Problem 11
Question
Changing dimensions in a rectangle The length \(l\) of a rectangle is decreasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec}\) while the width \(w\) is in- creasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec}\) . When \(l=12 \mathrm{cm}\) and \(w=5 \mathrm{cm}\) , find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing?
Step-by-Step Solution
Verified Answer
The area is increasing, the perimeter is constant, and the diagonal is decreasing.
1Step 1: Understanding the Problem
We start by summarizing the given information: The length \( l \) is decreasing at \( 2 \, \text{cm/sec} \) and the width \( w \) is increasing at \( 2 \, \text{cm/sec} \). At a certain moment, \( l = 12 \, \text{cm} \) and \( w = 5 \, \text{cm} \). We aim to find the rates of change of the area, perimeter, and the diagonal of the rectangle, and identify if these quantities are increasing or decreasing.
2Step 2: Rate of Change of Area
The area \( A \) of the rectangle is given by \( A = l \times w \). To find the rate of change of the area, we differentiate with respect to time using the product rule:\[ \frac{dA}{dt} = \frac{dl}{dt} \cdot w + l \cdot \frac{dw}{dt} \]Substitute \( \frac{dl}{dt} = -2 \, \text{cm/sec} \), \( \frac{dw}{dt} = 2 \, \text{cm/sec} \), \( l = 12 \, \text{cm} \), and \( w = 5 \, \text{cm} \):\[ \frac{dA}{dt} = (-2)(5) + (12)(2) = -10 + 24 = 14 \, \text{cm}^2/\text{sec} \]
3Step 3: Rate of Change of Perimeter
The perimeter \( P \) is given by \( P = 2(l + w) \). Taking the derivative with respect to time:\[ \frac{dP}{dt} = 2 \left( \frac{dl}{dt} + \frac{dw}{dt} \right) \]Substitute \( \frac{dl}{dt} = -2 \, \text{cm/sec} \) and \( \frac{dw}{dt} = 2 \, \text{cm/sec} \):\[ \frac{dP}{dt} = 2(-2 + 2) = 2 \cdot 0 = 0 \, \text{cm/sec} \]
4Step 4: Rate of Change of Diagonal
The diagonal \( d \) of the rectangle is given by the Pythagorean theorem: \( d = \sqrt{l^2 + w^2} \). To find \( \frac{dd}{dt} \), first differentiate implicitly with respect to time:\[ d \cdot \frac{dd}{dt} = l \cdot \frac{dl}{dt} + w \cdot \frac{dw}{dt} \]Substitute values \( l = 12 \, \text{cm} \), \( w = 5 \, \text{cm} \), \( \frac{dl}{dt} = -2 \, \text{cm/sec} \), and \( \frac{dw}{dt} = 2 \, \text{cm/sec} \):First, calculate \( d = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13 \, \text{cm} \).Now, substitute\[ 13 \cdot \frac{dd}{dt} = 12(-2) + 5(2) = -24 + 10 = -14 \] \[ \frac{dd}{dt} = \frac{-14}{13} \, \text{cm/sec} \]
5Step 5: Determine Increasing or Decreasing
- The area is increasing as \( \frac{dA}{dt} = 14 \, \text{cm}^2/\text{sec} \).- The perimeter is constant since \( \frac{dP}{dt} = 0 \, \text{cm/sec} \).- The diagonal is decreasing because \( \frac{dd}{dt} = \frac{-14}{13} \, \text{cm/sec} \).
Key Concepts
RectanglesArea of a RectanglePerimeter of a RectanglePythagorean Theorem
Rectangles
A rectangle is a four-sided polygon, known in geometry as a quadrilateral, characterized by having opposite sides that are equal in length. Additionally, each internal angle measures 90 degrees, making it a right-angled shape. Understanding the properties of rectangles is essential for solving problems involving dimensions, areas, and perimeters.
Rectangles have a width (often labeled as \(w\)) and a length (\(l\)), which are the two adjacent sides of the rectangle. When solving problems related to dimensions changing over time, like in our exercise, understanding these terms helps calculate how properties like area and perimeter vary.
To summarize the core aspects of a rectangle:
Rectangles have a width (often labeled as \(w\)) and a length (\(l\)), which are the two adjacent sides of the rectangle. When solving problems related to dimensions changing over time, like in our exercise, understanding these terms helps calculate how properties like area and perimeter vary.
To summarize the core aspects of a rectangle:
- Four sides with opposite sides being equal.
- Four angles, each being 90 degrees.
- Defined by its length (\(l\)) and width (\(w\)).
Area of a Rectangle
The area of a rectangle represents the amount of space enclosed within its boundaries, expressed in square units. It is a crucial metric in geometry and is particularly useful in practical scenarios, such as designing a room or a garden.
The formula to calculate the area is straightforward: \[ A = l \times w \] where \( A \) stands for area, \( l \) is the length, and \( w \) is the width of the rectangle. This formula arises from the concept that area can be found by multiplying two perpendicular dimensions.
In the context of changing dimensions, as presented in the related exercise, the area can vary as length and width change over time. To calculate this change, we use the derivative of the area concerning time. This process involves applying the product rule of differentiation, due to the multiplication of two variable quantities:\[ \frac{dA}{dt} = \frac{dl}{dt} \cdot w + l \cdot \frac{dw}{dt} \]
This derivation helps determine whether the area is increasing or decreasing based on how \( l \) and \( w \) change concerning time.
The formula to calculate the area is straightforward: \[ A = l \times w \] where \( A \) stands for area, \( l \) is the length, and \( w \) is the width of the rectangle. This formula arises from the concept that area can be found by multiplying two perpendicular dimensions.
In the context of changing dimensions, as presented in the related exercise, the area can vary as length and width change over time. To calculate this change, we use the derivative of the area concerning time. This process involves applying the product rule of differentiation, due to the multiplication of two variable quantities:\[ \frac{dA}{dt} = \frac{dl}{dt} \cdot w + l \cdot \frac{dw}{dt} \]
This derivation helps determine whether the area is increasing or decreasing based on how \( l \) and \( w \) change concerning time.
Perimeter of a Rectangle
The perimeter is the total length around the rectangle, an important measure for tasks like fencing a garden or framing a picture.
To determine the perimeter of a rectangle, you sum the lengths of all four sides. The formula is given by:\[ P = 2(l + w) \] where \( P \) stands for perimeter, and \( l \) and \( w \) are the rectangle’s length and width, respectively.
In problems involving time-varying dimensions, the rate of change of the perimeter can be determined. This is done by differentiating the perimeter with respect to time:\[ \frac{dP}{dt} = 2\left(\frac{dl}{dt} + \frac{dw}{dt}\right) \] Given the context where a rectangle's length decreases, and width increases, the formula helps find out whether the overall perimeter increases, decreases, or remains constant. In the exercise example, we found that the perimeter remains constant since the changes in length and width balance each other.
To determine the perimeter of a rectangle, you sum the lengths of all four sides. The formula is given by:\[ P = 2(l + w) \] where \( P \) stands for perimeter, and \( l \) and \( w \) are the rectangle’s length and width, respectively.
In problems involving time-varying dimensions, the rate of change of the perimeter can be determined. This is done by differentiating the perimeter with respect to time:\[ \frac{dP}{dt} = 2\left(\frac{dl}{dt} + \frac{dw}{dt}\right) \] Given the context where a rectangle's length decreases, and width increases, the formula helps find out whether the overall perimeter increases, decreases, or remains constant. In the exercise example, we found that the perimeter remains constant since the changes in length and width balance each other.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone of geometry, particularly useful for calculating the diagonal length of rectangles and other right-angled shapes. It states:\[ a^2 + b^2 = c^2 \] where \( a \) and \( b \) are the legs of the right triangle, and \( c \) is the hypotenuse.
In the context of our rectangle, this theorem helps calculate the diagonal \( d \) by treating it as the hypotenuse of a right triangle with the length and width of the rectangle as its legs:\[ d = \sqrt{l^2 + w^2} \]
For problems involving time-dependent changes, the Pythagorean theorem is crucial for determining how the diagonal changes. To find the rate of change of the diagonal, we differentiate the formula with respect to time, involving implicit differentiation:\[ d \cdot \frac{dd}{dt} = l \cdot \frac{dl}{dt} + w \cdot \frac{dw}{dt} \] This equation helps ascertain whether the diagonal length is increasing or decreasing as the rectangle's dimensions vary over time.
In the context of our rectangle, this theorem helps calculate the diagonal \( d \) by treating it as the hypotenuse of a right triangle with the length and width of the rectangle as its legs:\[ d = \sqrt{l^2 + w^2} \]
For problems involving time-dependent changes, the Pythagorean theorem is crucial for determining how the diagonal changes. To find the rate of change of the diagonal, we differentiate the formula with respect to time, involving implicit differentiation:\[ d \cdot \frac{dd}{dt} = l \cdot \frac{dl}{dt} + w \cdot \frac{dw}{dt} \] This equation helps ascertain whether the diagonal length is increasing or decreasing as the rectangle's dimensions vary over time.
Other exercises in this chapter
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In Exercises \(7-12,\) find the indicated derivatives. $$ \frac{d v}{d t} \quad \text { if } \quad v=t-\frac{1}{t} $$
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Find the first derivatives of the functions in Exercises \(11-18\) . $$ s=\sqrt[7]{t^{2}} $$
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In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$ y=\left(1-\frac{x}{7}\righ
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