Problem 5
Question
Area The radius \(r\) of a circle is increasing at a rate of 3 inches per minute. Find the rates of change of the area when (a) \(r=6\) inches and (b) \(r=24\) inches.
Step-by-Step Solution
Verified Answer
The rates of change of the area are (a) \(36 \pi\) square inches per minute when \(r=6\) inches and (b) \(144 \pi\) square inches per minute when \(r=24\) inches.
1Step 1: Identify the Basic Formulas and Known Quantities
The area \(A\) of a circle is given by \(A = \pi r^2\). We also know that \(\frac{dr}{dt} = 3\) inches/minute, indicating that the radius is increasing at a constant rate.
2Step 2: Differentiate, Underlining the Related Rates Concept
Differentiate both sides of the area formula with respect to time \(t\). This usage of implicit differentiation results in \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
3Step 3: Substitute Known Quantities for r=6 inches
Substitute \(r = 6\) and \(\frac{dr}{dt} = 3\) into the equation to find \(\frac{dA}{dt}\) when \(r = 6\). This results in \(\frac{dA}{dt} = 2 \pi \times 6 \times 3 = 36 \pi\) inches sq per minute.
4Step 4: Substitute Known Quantities for r=24 inches
Substitute \(r = 24\) and \(\frac{dr}{dt} = 3\) into the equation to find \(\frac{dA}{dt}\) when \(r = 24\). This results in \(\frac{dA}{dt} = 2 \pi \times 24 \times 3 = 144 \pi\) inches sq per minute.
Key Concepts
Implicit DifferentiationRate of ChangeCircle Area Formula
Implicit Differentiation
Implicit differentiation is a technique used in calculus to find the derivative of an equation that is not explicitly solved for one variable in terms of another. For example, in our exercise, the area of a circle is expressed implicitly as a function of the radius through the formula \(A = \pi r^2\). Although \(A\) is not directly given in terms of the time \(t\), we need to find its rate of change with respect to \(t\).
To do this, we perform implicit differentiation, where we differentiate both sides of the equation \(A = \pi r^2\) with respect to \(t\). Using the chain rule, the differentiation yields \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
Here's why this works: although \(A\) is initially described as a function of \(r\), both \(A\) and \(r\) change with time \(t\), making \(r\) and subsequently \(A\), implicitly dependent on \(t\). Therefore, using implicit differentiation, we can relate the changes in \(A\) and \(r\) over time.
To do this, we perform implicit differentiation, where we differentiate both sides of the equation \(A = \pi r^2\) with respect to \(t\). Using the chain rule, the differentiation yields \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
Here's why this works: although \(A\) is initially described as a function of \(r\), both \(A\) and \(r\) change with time \(t\), making \(r\) and subsequently \(A\), implicitly dependent on \(t\). Therefore, using implicit differentiation, we can relate the changes in \(A\) and \(r\) over time.
Rate of Change
Rate of change is a crucial concept in calculus that describes how quickly a quantity is changing over time. In this particular exercise, we're interested in how the area of a circle \(A\) changes as the radius \(r\) increases.
The rate of change of the radius is given by \(\frac{dr}{dt} = 3\) inches per minute, which means for every minute, the radius grows by 3 inches. This rate directly affects the rate of change of the area, \(\frac{dA}{dt}\).
After performing implicit differentiation, we've arrived at the formula \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). By substituting the specific values of \(r\) (such as when \(r = 6\) and \(r = 24\)) into this equation along with \(\frac{dr}{dt}\), we calculate exactly how fast the area is increasing at those specific moments.
The rate of change of the radius is given by \(\frac{dr}{dt} = 3\) inches per minute, which means for every minute, the radius grows by 3 inches. This rate directly affects the rate of change of the area, \(\frac{dA}{dt}\).
After performing implicit differentiation, we've arrived at the formula \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). By substituting the specific values of \(r\) (such as when \(r = 6\) and \(r = 24\)) into this equation along with \(\frac{dr}{dt}\), we calculate exactly how fast the area is increasing at those specific moments.
- At \(r = 6\): \(\frac{dA}{dt} = 36\pi\) square inches per minute.
- At \(r = 24\): \(\frac{dA}{dt} = 144\pi\) square inches per minute.
Circle Area Formula
The area of a circle is one of the most fundamental concepts in geometry. It gives us a way to quantify the space occupied by a circle. The formula for the area \(A\) of a circle is \(A = \pi r^2\), where \(r\) is the radius of the circle.
This formula is derived by integrating the circumference of the circle or visualizing the circle as a series of concentric rings. However, in problems involving changing rates, this formula allows us to see how variations in \(r\) affect the whole area \(A\).
When the radius \(r\) changes, the whole area consequently scales, and the formula \(A = \pi r^2\) tells us how sensitive the area is to changes in \(r\). By differentiating \(A = \pi r^2\) with respect to time, we form a link between the rate of change of the radius and the rate of change of the area. This relationship helps us solve real-world problems where physical attributes, like size, are constantly in flux.
This formula is derived by integrating the circumference of the circle or visualizing the circle as a series of concentric rings. However, in problems involving changing rates, this formula allows us to see how variations in \(r\) affect the whole area \(A\).
When the radius \(r\) changes, the whole area consequently scales, and the formula \(A = \pi r^2\) tells us how sensitive the area is to changes in \(r\). By differentiating \(A = \pi r^2\) with respect to time, we form a link between the rate of change of the radius and the rate of change of the area. This relationship helps us solve real-world problems where physical attributes, like size, are constantly in flux.
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