Problem 57 - Essential Calculus

Question

The force \(F\) (in newtons) of a hydraulic cylinder in a press is proportional to the square of \(\sec x,\) where \(x\) is the distance (in meters) that the cylinder is extended in its cycle. The domain of \(F\) is \([0, \pi / 3],\) and \(F(0)=500\). (a) Find \(F\) as a function of \(x\). (b) Find the average force exerted by the press over the interval \([0, \pi / 3]\)

Step-by-Step Solution

Answer

The function representing force in terms of \(x\) is \(F(x) = 500 \cdot \sec^2x\) and the average force over the interval \([0, \(\pi / 3\)]\) needs to be evaluated by using the definite integral of the function \(F(x)\) over this interval.

Step 1: Derive \(F(x)\)

We know that \(F(x) = k \cdot \sec^2x\), where \(k\) is the constant of proportionality. Using the given condition, \(F(0)=500\), we can solve for \(k\):\nSo, when \(x = 0\), \(\sec 0 = 1\). Substituting these values into \(F(x) = k \cdot \sec^2x\), we get \(500 = k \cdot (1^2) = k\). Thus, \(k = 500\). So, the function \(F(x)\) is \(F(x) = 500 \cdot \sec^2x\).

Step 2: Compute the interval length

The length of the interval [0, \(\pi / 3\)] is \(\pi / 3 - 0 = \pi / 3\).

Step 3: Calculate the average force

The formula for the average value of a function \(f(x)\) over an interval [a, b] is \(f_{ave} = \frac{1}{b-a}\int_a^b f(x) dx.\) Let \(f(x) = F(x)\), \(a = 0\), and \(b = \pi / 3\). Substituting these values into the formula, we need to evaluate \(F_{ave} = \frac{1}{\pi / 3}\int_0^{\pi / 3} 500 \cdot \sec^2x dx.\)

Key Concepts

Force Function

A force function helps describe how much force is exerted by a system over a particular variable. In this problem, the force \( F(x) \) exerted by a hydraulic cylinder is proportional to the square of the secant of the angle \( x \). This means that as the angle or displacement changes, the force applied does too. We express this relationship as:

• \( F(x) = k \cdot \sec^2x \)

where \( k \) is the constant of proportionality. To find this constant, we use the provided information that \( F(0) = 500 \). Since \( \sec 0 = 1 \), we substitute these values to get \( 500 = k \cdot 1^2 \), solving for \( k \), we find \( k = 500 \). Consequently, the force function becomes \( F(x) = 500 \cdot \sec^2x \). This means as the cylinder extends, the force changes according to the secant function, with the amplitude scaled by 500.

Hydraulic Systems

Hydraulic systems use fluids to increase force, often seen in machinery like presses and lifts. They work by transferring force applied at one point to another via incompressible fluids, usually oil or water. Thus, they allow for the magnification of force in mechanical systems. In the given problem, the hydraulic system in question involves a cylinder where force exertion is a key component. The force exerted by the hydraulic cylinder depends on the position of the cylinder, which is an example of how hydraulic systems apply force dynamically depending on mechanical configurations. This type of application is critical in industrial settings as it provides precise control over movement and force exertion, facilitating tasks like lifting heavy objects or compressing materials efficiently. This dynamic force application is why understanding the force function is integral for designing and utilizing hydraulic systems.

Average Value of Function

The average value of a function gives us an idea of the "typical" value of a function over a certain interval. It is particularly useful in contexts like physics or engineering where we want to know a representative quantity over time or a spatial domain.To calculate this, you take the integral of the function over the interval and divide by the interval's length. The formula is:

• \( f_{ave} = \frac{1}{b-a}\int_a^b f(x) \, dx \)

Applying this to our force function \( F(x) = 500 \cdot \sec^2x \) over the interval \([0, \pi / 3]\), the length of this interval is \( \pi / 3 \). Therefore, the average force \( F_{ave} \) exerted is:\[F_{ave} = \frac{1}{\pi / 3}\int_0^{\pi / 3} 500 \cdot \sec^2x \, dx\]Evaluating this integral will provide the average force, representing a constant equivalent force exerted by the hydraulic cylinder over the specified extension. This measure helps for a clearer understanding of the overall mechanical performance within that range.