Problem 40 A ladder 26 ft long leans agains... [FREE SOLUTION]

Chapter 3: Problem 40

A ladder 26 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of \(5 \mathrm{ft} / \mathrm{sec}\), how fast is the height of the top decreasing (this will be a negative rate) when the lower end is \(10 \mathrm{ft}\) from the wall?

Short Answer

Expert verified

dy/dt = -25/12 ft/sec

Step by step solution

Understand the setup

Visualize the situation where a ladder is leaning against a wall. Denote the length of the ladder by 26 ft. Let the distance from the wall to the bottom of the ladder be denoted as x and the height of the ladder on the wall be denoted as y.

Apply the Pythagorean theorem

Since the ladder, wall, and ground form a right triangle, apply the Pythagorean theorem: y^2 + x^2 = 26^2This can be rewritten as:y^2 + x^2 = 676

Differentiate with respect to time

Differentiate both sides of the equation with respect to time (t). Remember that x and y are both functions of t.d/dt(y^2 + x^2) = d/dt(676)Apply the chain rule:2y * dy/dt + 2x * dx/dt = 0

Substitute known values

We know that when x = 10 ft, y can be found using the Pythagorean theorem:y^2 + 10^2 = 26^2y^2 + 100 = 676y^2 = 576y = 24 ftAlso, we know that dx/dt = 5 ft/sec (since the lower end is moving away from the wall). Substitute these values into the derived equation:2 * 24 * dy/dt + 2 * 10 * 5 = 0

Solve for dy/dt

Simplify and solve for dy/dt:48 * dy/dt + 100 = 048 * dy/dt = -100dy/dt = -100 / 48dy/dt = -25/12 ft/sec

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Key Concepts

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

Pythagorean Theorem

The Pythagorean theorem is a fundamental concept in geometry that relates the sides of a right triangle. In our ladder problem, the ladder, the wall, and the ground make a right triangle. The theorem states that:

\( a^2 + b^2 = c^2 \)

Here, \( a\) is the height y of the ladder on the wall, \( b\) is the distance x from the wall to the bottom of the ladder, and \( c\) is the length of the ladder, which is 26 feet. So, when we substitute, we get:

\[ y^2 + x^2 = 26^2 \]

By knowing this theorem, we can more easily find unknown lengths, such as the height of the ladder (y) when the bottom is a certain distance from the wall (x).
When x = 10 ft, the height y can be found using \[ y^2 + 10^2 = 676 \]

\[ y^2 = 576 \]
y = 24 \text{ ft}

Differentiation

Differentiation is a key tool used in calculus to find the rate at which one quantity changes with respect to another. In this problem, after establishing the relationship \[ y^2 + x^2 = 676 \], we need to differentiate both sides with respect to time (t) to understand how the height of the ladder (y) changes as the bottom moves away from the wall. Remember, x and y are both functions of t.
Using the differentiation rules, we get:

\[ \frac{d}{dt}(y^2 + x^2) = \frac{d}{dt}(676) \]

This is where the chain rule comes into play, breaking down the differentiation of composite functions step by step.

Chain Rule

The chain rule is a formula used to differentiate composite functions. In our exercise, both y and x are functions of time (t), meaning when we differentiate \[ y^2 + x^2 \] with respect to t, we need to apply the chain rule. We used the following steps:

Differentiate the left side: \ d/dt(y^2) + d/dt(x^2) = \frac{d(y^2 + x^2)}{dt} \
Apply the chain rule: \ 2y \frac{dy}{dt} + 2x \frac{dx}{dt} \

Remember, the right side of the equation remains zero since the derivative of a constant (676) is zero. This can be written as:

\[ 2y \frac{dy}{dt} + 2x \frac{dx}{dt} = 0 \]

The chain rule is crucial for breaking down the differentiation process into manageable parts, making it easier to solve complex calculus problems.

Calculus Problem-Solving

Solving calculus problems involves applying different concepts step by step. Here, to find the speed at which the height of a ladder decreases, we followed these steps:

First, we visualized the problem and set up the Pythagorean relationship.
Next, we differentiated both sides of the equation with respect to time, using the chain rule: \[ 2y \frac{dy}{dt} + 2x \frac{dx}{dt} = 0 \]
We substituted known values: \ y = 24 \ ft, \ dx/dt = 5 \ ft/s
Simplified the equation to solve for \ \frac{dy}{dt} \: \[ 48 \frac{dy}{dt} + 100 = 0 \]
Finally, we solved to find \ \frac{dy}{dt} \: \[ \frac{dy}{dt} = - \frac{100}{48} = - \frac{25}{12} \text{ ft/s} \]

This step-by-step process is crucial for tackling related rates problems in calculus. Each step builds on the previous one to arrive at a comprehensive solution.

91影视

A ladder 26 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of \(5 \mathrm{ft} / \mathrm{sec}\), how fast is the height of the top decreasing (this will be a negative rate) when the lower end is \(10 \mathrm{ft}\) from the wall?

Short Answer

Step by step solution

Understand the setup

Apply the Pythagorean theorem

Differentiate with respect to time

Substitute known values

Solve for dy/dt

Key Concepts

Pythagorean Theorem

Differentiation

Chain Rule

Calculus Problem-Solving

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