Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

An 18 -foot ladder leans against a 12 -foot vertical wall, its top extending over the wall. The bottom end of the ladder is pulled along the ground away from the wall at 2 feet per second. (a) Find the vertical velocity of the top end when the ladder makes an angle of \(60^{\circ}\) with the ground. (b) Find the vertical acceleration at the same instant.

Step-by-Step Solution

Answer

(a) \(-\frac{1}{\sqrt{3}}\) feet per second, (b) \(0\) feet per second squared.

Step 1: Set up the Problem with Known Values

Let's denote:- The length of the ladder as \( L = 18 \) feet (this is constant).- \( x \) as the horizontal distance from the wall to the bottom of the ladder.- \( y \) as the vertical height from the ground to the top of the ladder.- The rate at which the bottom of the ladder is being pulled away from the wall as \( \frac{dx}{dt} = 2 \) feet per second.We need to find \( \frac{dy}{dt} \) when the ladder makes an angle of \(60^{\circ}\) with the ground, and \( \frac{d^2y}{dt^2} \) for the same condition.

Step 2: Express Relationship Using Trigonometry

Use trigonometric relations for the angle \( \theta = 60^{\circ} \).The cosine of the angle gives:\[ \cos(60^{\circ}) = \frac{x}{18} \]\[ \frac{1}{2} = \frac{x}{18} \]Solving gives \( x = 9 \) feet.The sine of the angle gives:\[ \sin(60^{\circ}) = \frac{y}{18} \]\[ \frac{\sqrt{3}}{2} = \frac{y}{18} \]Solving gives \( y = 9\sqrt{3} \) feet.

Step 3: Use Pythagorean Theorem

Using the Pythagorean theorem, for any position of the ladder:\[ x^2 + y^2 = 18^2 \]Differentiating both sides with respect to \( t \) gives:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

Step 4: Solve for Vertical Velocity \( \frac{dy}{dt} \)

Substitute known values from earlier:\[ 2(9)(2) + 2(9\sqrt{3}) \frac{dy}{dt} = 0 \]Solving for \( \frac{dy}{dt} \) gives:\[ 18 + 18\sqrt{3} \frac{dy}{dt} = 0 \]\[ \frac{dy}{dt} = -\frac{1}{\sqrt{3}} \text{ feet per second} \]So the top of the ladder is descending at approximately \( -1.732 \) feet per second.

Step 5: Calculate Vertical Acceleration \( \frac{d^2y}{dt^2} \)

To find acceleration, differentiate \( \frac{dy}{dt} \) with respect to time:We have:\[ y \frac{d^2y}{dt^2} = -x \frac{d^2x}{dt^2} \]For \( \frac{d^2x}{dt^2} = 0 \) because \( \frac{dx}{dt}\) is constant at \( 2 \) ft/s:\[ y \frac{d^2y}{dt^2} = 0 \]Therefore, \( \frac{d^2y}{dt^2} = 0 \text{ feet per second squared} \).

Key Concepts

Trigonometry

Trigonometry plays a key role in this problem when determining the position of the ladder. When the ladder makes an angle of \(60^{\circ}\) with the ground, trigonometric functions help us establish relationships between the ladder's length, the height it reaches on the wall, and its distance from the wall.

• Cosine: The cosine of the angle helps us find the horizontal distance \(x\) from the wall. Using the relation \(\cos(60^{\circ}) = \frac{x}{18}\) and knowing \(\cos(60^{\circ}) = \frac{1}{2}\), we calculate that \(x = 9\) feet.


• Sine: Similarly, the sine function helps determine the vertical height \(y\) of the ladder on the wall. Using \(\sin(60^{\circ}) = \frac{y}{18}\) and the known value \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\), we find \(y = 9\sqrt{3}\) feet.

These trigonometric equations are essential for solving the exercise as they provide the necessary geometric parameters to proceed with calculus.

Pythagorean Theorem

The Pythagorean Theorem is fundamental when dealing with right triangles, such as the one formed by the ladder, the wall, and the ground. This theorem expresses the relation \(x^2 + y^2 = L^2\), where \(L\) is the constant length of the ladder.

• Equation Setup: Using the theorem, we get \(x^2 + y^2 = 18^2\), implying \(x^2 + y^2 = 324\).


• Differentiation: To connect changes over time, we differentiate with respect to \(t\), giving: \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\).

The Pythagorean theorem helps transition from static geometric relationships to dynamic contexts where rates of change need to be calculated.

Vertical Velocity

Vertical Velocity relates to how fast the top of the ladder is moving along the wall. Given that the bottom moves horizontally, the focus is on the vertical speed at the moment the angle is \(60^{\circ}\).

• Substitution: Use known values \(x = 9\), \(y = 9\sqrt{3}\), and \(\frac{dx}{dt} = 2\) ft/s into the differentiated Pythagorean equation \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\).


• Solving for Velocity: We substitute these into the equation, yielding: \(18 + 18\sqrt{3} \frac{dy}{dt} = 0\), leading to \(\frac{dy}{dt} = -\frac{1}{\sqrt{3}}\) ft/s which approximates to \(-1.732\) ft/s.

This negative sign indicates that the top of the ladder is descending, which clarifies the real-world motion observed.

Calculus Problem Solving

Calculus provides the tools necessary to resolve this related rates problem efficiently. Given the continuous change in the ladder's position, calculus helps express how speeds are interrelated.

• Related Rates: We use derivatives to link the rate at which the ladder's bottom is pulled away (horizontal velocity) to the rate at which the top descends (vertical velocity).


• Acceleration Analysis: By differentiating vertical velocity, we understand that vertical acceleration is zero, given that \(\frac{dx}{dt}\) is constant and \(\frac{d^2x}{dt^2} = 0\). Thus, \(\frac{d^2y}{dt^2} = 0\), implying the top's speed remains consistently descending without further acceleration.

Calculus techniques enable a deeper understanding beyond static calculations, allowing both velocity and acceleration assessments for complete problem descriptions.