Chapter 18: Problem 46
Solve the given applied problems involving variation. The \(f\) -number lens setting of a camera varies directly as the square root of the time \(t\) that the film is exposed. If the \(f\) -number is 8 (written as \(f / 8\) ) for \(t=0.0200\) s, find the \(f\) -number for \(t=0.0098\) s.
Short Answer
Expert verified
The f-number for \( t=0.0098 \) s is approximately 5.60.
Step by step solution
01
Understand the Direct Variation
In problems involving direct variation, we have the relationship that one quantity varies directly as another quantity. Here, the f-number varies directly as the square root of time, which can be expressed mathematically as \( f = k \sqrt{t} \), where \( k \) is a constant of variation.
02
Use Given Information to Find the Constant
We know that for \( t = 0.0200 \) seconds, the f-number is 8. Substitute these values into the equation \( f = k \sqrt{t} \) to find \( k \): \[ 8 = k \sqrt{0.0200} \]Solve for \( k \) by dividing both sides by \( \sqrt{0.0200} \): \[ k = \frac{8}{\sqrt{0.0200}} \].
03
Calculate the Constant of Variation
Compute \( \sqrt{0.0200} \) which is approximately 0.1414. Therefore, \[ k = \frac{8}{0.1414} \approx 56.56 \].
04
Find the New f-number
Now that we have \( k \), use it to find the f-number for \( t = 0.0098 \) seconds. Substitute \( t = 0.0098 \) into the equation \( f = k \sqrt{t} \):\[ f = 56.56 \times \sqrt{0.0098} \].
05
Calculate the New f-number
Compute \( \sqrt{0.0098} \) which is approximately 0.099. Multiply by \( k \):\[ f = 56.56 \times 0.099 \approx 5.60 \]. The f-number for \( t = 0.0098 \) seconds is approximately 5.60.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direct Variation
In mathematics, direct variation describes a linear relationship between two variables where one variable is a constant multiple of the other. This is expressed in the form \( y = kx \), where \( y \) varies directly as \( x \), and \( k \) is the constant of variation. If \( x \) increases, \( y \) increases proportionally, which means they either both rise or fall together.
For example, in the context of photography and the given problem, the f-number of a lens varies directly with the square root of the time that the film is exposed to light. This can be captured by the equation \( f = k \sqrt{t} \). It signifies that if the exposure time increases, the f-number increases in proportion to the square root of the time. This relationship helps photographers determine the appropriate f-number needed for different exposure times.
For example, in the context of photography and the given problem, the f-number of a lens varies directly with the square root of the time that the film is exposed to light. This can be captured by the equation \( f = k \sqrt{t} \). It signifies that if the exposure time increases, the f-number increases in proportion to the square root of the time. This relationship helps photographers determine the appropriate f-number needed for different exposure times.
- Direct variation involves linear relationships.
- This concept supports calculations in real-life applications, like determining lens settings in photography.
F-number Calculation
F-number, often seen in the format \( f/x \), is a critical concept in photography that influences image brightness and depth of field. It is determined by the formula \( f = k \sqrt{t} \) in scenarios of direct variation involving time \( t \) as in the given problem.
This formula not only helps compute the f-number when given the time but also enables you to predict how a change in exposure time will affect the f-number. In the exercise, for an initial time of 0.0200 seconds with an f-number of 8, the constant \( k \) is computed first. Then, this constant allows calculation of the new f-number for a different time, 0.0098 seconds.
The key steps in calculating the f-number:
This formula not only helps compute the f-number when given the time but also enables you to predict how a change in exposure time will affect the f-number. In the exercise, for an initial time of 0.0200 seconds with an f-number of 8, the constant \( k \) is computed first. Then, this constant allows calculation of the new f-number for a different time, 0.0098 seconds.
The key steps in calculating the f-number:
- Identify the exposure time \( t \) and corresponding f-number.
- Solve for the constant \( k \).
- Use \( k \) to find the f-number for any other given time \( t \).
Constant of Variation
The constant of variation, denoted as \( k \), remains consistent in a direct variation problem and is essential for converting an equation into a practical tool for solving problems.
In our camera setting problem, \( k \) was determined with known values of the f-number and exposure time. Once calculated, \( k \) is used to find the f-number for any other exposure time by substituting back into the equation \( f = k \sqrt{t} \).
By solving \( 8 = k \sqrt{0.0200} \) and finding \( k \approx 56.56 \), the equation confidently adjusts to find an accurate f-number for a different time. This approach exemplifies the power of constant \( k \) in applying theoretical mathematics to practical scenarios.
In our camera setting problem, \( k \) was determined with known values of the f-number and exposure time. Once calculated, \( k \) is used to find the f-number for any other exposure time by substituting back into the equation \( f = k \sqrt{t} \).
By solving \( 8 = k \sqrt{0.0200} \) and finding \( k \approx 56.56 \), the equation confidently adjusts to find an accurate f-number for a different time. This approach exemplifies the power of constant \( k \) in applying theoretical mathematics to practical scenarios.
- Constant \( k \) links the mathematical model to real-world application.
- Knowing \( k \) allows seamless transitions between different scenarios.