The following table represents the diameter of the cross section of a wire at continuous heights (feet) above the ground. Assume that each cross section is circular.
Height(ft) 2 6 10 14 18 22 26 30
Diameter 2 2 2.0 1.8 1.6 1.5 1.3 1.2 (feet)
(those should all line up)
a) Approximate how fast the diameter of the tree is changing 22 ft. above the ground. indicate units of measure
b) Use the trapezoid rule to approximate the volume of the tree from 14 ft to 30 ft above the ground. Indicate units of measure
C) the section of the tree from 2 ft to 8 ft is used to make a rectangular beam of length 6 ft. The strength of the beam varies jointly as its width and the square of its height. What should be the width and height of the beam in order to have the strongest beam?
a) To approximate how fast the diameter of the tree is changing 22 ft above the ground, we can use the concept of average rate of change. We need to find the difference in diameter at heights 22 ft and 26 ft.
The difference in diameter at these two heights is 1.3 - 1.5 = -0.2 ft.
To find the rate of change, we divide this difference by the corresponding change in height. The change in height is 26 ft - 22 ft = 4 ft.
Therefore, the approximate rate of change of the diameter at 22 ft above the ground is -0.2 ft / 4 ft = -0.05 ft/ft. This represents a decrease in diameter of 0.05 ft for every 1 ft increase in height.
b) To approximate the volume of the tree from 14 ft to 30 ft above the ground using the trapezoid rule, we need to sum up the areas of trapezoids formed by the consecutive pairs of heights and diameters.
First, we calculate the areas of each trapezoid:
Area1 = (1/2) * (2 + 2) * (6 - 2) = 4 sq. ft
Area2 = (1/2) * (2 + 2.0) * (10 - 6) = 6 sq. ft
Area3 = (1/2) * (2.0 + 1.8) * (14 - 10) = 8 sq. ft
Area4 = (1/2) * (1.8 + 1.6) * (18 - 14) = 6 sq. ft
Area5 = (1/2) * (1.6 + 1.5) * (22 - 18) = 5.5 sq. ft
Area6 = (1/2) * (1.5 + 1.3) * (26 - 22) = 5.2 sq. ft
Area7 = (1/2) * (1.3 + 1.2) * (30 - 26) = 5.0 sq. ft
Now, we sum up these areas: Total Area = Area1 + Area2 + Area3 + Area4 + Area5 + Area6 + Area7
Total Area = 4 + 6 + 8 + 6 + 5.5 + 5.2 + 5.0 = 40.7 sq. ft
Therefore, the approximate volume of the tree from 14 ft to 30 ft above the ground is 40.7 cubic ft.
c) To find the width and height of the beam to have the strongest beam, we need to maximize the strength of the beam. The strength varies jointly as the width and the square of the height, so we need to maximize the product of the width and the square of the height.
Let the width of the beam be w ft and the height be h ft.
The product of the width and the square of the height is w * h^2.
Given that w = 6 ft, we need to find the value of h that maximizes the expression w * h^2.
To find the maximum value, we can take the derivative of the expression with respect to h, set it equal to zero, and solve for h.
d/dh (w * h^2) = 2w * h
Setting 2w * h = 0, we find that h = 0.
However, a height of 0 ft does not make sense in this context, so there might be a mistake in the problem statement. Please double-check the details or clarify any additional constraints for finding the maximum strength of the beam.