The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A(x) = 4x(sqrt{1 - x^2}), where x represents the length of half the base of the beam. Determine the cross-sectional area of the beam if the length of half the base of the beam is 1/3 of a foot.
A = 4 (1/3)sqrt (1 - 1/9)
= (4/3)sqrt (8/9)
= (4/3)sqrt (4 *2/9)
= (4/3)(2/3) sqrt(2)
= (8/9)sqrt(2)
To determine the cross-sectional area of the beam when the length of half the base is 1/3 of a foot, substitute x = 1/3 into the function A(x) = 4x(sqrt{1 - x^2}).
A(x) = 4(1/3)(sqrt{1 - (1/3)^2})
Simplifying this expression:
A(x) = 4/3(sqrt{1 - 1/9})
A(x) = 4/3(sqrt{9/9 - 1/9})
A(x) = 4/3(sqrt{8/9})
Now, simplify the square root:
A(x) = 4/3(sqrt{8}/sqrt{9})
Since sqrt{9} = 3:
A(x) = 4/3(sqrt{8}/3)
Next, simplify the square root of 8:
A(x) = 4/3(2sqrt{2}/3)
Finally, multiply:
A(x) = 8sqrt{2}/9
Therefore, the cross-sectional area of the beam when the length of half the base is 1/3 of a foot is 8sqrt{2}/9 square feet.
To determine the cross-sectional area of the beam, we need to substitute the value of x into the function A(x) = 4x(sqrt{1 - x^2}).
Given that x represents half the length of the base of the beam and is 1/3 of a foot, we can substitute x = 1/3 into the function.
A(1/3) = 4(1/3)(sqrt{1 - (1/3)^2})
Now let's simplify this expression step by step:
1. Simplify the term inside the square root:
1 - (1/3)^2 = 1 - (1/9)
= 9/9 - 1/9
= 8/9
2. Take the square root of 8/9:
sqrt(8/9) = sqrt(8)/sqrt(9)
= sqrt(8)/3
3. Substitute this value into the function:
A(1/3) = 4(1/3)(sqrt{8}/3)
Now, let's simplify further:
4/3 * sqrt{8}/3
= (4 * sqrt{8})/9
To simplify further, let's break down sqrt{8} into its prime factors:
sqrt{8} = sqrt{4 * 2)
= sqrt{4} * sqrt{2}
= 2 * sqrt{2}
Substituting this back into the equation:
(4 * (2 * sqrt{2}))/9
= (8 * sqrt{2})/9
So, the cross-sectional area of the beam with a length of half the base of 1/3 foot is (8 * sqrt{2})/9 square feet.