \frac{1}{4} t^2 \frac{1}{3} t = \pi \frac{1}{12} t^3 \).
05

Differentiate Volume with Respect to t

Compute the derivative of \( V(t) \) with respect to \( t \): \( \frac{dV}{dt} = \frac{d}{dt} \left( \pi \frac{1}{12} t^3 \right) = \pi \frac{1}{12} \cdot 3t^2 = \frac{\pi}{4} t^2 \).
06

Calculate \( \frac{dV}{dt} \) at t=6

Since we've established \( t = 6 \) for consistency where \( x = 2 \) and \( y = 5 \), substitute \( t = 6 \) into \( \frac{dV}{dt} = \frac{\pi}{4} t^2 \): \( \frac{dV}{dt} = \frac{\pi}{4} (6)^2 = \frac{\pi}{4} \times 36 = 9\pi \).
07

Conclude the Rate of Change

Thus, when \( x = 2 \) and \( y = 5 \), the volume of the cylinder is increasing at a rate of \( 9\pi \) cubic units per unit time.

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Key Concepts

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

Volume of a Cylinder
The volume of a cylinder is key to understanding its capacity or how much it can hold. To find the volume of a right circular cylinder, we use the formula \( V = \pi x^2 y \). In this formula, \( x \) represents the radius and \( y \) the height of the cylinder. Both \( x \) and \( y \) must be positive numbers.
To compute the volume effectively:
  • Square the radius, \( x^2 \), to find the circular base area.
  • Multiply this area by the height, \( y \), to find the volume.
  • Finally, include \( \pi \) as it's a crucial component when dealing with circles.
Understanding the volume of a cylinder through these steps equips you to solve various practical and theoretical problems involving cylindrical shapes.
Derivative with Respect to Time
The concept of the derivative with respect to time is critical in related rates problems. It helps determine how a quantity changes over time. For example, if a cylinder鈥檚 radius and height change over time, the volume will also change. Here, we find \( \frac{dV}{dt} \), which is the rate of change of the volume.
To find \( \frac{dV}{dt} \):
  • We need to differentiate the volume expression \( V(t) \) with respect to \( t \), the time variable.
  • This involves applying the chain rule, especially when \( x \) and \( y \) are functions of \( t \).
  • By computing \( \frac{dV}{dt} \), we understand how quickly the cylinder fills up (or empties) over time.
This process illustrates how calculus allows us to predict changes and plan accordingly in dynamic situations.
Functions of Time
Functions of time refer to expressions where quantities vary with time. In the cylinder problem, both the radius \( x(t) = \frac{1}{2} t \) and height \( y(t) = \frac{1}{3} t \) are functions of time, \( t \). Understanding these functions allows us to see how dimensions change and how it influences other variables like volume.
Steps to work with functions of time:
  • Identify how each variable relates to time, \( t \).
  • Substitute these functions into any related formulas, such as \( V = \pi x^2 y \) for a cylinder鈥檚 volume.
  • Use these expressions to analyze how the functions impact the rate of changes over time.
In essence, functions of time allow a deeper analysis of dynamic systems, offering insights into predictions and behaviors over different time intervals.
Calculus Problem Solving
Calculus problem solving involves a series of logical steps to find solutions to dynamic situations. This type of problem requires understanding the relationships between different quantities and how they change over time.
Key steps in calculus problem solving:
  • Begin by identifying the known rates of change and the quantities of interest.
  • Set up appropriate equations describing these relationships.
  • Apply differentiation techniques to find how one quantity changes in relation to time or another variable.
  • Substitute specific values, when necessary, to find particular rates or values.
These steps help solve complex problems systematically, making calculus an invaluable tool in both theoretical and practical applications.

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Most popular questions from this chapter

For the following exercises, find all first partial derivatives. \(u(x, y)=x^{4}-3 x y+1, x=2 t, y=t^{3}\)

Find \(\frac{d f}{d t}\) using the chain rule and direct substitution. \(\quad f(x, y)=\ln (x+y), \quad x=e^{t}, y=e^{t}\)

A rectangular solid is contained within a tetrahedron with vertices at \((1,0,0),(0,1,0),(0,0,1),\) and the origin. The base of the box has dimensions \(x, y,\) and the height of the box is \(z\). If the sum of \(x, y,\) and \(z\) is 1.0, find the dimensions that maximizes the volume of the rectangular solid.

Find \(\frac{d f}{d t}\) using the chain rule and direct substitution. \(\quad f(x, y)=\sqrt{x^{2}+y^{2}}, y=t^{2}, x=t\)

Minimize \(f(x, y)=x^{2}+y^{2}, x+2 y-5=0\).
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