/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A circle has an initial radius o... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 11

A circle has an initial radius of \(50 \mathrm{ft}\) when the radius begins decreasing at a rate of \(2 \mathrm{ft} / \mathrm{min}\). What is the rate of change of the area at the instant the radius is \(10 \mathrm{ft} ?\)

Short Answer

Expert verified
Answer: The rate of change of the area at the instant when the radius is 10 feet is \(-40\pi \ \mathrm{ft^2/min}\).

Step by step solution

01

Write down the formula of area of a circle

The formula for the area of a circle is given by: \(A = \pi r^2\) Where A is the area, r is the radius of the circle, and \(\pi\) is a constant approximately equal to 3.14159.
02

Determine the given variables

We are given: 1. The initial radius of the circle: \(r_0 = 50 \ \mathrm{ft}\). 2. The rate at which the radius decreases: \(\frac{dr}{dt} = -2 \ \mathrm{ft/min}\) (negative since it is decreasing). 3. We need to find the rate of change of the area at the instant \(r = 10 \ \mathrm{ft}\).
03

Differentiate the area formula with respect to time

To find the rate of change of area (A) with respect to time (t), we will differentiate the area formula with respect to time (t) using the chain rule. \(\frac{dA}{dt} = \frac{d (\pi r^2)}{dt} = \frac{d(\pi r^2)}{dr} \times \frac{dr}{dt} \) where \(\frac{dA}{dt}\) is the rate of change of area, \(\frac{d(\pi r^2)}{dr}\) is the derivative of the area function with respect to the radius, and \(\frac{dr}{dt}\) is the rate of change of the radius.
04

Calculate the derivative with respect to radius

Find the derivative of \(\pi r^2\) with respect to the radius (r): \(\frac{d(\pi r^2)}{dr} = 2\pi r\)
05

Substitute the given values and solve

We now have the necessary derivatives to solve the problem. Substitute the given values into the equation from step 3. \(\frac{dA}{dt} = (2\pi r) \times \frac{dr}{dt}\) When \(r = 10 \ \mathrm{ft}\), \(\frac{dr}{dt} = -2 \ \mathrm{ft/min}\): \(\frac{dA}{dt} = (2\pi (10)) \times (-2)\) \(\frac{dA}{dt} = (-40\pi) \ \mathrm{ft^2/min}\) The rate of change of the area at the instant when the radius is 10 feet is \(-40\pi \ \mathrm{ft^2/min}\).

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