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.

Short answer

(a) Vertical velocity is \(-\frac{2}{\sqrt{3}}\) ft/sec. (b) Vertical acceleration is \(-\frac{16}{27\sqrt{3}}\) ft/sec².

Step-by-step solution

Understand the Problem

We have a ladder leaning against a wall, forming a right triangle. The ladder's length is fixed at 18 feet as the hypotenuse. The vertical wall is 12 feet high, and we need to find how the vertical position (top of the ladder) and its rate changes as the base is moved away.

Establish Relationships

When the ladder makes an angle of \(60^\circ\) with the ground, we need to determine the length of the base (x) and the height (y). Using the trigonometric relationship: \(\cos\theta = \frac{x}{h}\), where \(h = 18\). We have:\[ x = h \cos(60^\circ) = 18 \times \frac{1}{2} = 9 \, \text{feet} \]Use \(\sin\theta = \frac{y}{h}\) to find \( y \):\[ y = h \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \, \text{feet} \]

Differentiate and Find Vertical Velocity

The rate of change of the base \(\frac{dx}{dt} = 2\, \text{ft/sec}\). We use implicit differentiation on \(x^2 + y^2 = h^2\). Differentiating:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]Solve for \(\frac{dy}{dt}\):\[ y \frac{dy}{dt} = -x \frac{dx}{dt} \quad \Rightarrow \quad \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]Substitute \(x = 9\), \(y = 9\sqrt{3}\), and \(\frac{dx}{dt} = 2\):\[ \frac{dy}{dt} = -\frac{9}{9\sqrt{3}} \times 2 = -\frac{2}{\sqrt{3}} \, \text{ft/sec} \]

Differentiate Again for Vertical Acceleration

Differentiate \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\) again. The goal is to find \(\frac{d^2y}{dt^2}\):\[ \frac{d}{dt}(y \frac{dy}{dt}) = \frac{d}{dt}( -x \frac{dx}{dt}) \]Using product rule, on both sides, we get:\[ y \frac{d^2y}{dt^2} + \left(\frac{dy}{dt}\right)^2 = - (\frac{dx}{dt})^2 - x \frac{d^2x}{dt^2} \]Since \( \frac{d^2x}{dt^2} = 0\) (constant velocity), solve for \(\frac{d^2y}{dt^2}\):\[ y \frac{d^2y}{dt^2} = - (\frac{dx}{dt})^2 - \left(\frac{dy}{dt}\right)^2 \]Substitute the known values and solve:\[ 9\sqrt{3} \frac{d^2y}{dt^2} = - (2)^2 - \left(-\frac{2}{\sqrt{3}}\right)^2 \]\[ 9\sqrt{3} \frac{d^2y}{dt^2} = -4 - \frac{4}{3} \]\[ 9\sqrt{3} \frac{d^2y}{dt^2} = -\frac{16}{3} \]\[ \frac{d^2y}{dt^2} = -\frac{16}{27\sqrt{3}} \, \text{ft/sec}^2 \]

Conclusion

The vertical velocity of the top of the ladder when the angle is \(60^\circ\) is \(-\frac{2}{\sqrt{3}}\) ft/sec, and the vertical acceleration is \(-\frac{16}{27\sqrt{3}}\) ft/sec².

Key concepts

Trigonometry

Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. In this problem, trigonometric functions like cosine and sine help us relate the lengths of the ladder, the distance from the wall, and the height against the wall.

The cosine function is particularly useful here because it relates the adjacent side (the base of the ladder) to the hypotenuse (the ladder itself). When given an angle, such as \(60^\circ\), we can quickly find how far from the wall the base of the ladder is by calculating \(x = h \cos(60^\circ)\). This tells us that when the ladder makes a \(60^\circ\) angle with the ground, the base of the ladder is 9 feet away from the wall.

Similarly, the sine function helps us with the height along the wall, using the formula \(y = h \sin(60^\circ)\). With the given angle, this height comes out to be \(9\sqrt{3}\) feet. In problems like these, identifying which trigonometric function to use based on known quantities and desired measurements is crucial.

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find derivatives of functions that are not isolated on one side of the equation. It comes in handy when dealing with relationships between variables like in this problem of the ladder and the wall.

In the equation \(x^2 + y^2 = h^2\), where \(h\) is the constant length of the ladder, we differentiate both sides with respect to time \(t\). This gives rise to terms involving \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), which signify the rates of change of the base and height of the triangle formed by the ladder.

Applying the concept of implicit differentiation, we obtain \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\). Solving for \(\frac{dy}{dt}\) allows us to find how quickly the vertical position of the ladder's top changes as it slides away from the wall. Understanding this technique is key to analyzing the dynamics of systems where change happens simultaneously across multiple dimensions or constraints.

Kinematics

Kinematics is the study of motion without considering the forces that cause the motion. In this problem, kinematics is used to analyze how the ladder moves as its base is pulled away from the wall.

We apply kinematic principles to understand the relationship between the horizontal and vertical velocities of the ladder's ends. With the base moving at a constant speed of \(2\,\text{ft/sec}\), we examined how this affects the vertical speed of the ladder's top, which is found to be \(-\frac{2}{\sqrt{3}}\, \text{ft/sec}\). This negative value indicates that as the ladder slides away, the top is moving downward.

Similarly, for acceleration, we differentiate the velocity to find the vertical acceleration \(\frac{d^2y}{dt^2}\), which gives us insight into how the vertical motion is changing over time. Kinematics allows us to predict future states or positions by understanding these rates of change.

Ladder Problems

Ladder problems, like the one described here, are classic examples in calculus and physics, focusing on the concepts of related rates and right triangle geometry. They highlight how a change in one dimension can impact another when variables are related geometrically.

The ladder is treated as a hypotenuse of a right triangle, where movements of its base or top can be analyzed through a set of equations. By employing trigonometric identities and differentiation, one determines how shifting the ladder's base, for instance, affects the height it reaches against a wall.

These problems often involve the application of trigonometry, implicit differentiation, and kinematics, making them an excellent practice for understanding real-world applications of mathematics, as seen in structures and mechanical movements.