(a) what radius produces such a disk? (round your answer to four decimal places.) cm\n(b) if the machinist…

(a) what radius produces such a disk? (round your answer to four decimal places.) cm\n(b) if the machinist is allowed an error tolerance of ±8 cm² in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (round your answers to four decimal places.) cm < r < cm\n(c) in terms of the ε, δ definition of lim f(x) = l, what is x?\n○ area\n○ target radius\n○ radius\n○ target area\n○ tolerance in the area\nwhat is f(x)?\n○ area\n○ target radius\n○ radius\n○ target area\n○ tolerance in the area\nwhat is a?\n○ area\n○ target radius\n○ radius\n○ target area

(a) what radius produces such a disk? (round your answer to four decimal places.) cm\n(b) if the machinist is allowed an error tolerance of ±8 cm² in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (round your answers to four decimal places.) cm < r < cm\n(c) in terms of the ε, δ definition of lim f(x) = l, what is x?\n○ area\n○ target radius\n○ radius\n○ target area\n○ tolerance in the area\nwhat is f(x)?\n○ area\n○ target radius\n○ radius\n○ target area\n○ tolerance in the area\nwhat is a?\n○ area\n○ target radius\n○ radius\n○ target area

Answer

  1. (a) Solution:
    • The area formula of a disk is (A = \pi r^{2}). Assume the area (A = 600\ cm^{2}).

    • Explanation:

      • Step1: Rearrange the area - formula for radius

        • Given (A=\pi r^{2}), we can solve for (r) as (r=\sqrt{\frac{A}{\pi}}).

      • Step2: Substitute (A = 600) into the formula

        • (r=\sqrt{\frac{600}{\pi}}\approx\sqrt{\frac{600}{3.14159}}\approx\sqrt{190.9859}\approx13.8197\ cm).

    • Answer:

      • (13.8197)
  2. (b) Solution:
    • The area formula is (A=\pi r^{2}). We know that the error in area (\Delta A=\pm8\ cm^{2}).
    • First, when (A_1 = 600 - 8=592\ cm^{2}), then (r_1=\sqrt{\frac{592}{\pi}}\approx\sqrt{\frac{592}{3.14159}}\approx\sqrt{188.4407}\approx13.7274\ cm).
    • Second, when (A_2 = 600 + 8 = 608\ cm^{2}), then (r_2=\sqrt{\frac{608}{\pi}}\approx\sqrt{\frac{608}{3.14159}}\approx\sqrt{193.5326}\approx13.9116\ cm).

    • Explanation:

      • Step1: Find the lower - bound radius

        • Set (A = 592) in (r=\sqrt{\frac{A}{\pi}}), so (r_1=\sqrt{\frac{592}{\pi}}).

      • Step2: Find the upper - bound radius

        • Set (A = 608) in (r=\sqrt{\frac{A}{\pi}}), so (r_2=\sqrt{\frac{608}{\pi}}).

    • Answer:

      • (13.7274)
      • (13.9116)
  3. (c) Solution:
    • In the (\epsilon-\delta) definition of (\lim_{x\rightarrow a}f(x)=L), for the disk - area problem where (A=\pi r^{2}).
    • The variable (x) represents the radius (r), (f(x)) represents the area (A) (since the area is a function of the radius (A = f(r)=\pi r^{2})), and (a) represents the target radius (the radius corresponding to the ideal area).

    • Brief Explanations:

      • The (\epsilon-\delta) definition is used for limits. Here, the area (A) depends on the radius (r). The input variable for the function that gives the area is the radius.

    • Answer:

      • For "What is (x)?" - radius
      • For "What is (f(x))?" - area
      • For "What is (a)?" - target radius