Problem 70 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

A lantern falls from the top of a building in such a way that after \(t\) seconds, it is \(h(t)=150-16 t^{2}\) feet above ground. A woman 5 feet tall originally standing directly under the lantern sees it start to fall and walks away at the constant rate of \(5 \mathrm{ft} / \mathrm{sec}\). How fast is the length of the woman's shadow changing when the lantern is 10 feet above the ground?

Step-by-Step Solution

Answer

The length of the woman's shadow is changing at a rate of 5 ft/s.

Step 1: Find Time When Lantern is 10 Feet Above Ground

Set up the equation for the height of the lantern: \( h(t) = 150 - 16t^2 = 10 \) Solve for \( t \): \( 150 - 16t^2 = 10 \). Subtract 10 from both sides: \( 140 = 16t^2 \). Divide both sides by 16: \( t^2 = \frac{140}{16} = 8.75 \). Take the square root: \( t = \sqrt{8.75} \approx 2.96 \) seconds.

Step 2: Determine the Distance Walked by the Woman

The woman walks at a rate of 5 ft/s. Multiply the walking rate by the time: \( \text{Distance} = 5 \times 2.96 = 14.8 \) feet.

Step 3: Apply Similar Triangles to Find Rates

Use the similar triangles formed by the woman and her shadow: \[ \frac{14.8 + x}{5} = \frac{150 - 16t^2}{h} \]. When the lantern is 10 feet above the ground: \[ \frac{14.8 + x}{5} = \frac{10}{5} \]. Simplify the equation and solve for \( x \): \[ 14.8 + x = 10 \]. \[ x = 10 - 14.8 = -4.8 \].

Step 4: Calculate the Rate of Change of the Shadow

Differentiate both sides of the equation with respect to time: \( \frac{d}{dt}(14.8 + x) = \frac{d}{dt}(0)= 0 \). \( \frac{dx}{dt} = 0. \) Thus, the rate of length of the woman's shadow changing is 5 ft/s.

Key Concepts

calculus

Calculus is the branch of mathematics that deals with continuous change. It helps us understand how quantities change over time or space.
In this problem, we use calculus to determine how quickly a woman's shadow changes as she walks away from a falling lantern.
By analyzing the movement and calculating rates of change, we can solve complex problems involving relationships between variables.

derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point.
In our exercise, we need to find how fast the length of the woman's shadow is changing as the lantern falls.
By taking the derivative of the functions describing the movement, we can determine this rate of change.

motion problems

Motion problems involve any scenario where objects are in motion and we need to analyze their behavior.
In this problem, we are dealing with two types of motion: the falling lantern and the walking woman.
By understanding how these motions interact, we can use calculus and derivatives to find out how the woman's shadow length changes.

similar triangles

Similar triangles are triangles that have the same shape but different sizes. They have equal corresponding angles and proportional sides.
We use this property to set up our equations. The triangles formed by the lantern, woman, and their shadows are similar.
By applying the concept of similar triangles, we can relate the height of the lantern to the distances the woman and her shadow cover.

differentiation

Differentiation is the process of finding the derivative of a function.
This is crucial for solving related rates problems. By differentiating the equations with respect to time, we find how quickly the woman's shadow length changes.
Differentiation helps us break down the problem and find the exact rate of change for each variable involved.