/*! 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 35 A hot-air balloon is \(150 \math... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 35

A hot-air balloon is \(150 \mathrm{ft}\) above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going \(40 \mathrm{mi} / \mathrm{hr}\) \((58.67 \mathrm{ft} / \mathrm{s}) .\) If the balloon rises vertically at a rate of \(10 \mathrm{ft} / \mathrm{s}\) what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?

Short Answer

Expert verified
Answer: The rate of change of the distance between the motorcycle and the hot-air balloon after 10 seconds is approximately 60.44 ft/s.

Step by step solution

01

1. Determine the positions of the balloon and motorcycle at t=10 seconds

In order to find the rate of change of the distance between the motorcycle and the balloon at t=10 seconds, we first need to find their positions at that time. After 10 seconds, the height of the balloon above the ground will be: Initial height + (rate of ascent 脳 time elapsed) = 150 ft + (10 ft/s 脳 10 s) = 250 ft The horizontal distance traveled by the motorcycle after 10 seconds will be: Horizontal speed 脳 time elapsed = 58.67 ft/s 脳 10 s = 586.7 ft
02

2. Relate the distances using the Pythagorean theorem

Let d(t) be the distance between the balloon and the motorcycle at time t. At t=10 seconds, we can use the Pythagorean theorem to relate the horizontal distance (x), the vertical distance (y), and the distance between the two objects (d): x^2 + y^2 = d^2 At t=10 seconds, x=586.7 ft and y=250 ft. Plug in these values into the equation to get: (586.7)^2 + (250)^2 = d^2
03

3. Differentiate the equation with respect to time

To find the rate of change of the distance d with respect to time, we need to differentiate both sides of the equation with respect to time t: \frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(d^2) Using the chain rule, we get: 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2d\frac{dd}{dt} Notice that the rates \frac{dx}{dt} and \frac{dy}{dt} are the horizontal speed of the motorcycle (58.67 ft/s) and the vertical speed of the balloon (10 ft/s), respectively.
04

4. Plug in the values and solve for the rate of change of distance

Now we can plug in the values of x, y, \frac{dx}{dt}, and \frac{dy}{dt}, and solve for the rate of change of the distance between the motorcycle and the balloon, \frac{dd}{dt}: 2(586.7)(58.67) + 2(250)(10) = 2d\frac{dd}{dt} First, let's calculate d from our earlier equation: d^2 = (586.7)^2 + (250)^2 d = sqrt((586.7)^2 + (250)^2) 鈮� 633.49 ft Now we can plug this value into the equation: 2(586.7)(58.67) + 2(250)(10) = 2(633.49)\frac{dd}{dt} Solve for \frac{dd}{dt}: \frac{dd}{dt} = \frac{2(586.7)(58.67) + 2(250)(10)}{2(633.49)} 鈮� 60.44 ft/s The rate of change of the distance between the motorcycle and the balloon after 10 seconds is approximately 60.44 ft/s.

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