Question: EXAMPLE 1 A plate cam for opening and closing a valve is shown below. Widths of the face of…

EXAMPLE 1

A plate cam for opening and closing a valve is shown below. Widths of the face of the cam are shown at 2.00-cm intervals from one end of the cam to the other. Find the area of the face of the cam using the best rule for the given measurements and the shape. Round to 3 significant digits.

[Diagram of the plate cam with measurements]

• Widths at 2.00-cm intervals: 2.00 cm, 3.82 cm, 3.25 cm, 2.95 cm, 1.85 cm, 0.00 cm

Solution

To find the area of the cam, we can use the trapezoidal rule since the measurements are at regular intervals. The trapezoidal rule is used to approximate the area under a curve by dividing it into trapezoids. Given intervals and widths: - \( x_0 = 2.56 \, \text{cm} \) - \( x_1 = 3.82 \, \text{cm} \) - \( x_2 = 3.25 \, \text{cm} \) - \( x_3 = 2.95 \, \text{cm} \) - \( x_4 = 1.85 \, \text{cm} \) - \( x_5 = 0.00 \, \text{cm} \) The interval width, \( h \), is 2 cm. Apply the trapezoidal rule formula: \[ A = \frac{h}{2} \left( x_0 + 2(x_1 + x_2 + x_3 + x_4) + x_5 \right) \] Calculate: \[ A = \frac{2}{2} \left( 2.56 + 2(3.82 + 3.25 + 2.95 + 1.85) + 0.00 \right) \] \[ A = 1 \times \left( 2.56 + 2(11.87) \right) \] \[ A = 2.56 + 23.74 \] \[ A = 26.30 \, \text{cm}^2 \] Rounded to three significant digits, the area is \( 26.3 \, \text{cm}^2 \).