Math Problem Statement

A factory is making a machine part whose shape is bounded by the curves f(x) = cosh(x), g(x) = x, x = 0, and x = 3. Sketch these curves and shade the area of the part. Use integration to determine the area of this part. If the part’s mass density is 7 grams/(unit area), what is the mass of the part?

Solution

To solve this problem, let's break it down step-by-step:

1. Understanding the Curves:

  • We are given two functions: f(x)=cosh(x)andg(x)=xf(x) = \cosh(x) \quad \text{and} \quad g(x) = x as well as the vertical boundaries x=0x = 0 and x=3x = 3.

The region of interest is the area bounded by these curves between x=0x = 0 and x=3x = 3.

2. Sketching the Curves:

  • The function cosh(x)\cosh(x) (the hyperbolic cosine) is symmetric and grows exponentially for large values of xx. It looks similar to an upward-opening parabola but grows faster.
  • The function g(x)=xg(x) = x is a straight line passing through the origin with a slope of 1.
  • You are asked to find the area between these two curves from x=0x = 0 to x=3x = 3.

3. Setting Up the Integral:

The area between two curves is given by the formula: Area=ab(f(x)g(x))dx\text{Area} = \int_{a}^{b} \left( f(x) - g(x) \right) \, dx Here, f(x)=cosh(x)f(x) = \cosh(x) and g(x)=xg(x) = x, and the bounds are x=0x = 0 to x=3x = 3. Therefore, the area is: Area=03(cosh(x)x)dx\text{Area} = \int_{0}^{3} \left( \cosh(x) - x \right) \, dx

4. Computing the Integral:

We can now compute the integral: Area=03cosh(x)dx03xdx\text{Area} = \int_{0}^{3} \cosh(x) \, dx - \int_{0}^{3} x \, dx

The first integral is straightforward because the derivative of sinh(x)\sinh(x) is cosh(x)\cosh(x): cosh(x)dx=sinh(x)\int \cosh(x) \, dx = \sinh(x) So: 03cosh(x)dx=sinh(3)sinh(0)=sinh(3)\int_{0}^{3} \cosh(x) \, dx = \sinh(3) - \sinh(0) = \sinh(3)

The second integral is a basic polynomial integral: xdx=x22\int x \, dx = \frac{x^2}{2} So: 03xdx=322022=92\int_{0}^{3} x \, dx = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2}

Putting these together: Area=sinh(3)92\text{Area} = \sinh(3) - \frac{9}{2}

Using the value sinh(3)10.0179\sinh(3) \approx 10.0179, we get: Area10.01794.5=5.5179square units\text{Area} \approx 10.0179 - 4.5 = 5.5179 \, \text{square units}

5. Finding the Mass:

The mass density is given as 7 grams per unit area. Therefore, the mass of the part is: Mass=density×area=7×5.517938.6253grams\text{Mass} = \text{density} \times \text{area} = 7 \times 5.5179 \approx 38.6253 \, \text{grams}

Conclusion:

  • The area of the machine part is approximately 5.518 square units.
  • The mass of the machine part is approximately 38.63 grams.

Would you like more details on the integration steps or visual representation of the curves?

Here are 5 related questions:

  1. How does the graph of cosh(x)\cosh(x) differ from exe^x?
  2. How do you calculate the mass of an object with variable density?
  3. What are the properties of hyperbolic functions like cosh(x)\cosh(x) and sinh(x)\sinh(x)?
  4. How do you find the volume of a 3D solid using integration?
  5. What is the relationship between area under a curve and definite integrals?

Tip: When integrating, it's helpful to break complex regions into smaller, more manageable integrals.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Hyperbolic Functions

Formulas

Area between two curves: A = ∫[a,b] (f(x) - g(x)) dx
Integral of cosh(x): ∫cosh(x) dx = sinh(x)
Mass of a region: Mass = density × area

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or Introductory College Level