Problem 4 - Calculus: Early Transcendentals

Question

A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of \(0.6 \mathrm{~m} / \mathrm{sec} .\) How fast is the top sliding down the wall when the foot of the ladder is \(5 \mathrm{~m}\) from the wall?

Step-by-Step Solution

Answer

The top of the ladder slides down at a rate of \(-\frac{1}{4}\) m/s.

Step 1: Visualizing the Problem

Imagine a right triangle formed by the ladder, the ground, and the wall. The ladder acts as the hypotenuse, the distance from the wall to the foot of the ladder forms one leg, and the height up the wall forms the other leg.

Step 2: Assign Variables

Let \( x \) be the distance from the wall to the foot of the ladder on the ground, and \( y \) be the height of the ladder on the wall. Given that \( \frac{dx}{dt} = 0.6 \text{ m/s} \) and \( x = 5 \text{ m} \), we want to find \( \frac{dy}{dt} \) when \( x = 5 \).

Step 3: Use Pythagorean Theorem

According to the Pythagorean theorem, you have \( x^2 + y^2 = 13^2 \). Differentiate this equation with respect to time \( t \) to create a relationship between the rates of change of \( x \) and \( y \).

Step 4: Differentiate with Respect to Time

Differentiate \( x^2 + y^2 = 169 \) with respect to \( t \): \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0. \] Simplify to \( x \frac{dx}{dt} + y \frac{dy}{dt} = 0. \)

Step 5: Substitute Known Values

Substitute \( x = 5 \), \( \frac{dx}{dt} = 0.6 \), and find \( y \) using the Pythagorean theorem with \( x = 5 \): \( y^2 = 169 - 25 = 144 \) so \( y = 12 \). Substitute into the differentiated equation: \( 5(0.6) + 12 \frac{dy}{dt} = 0. \)

Step 6: Solve for \( \frac{dy}{dt} \)

Rearrange to solve for \( \frac{dy}{dt} \): \[ 3 + 12 \frac{dy}{dt} = 0 \] \( \Rightarrow \frac{dy}{dt} = -\frac{3}{12} = -\frac{1}{4} \) m/s. The negative sign indicates that \( y \), the height, is decreasing.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in mathematics, especially in geometry. It deals with right triangles and establishes the relationship between the lengths of the sides. In every right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be mathematically expressed as:\[ x^2 + y^2 = z^2 \]where:

• \( x \) and \( y \) are the lengths of the two legs of the triangle.
• \( z \) is the length of the hypotenuse.

In the context of our problem, the ladder acts as the hypotenuse \( z = 13 \text{ m} \), one leg \( x \) is the distance from the wall to the bottom of the ladder, and the other leg \( y \) is how high the ladder reaches up the wall. To find any of these, knowing two values and applying the theorem helps bridge the gap by making the equation solvable. The ability to realize such a relationship is key in solving many spatial problems.

differentiation with respect to time

Differentiation with respect to time is a technique used in calculus to find how a quantity changes over time. This is vital for solving problems involving related rates, where two or more quantities change at once. In our example, we have distance \( x \) and height \( y \), both dependent on time.To perform differentiation with respect to time on an equation like:\[ x^2 + y^2 = 169 \]we apply the chain rule:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]Here:

• \( \frac{dx}{dt} \) is the rate of change of \( x \), given as 0.6 m/s.
• \( \frac{dy}{dt} \) is the rate of change of \( y \), which we want to find.

This equation allows us to relate the rates at which \( x \) and \( y \) change over time. With known measurements, you can plug values into this differentiated equation to find unknowns such as \( \frac{dy}{dt} \).

right triangle geometry

Right triangle geometry is a special area of geometry focused on triangles with one 90-degree angle. Understanding the properties of right triangles efficiently unlocks the ability to solve various geometrical problems. Key components include:

• **Hypotenuse**: The longest side, opposite the right angle.
• **Legs**: The two shorter sides forming the right angle.

For any right triangle:

• The Pythagorean theorem holds true, helping relate all side lengths.
• Trigonometric functions such as sine, cosine, and tangent also spring from relationships in right triangles.

In our specific problem, placing the ladder correctly uses right triangle geometry to visualizing it, where the ladder stands as the hypotenuse with its foot placed horizontally and the top against the wall. Being clear about such spatial arrangements allows accurate setting up of equations, offering paths to quickly reach solutions.