/*! 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 25 A 5-foot-tall woman walks at 8 f... [FREE SOLUTION] | 91Ó°ÊÓ

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A 5-foot-tall woman walks at 8 ft/s toward a streetlight that is \(20 \mathrm{ft}\) above the ground. What is the rate of change of the length of her shadow when she is \(15 \mathrm{ft}\) from the streetlight? At what rate is the tip of her shadow moving?

Short Answer

Expert verified
Answer: When the woman is 15 ft away from the streetlight, the rate of change of the length of her shadow is \(\frac{32}{3}\) ft/s, and the tip of her shadow is moving at a rate of \(\frac{8}{3}\) ft/s.

Step by step solution

01

Draw a diagram and label given information

Begin by drawing a diagram of the streetlight, the woman, and her shadow. Label the height of the streetlight as 20 ft and the height of the woman as 5 ft. Label the distance between the woman and the streetlight as x ft (initially 15). Now, label the length of the shadow as s ft. We know that the woman is walking at a rate of 8 ft/s towards the streetlight, or dx/dt = -8 ft/s, as it's a decreasing distance.
02

Set up the similar triangles and an equation

In this situation, we have two similar triangles: the larger triangle formed by the streetlight, the ground, and the tip of the shadow, and the smaller triangle formed by the woman, the ground, and her shadow. Set up a proportion using the height and shadow length: \[\frac{5}{s} = \frac{20}{x+s}\]
03

Solve the equation for s

In order to find the rate of change of the length of her shadow (ds/dt), we must first find the value of s when x = 15 ft. Start by plugging x = 15 into the proportion derived from the similar triangles: \[\frac{5}{s} = \frac{20}{15+s}\] Next, cross-multiply and simplify to isolate s: \[5(15+s) = 20s\] \[75+5s = 20s\] \[15s = 75\] \[s = 5\] So, the length of her shadow when she is 15 feet away is 5 ft.
04

Differentiate the proportion with respect to time

Now, we differentiate both sides of the proportion equation with respect to time, t: \[\frac{d}{dt}(\frac{5}{s}) = \frac{d}{dt}(\frac{20}{x+s})\] This yields: \[\frac{-5}{s^2} \frac{ds}{dt} = \frac{-20}{(x+s)^2} (\frac{dx}{dt} + \frac{ds}{dt})\]
05

Plug in values and solve for ds/dt

Substitute x = 15, s = 5, and dx/dt = -8 into the equation: \[\frac{-5}{5^2} \frac{ds}{dt} = \frac{-20}{(15+5)^2} (-8 + \frac{ds}{dt})\] Now, solve for ds/dt: \[\frac{-1}{5}\frac{ds}{dt} = \frac{-20}{400}(8+\frac{ds}{dt})\] Simplify and solve for ds/dt: \[\frac{1}{5}ds/dt = \frac{1}{2}(8+\frac{ds}{dt})\] \[\frac{ds}{dt} = 4(8+\frac{ds}{dt})\] \[\frac{ds}{dt} = 32+4\frac{ds}{dt}\] \[3\frac{ds}{dt} = 32\] \[\frac{ds}{dt} = \frac{32}{3}\] The rate of change of the length of her shadow is \(\frac{32}{3}\) ft/s when she is 15 ft from the streetlight.
06

Find the rate at which the tip of her shadow is moving

To find the rate at which the tip of her shadow is moving, we need to consider both how fast the woman is walking and how fast her shadow is getting shorter. The tip of her shadow is moving as the sum of the speeds of the woman and the shadow: \[Rate_{tip} = \frac{dx}{dt} + \frac{ds}{dt}\] Substituting the given and calculated rates: \[Rate_{tip} = -8 + \frac{32}{3}\] Thus, the tip of her shadow is moving at a rate of \(\frac{8}{3}\) ft/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Similar Triangles
In this shadow problem, similar triangles play an essential role. When two triangles have the same shape but different sizes, they are referred to as similar triangles. This means that their corresponding angles are equal, and the ratios of their corresponding sides are proportional.
In our exercise, we have two triangles:
  • The larger triangle, formed by the streetlight's height, the ground, and the tip of the woman's shadow.
  • The smaller triangle, formed by the woman's height, the ground, and the end of her shadow.
By establishing a proportion based on these triangles, we can precisely calculate shadow lengths and determine how variables change with time.
Given the proportion \(\frac{5}{s} = \frac{20}{x+s}\), where 5 feet is the woman's height, 20 feet is the streetlight's height, and \(s\) and \(x\) are the shadow and distance she travels toward the light, respectively.
Differentiation
Differentiation is a core concept in calculus used to determine how a function changes as its input changes. It's incredibly useful in problems involving rates because it helps us calculate how quickly something is increasing or decreasing over time.
In this scenario, after establishing a formula using similar triangles, the next step involves differentiating that equation with respect to time, \(t\). This process allows us to find the rate at which the length of the shadow \(s\) changes (\(\frac{ds}{dt}\)).
The derivative of both sides of our equation in Step 4 gives us:\[\frac{-5}{s^2} \frac{ds}{dt} = \frac{-20}{(x+s)^2} \left(\frac{dx}{dt} + \frac{ds}{dt}\right)\]
By substituting the initial conditions and known rates, we solve this equation for \(\frac{ds}{dt}\), illustrating how differentiation helps highlight dynamic aspects in real-world situations.
Shadow Problems
Shadow problems are classic related rates exercises in calculus, often involving moving objects and their shadows cast by a light source. These problems typically examine the interplay of various distances, rates, and angles as objects move relative to a light source.
In our case, the shadow length changes dynamically as the woman walks toward the streetlight. By setting up the problem with a clear understanding of similar triangles, differentiating appropriately, and substituting known values, we can find the precise rate at which the shadow's length changes over time.
Key elements include:
  • Understanding of proportions related to the light source, the person, and their shadow.
  • Using related rates to connect how these quantities change as time progresses.
  • Simplification through calculus to solve for desired rates, such as the shadow's shrinking rate or the tip's moving speed.
By breaking down the problem in logical steps, shadow problems elegantly illustrate the application of calculus in real-world situations, aiding students in mastering related rates concepts.

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