Problem 6 - Calculus Volume 1

Question

Sketch the situation if necessary and used related rates to solve for the quantities. A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?

Step-by-Step Solution

Answer

The ladder moves up the wall at \( \frac{3}{4} \) ft/sec after 5 seconds.

Step 1: Understand the Problem

We have a ladder leaning against a wall forming a right triangle with the ground and the wall. The length of the ladder is 25 ft, which is the hypotenuse, and the distance between the base of the ladder and the wall is decreasing at a rate of 1 ft/sec.

Step 2: Set Up the Variables

Let \( x \) be the distance from the wall to the bottom of the ladder, and \( y \) be the height of the ladder on the wall. The relation between \( x \), \( y \), and the hypotenuse (25 ft) is \( x^2 + y^2 = 25^2 \). We know \( \frac{dx}{dt} = -1 \) ft/sec.

Step 3: Use the Pythagorean Theorem

Apply the Pythagorean theorem at the start: \( x = 20 \). So initially, we have \( 20^2 + y^2 = 25^2 \). Thus, \( y = \sqrt{625 - 400} = 15 \) ft initially.

Step 4: Differentiate the Equation with Respect to Time

Differentiate \( x^2 + y^2 = 625 \) with respect to \( t \):\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0. \]

Step 5: Substitute Known Values

Five seconds after pushing, \( x = 20 - 5 = 15 \) ft. Substitute \( x = 15 \), \( y = \sqrt{625 - 225} = 20 \) ft, and \( \frac{dx}{dt} = -1 \) into the differentiated equation:\[ 2(15)(-1) + 2(20) \frac{dy}{dt} = 0. \]

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

Simplify and solve:\[ -30 + 40 \frac{dy}{dt} = 0, \]\[ 40 \frac{dy}{dt} = 30, \]\[ \frac{dy}{dt} = \frac{30}{40} = \frac{3}{4} \text{ ft/sec}. \]

Step 7: Conclusion

The rate at which the top of the ladder moves up the wall 5 seconds after we start pushing the ladder is \( \frac{3}{4} \) ft/sec.

Key Concepts

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry. It relates the sides of a right triangle. In simple terms, it states that for any 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.
For a triangle with sides of length \( a \), \( b \), and hypotenuse \( c \), the theorem is written as: \[ a^2 + b^2 = c^2. \]
In the context of our ladder problem, the ladder acts as the hypotenuse of the right triangle formed with the ground and the wall. Here:

• The ladder length is \( 25 \) ft, so \( c = 25 \).
• The distance from the wall to the bottom of the ladder is \( x \).
• The height of the point on the wall where the ladder touches is \( y \).

Initially, when the base of the ladder is 20 ft from the wall, we used the Pythagorean theorem to calculate the height \( y \) where the ladder connects to the wall: \[ 20^2 + y^2 = 25^2, \] giving us \( y = 15 \) ft.

Differentiation

Differentiation is a tool used in calculus to find the rate at which one quantity changes concerning another. In problems involving related rates, such as this ladder problem, differentiation helps us connect how one part's movement affects another part's movement in real time.
To apply differentiation to this problem:

• We identified \( x \) and \( y \) as functions of time \( t \).
• Differentiating the equation \( x^2 + y^2 = 625 \) with respect to time \( t \) gives:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0. \]

This expression tells us how changes in \( x \) and \( y \) are related over time. It's a way to "see" how pushing the ladder base inward at \(-1\) ft/sec (meaning it's getting closer to the wall) affects how fast the top of the ladder (\( y \)) climbs the wall. Differentiation is essential here because it's not just about the change itself, but about how fast that change happens.

Right Triangles

Understanding right triangles is critical for solving problems that involve the Pythagorean theorem. A right triangle has one angle exactly equal to 90 degrees. This special angle formation makes computation straightforward when applying the theorem.
Regarding the ladder scenario, these right triangles have some characteristics:

• The triangle comprises the wall (height \( y \)), the ground (base \( x \)), and the ladder (hypotenuse \( c = 25 \)).
• The problem initially sets the base at 20 ft and changes over time as we push the ladder base towards the wall.

By understanding these triangles, we're able to visualize and set up equations directly relating to the problem's conditions and variables. When dealing with related rates, comprehending the consistent nature of right triangles helps in visualizing physical changes, making it easier to solve for unknown rates like \( \frac{dy}{dt} \), the climb rate of the ladder on the wall.