Problem 15 - The Calculus with Analytic Geometry

Question

A ladder \(20 \mathrm{ft}\) long is leaning against an embankment inclined \(60^{\circ}\) to the horizontal. If the bottom of the ladder is being moved horizontally toward the embankment at \(1 \mathrm{ft} / \mathrm{sec}\), how fast is the top of the ladder moving when the bottom is \(4 \mathrm{ft}\) from the embankment?

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \frac{\sqrt{6}}{12} \approx 0.204 \mathrm{ft/sec} \)

Step 1: Define Variables and Given Information

Let the bottom of the ladder be point A and the top of the ladder be point B. Let the distance from the bottom of the ladder (point A) to the embankment be x(t) and the height of the ladder on the embankment (point B) be y(t). Given: length of the ladder (L) is 20 ft, the angle of inclination is 60°, and the rate at which the bottom is moving toward the embankment is \( \frac{dx}{dt} = -1 \).

Step 2: Write the Relationship between the Variables

Since the ladder forms a right triangle with the embankment, use the Pythagorean theorem: \( x^2 + y^2 = L^2 \). Substituting L = 20 ft: \( x^2 + y^2 = 400 \).

Step 3: Implicit Differentiation

Differentiate both sides of \( x^2 + y^2 = 400 \) with respect to t: \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \). Simplify to get: \( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \).

Step 4: Solve for the Rate of Change of y

Solve for \( \frac{dy}{dt} \): \( \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \).

Step 5: Find y when x = 4 ft

Substitute x = 4 ft into \( x^2 + y^2 = 400 \) to find y: \[ 4^2 + y^2 = 400 \] \[ 16 + y^2 = 400 \] \[ y^2 = 384 \] \[ y = \sqrt{384} = 8\sqrt{6} \].

Step 6: Substitute Values and Solve

Substitute x = 4 ft, \( y = 8 \sqrt{6} \), and \( \frac{dx}{dt} = -1 \) into the differentiated equation: \[ \frac{dy}{dt} = -\frac{4}{8\sqrt{6}} \times (-1) \] \[ \frac{dy}{dt} = \frac{1}{2 \sqrt{6}} \].

Step 7: Simplify the Result

Simplify \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = \frac{\sqrt{6}}{12} \approx 0.204 \mathrm{ft/sec} \].

Key Concepts

Pythagorean Theorem

The Pythagorean theorem is an essential principle in geometry that relates the sides of a right-angled triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The formula is \[ a^2 + b^2 = c^2 \] where \( c \) is the hypotenuse and \( a \) and \( b \) are the other two sides. In this problem, our ladder forms a right triangle with the embankment, allowing us to apply this theorem. The ladder itself is the hypotenuse with a length of 20 feet, and the sides are distances \( x(t) \) (the distance from the bottom to the embankment) and \( y(t) \) (the height on the embankment). By substituting these into the theorem, we get \[ x^2 + y^2 = 20^2 = 400 \]. This relationship is crucial for connecting the distances as the ladder moves.

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used when dealing with equations where the variables are interdependent. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to a third variable, usually time \( t \). In the case of the ladder problem, we used implicit differentiation on \[ x^2 + y^2 = 400 \]. Differentiating both sides with respect to \( t \), using the chain rule, we obtain:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]. Here, \( \frac{dx}{dt} \) is the rate at which the bottom of the ladder moves, and \( \frac{dy}{dt} \) is the rate at which the top moves. This step allows us to relate the two rates of change and solve for unknowns easily.

Rate of Change

The rate of change tells us how one quantity changes in relation to another. In this exercise, we're interested in how fast the top of the ladder is moving down the embankment as the bottom is pushed towards it. The given rate of movement of the bottom is \( \frac{dx}{dt} = -1 \) (since it is moving towards the embankment). We need to find \( \frac{dy}{dt} \) when \( x = 4 \) feet. First, we used the Pythagorean theorem to find \( y \) at \( x = 4 \): \[ 4^2 + y^2 = 400 \] \[ y = 8\sqrt{6} \]. Then, substituting into our differentiated equation, we found:\[ \frac{dy}{dt} = -\frac{4}{8\sqrt{6}} \times -1 = \frac{1}{2 \sqrt{6}} \approx 0.204 \text{ ft/sec} \]. This tells us that at the moment when the ladder's bottom is 4 feet from the embankment, the top is sliding down at approximately 0.204 feet per second.