Problem 2 - Active Calculus

Question

A street light is at the top of a 12 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of \(8 \mathrm{ft} /\) sec along a straight path. How fast is the tip of her shadow moving when she is 50 feet from the base of the pole? The tip of the shadow is moving at ______ \(\mathrm{ft} / \mathrm{sec}\)

Step-by-Step Solution

Answer

16 ft/sec

Step 1: Understand the problem

We need to find how fast the tip of a woman's shadow is moving as she walks away from a 12-foot tall street light. The woman is 6 feet tall and is moving at a speed of 8 ft/sec.

Step 2: Setup the similar triangles

Since the 12-foot pole, the woman, and their shadows form similar triangles, we can use the properties of similar triangles to set up a relationship. Let's denote:- The distance of the woman from the pole as x.- The length of her shadow as s.- Since the triangles are similar, we have the ratio: \[ \frac{12}{x+s} = \frac{6}{s} \]

Step 3: Solve for s in terms of x

From the ratio \[ \frac{12}{x+s} = \frac{6}{s} \], cross-multiply to get the equation:\[ 12s = 6(x + s) \]which simplifies to \[ 12s = 6x + 6s \]Solving for s, we get:divide both sides by 6: \[ 2s = x + s \]Subtract s from both sides: \[ s = x \]

Step 4: Find the rate of change ds/dt

We are given that \[ \frac{dx}{dt} = 8 \text{ ft/sec} \]. We need to find \[ \frac{d}{dt} (s)\]. Using the relationship found earlier \[ s = x \], differentiate both sides with respect to t:\[ \frac{ds}{dt} = \frac{dx}{dt} \]

Step 5: Calculate the speed at which the tip of the shadow moves

Since the rate \[ \frac{dx}{dt} = 8 \text{ ft/sec} \], and \[ \frac{ds}{dt} = \frac{dx}{dt} \], we have:\[ \frac{ds}{dt} = 8 \text{ ft/sec} \]The rate at which the tip of the shadow moves is the sum of the rates at which the woman and the shadow move:\[ \frac{dx}{dt} + \frac{ds}{dt} = 8 \text{ ft/sec} + 8 \text{ ft/sec} = 16 \text{ ft/sec} \]

Key Concepts

Similar Triangles

In this shadow problem, understanding similar triangles is crucial. Similar triangles are triangles that have the same shape but may differ in size. This means their corresponding angles are equal, and their sides are proportional. In our problem, the 12-foot street light and the 6-foot tall woman form two sets of similar triangles with their respective shadows. By setting up a ratio comparing the height of the pole and the entire length of its shadow to the height of the woman and her shadow, we can establish a relationship that helps solve the problem. Mathematically, this relationship is written as \(\frac{12}{x+s} = \frac{6}{s}\). This ratio allows us to find how the woman's distance from the light pole (x) and the length of her shadow (s) are related.

Rate of Change

Rates of change help us understand how one quantity changes in relation to another over time. In our problem, we're interested in how the length of the shadow changes as the woman walks away from the pole. The speed at which she walks, 8 ft/sec, is her rate of change. We denote this as \( \frac{dx}{dt} \). To find how the shadow length s changes concerning time, we eventually need to differentiate the relationship between x and s with respect to time t. Our goal is to find \( \frac{ds}{dt} \) and understand that these rates will help us determine how fast the tip of her shadow moves, combining the woman's walking speed and the rate her shadow grows.

Differentiation

Differentiation is a calculus technique used to find how a function changes at any given point, essentially giving us the rate of change. Once we have the relationship between x and s from the similar triangles, i.e., \( s = x \), we can use differentiation to find \( \frac{ds}{dt} \). Differentiating both sides of \( s = x \) with respect to time t, we get \( \frac{ds}{dt} = \frac{dx}{dt} \). Given that the woman walks at \( 8 \text{ ft/sec} \), \( \frac{dx}{dt} = 8 \text{ ft/sec} \). Hence, \( \frac{ds}{dt} = 8 \text{ ft/sec} \) as well. The differentiation helps us directly link the change in shadow length to her walking speed.

Shadow Problems in Calculus

Shadow problems in calculus typically involve rates of change and often feature similar triangles, as we've seen. These problems require careful attention to the relationships between different parts of the scenario -- in this case, the woman's height, her shadow's length, and the streetlight height. By setting up the right equations and using differentiation, you can find how these elements change over time. The final step in these problems usually involves summing up different rates: here, it's the woman's walking speed and the rate her shadow lengthens. Both contribute equally to how fast the tip of her shadow moves, which is why we add \( 8 \text{ ft/sec} + 8 \text{ ft/sec} = 16 \text{ ft/sec} \). This kind of methodical approach can be applied to a wide range of related rates problems in calculus.