Question

(a) what radius produces such a disk? (round your answer to four decimal places.) cm
(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
(c) in terms of the ε, δ definition of lim f(x) = l, what is x?
○ area
○ target radius
○ radius
○ target area
○ tolerance in the area
what is f(x)?
○ area
○ target radius
○ radius
○ target area
○ tolerance in the area
what is a?
○ area
○ target radius
○ radius
○ target area

Explanation:

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:
• \(13.8197\)

1. (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:
• \(13.7274\)
• \(13.9116\)

1. (c) Solution:

• In the \(\epsilon-\delta\) definition of \(\lim_{x

ightarrow 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

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:
• \(13.8197\)

1. (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:
• \(13.7274\)
• \(13.9116\)

1. (c) Solution:

• In the \(\epsilon-\delta\) definition of \(\lim_{x

ightarrow 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