Problem 13 - Thomas Calculus

Question

A sliding ladder \(A 13\) -ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 \(\mathrm{ft}\) from the house, the base is moving at the rate of 5 \(\mathrm{ft} / \mathrm{sec}\) . a. How fast is the top of the ladder sliding down the wall then? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle \(\theta\) between the ladder and the ground changing then?

Step-by-Step Solution

Answer

a. -12 ft/sec, b. -34 ft²/sec, c. -5/12 radians/sec.

Step 1: Understand the setup

We're dealing with a right triangle formed by the ladder, the wall, and the ground. Let's define \( x \) as the distance of the base of the ladder from the wall, \( y \) as the height of the ladder on the wall, and \( L = 13 \) ft as the length of the ladder. Use the Pythagorean theorem: \( x^2 + y^2 = L^2 \).

Step 2: Differentiate the Pythagorean theorem

To find how fast the top of the ladder \( y \) is sliding down the wall, differentiate both sides of \( x^2 + y^2 = 13^2 \) with respect to time \( t \). The result is \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \).

Step 3: Solve for dy/dt

Substitute the known values into the differentiated equation. Here, \( x = 12 \), \( \frac{dx}{dt} = 5 \), and using the Pythagorean theorem, \( y = 5 \) (since \( 12^2 + 5^2 = 13^2 \)). Plug these into \( 2 \cdot 12 \cdot 5 + 2 \cdot 5 \cdot \frac{dy}{dt} = 0 \) to find \( \frac{dy}{dt} \). Simplifying gives \( \frac{dy}{dt} = -12 \) ft/sec. The negative sign means \( y \) is decreasing.

Step 4: Finding the rate of change of triangle area

The area of the triangle \( A = 0.5 \cdot x \cdot y \). Differentiate with respect to time: \( \frac{dA}{dt} = 0.5 \left(x \frac{dy}{dt} + y \frac{dx}{dt}\right) \). Substitute \( x = 12 \), \( y = 5 \), \( \frac{dy}{dt} = -12 \), and \( \frac{dx}{dt} = 5 \) into the formula to get the rate of area change: \( \frac{dA}{dt} = 0.5 \cdot (12 \cdot -12 + 5 \cdot 5) = -34 \) ft²/sec.

Step 5: Finding the rate of change of the angle θ

Use the relationship \( \cos(\theta) = \frac{x}{L} \). Differentiate to find \( \frac{d\theta}{dt} \), resulting in \( -\sin(\theta) \frac{d\theta}{dt} = \frac{1}{L} \frac{dx}{dt} \). Find \( \sin(\theta) \) using \( \sin(\theta) = \frac{y}{L} = \frac{5}{13} \) and rearranging gives \( \frac{d\theta}{dt} = -\frac{1}{\sin(\theta)} \frac{\frac{dx}{dt}}{L} \). Substitute the values to find \( \frac{d\theta}{dt} = -\frac{5}{12} \) radians/sec.

Key Concepts

Pythagorean theorem

The Pythagorean Theorem is a cornerstone in geometry, especially when dealing with right triangles. It illustrates a relationship between the lengths of a right triangle's sides. The theorem is expressed as: \[ x^2 + y^2 = L^2 \]Here, \(x\) is the base of the triangle, \(y\) is the height, and \(L\) is the hypotenuse, in our case, the length of the ladder. In this sliding ladder problem, the ladder leans against a wall, forming a right triangle with the ground. As the ladder's base moves, its height on the wall adjusts, maintaining the relationship dictated by the Pythagorean theorem. This equation allows us to determine unknown side lengths when the other two are known. For instance, knowing the hypotenuse and one leg, the triangle's other leg can be calculated. This theorem is foundational for tackling more complex derivative applications related to triangles.

derivative applications

Derivatives are invaluable in dynamics, providing insights into how quantities change over time. In this exercise, we use derivatives to assess how the height of the ladder (\(y\)) changes as its base (\(x\)) moves outward. By differentiating the Pythagorean theorem with respect to time \(t\), we determine the rate at which these variables are changing:\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]The equation tells us how the changes in \(x\) and \(y\) relate when the ladder moves. Given that \( dx/dt = 5 \) ft/sec (the base's speed), and substituting known values, we find the rate \( dy/dt \). The result, \( dy/dt = -12 \) ft/sec, implies the ladder's top drops at this rate, showing the coordinated changes in the triangle's dimensions.

triangle geometry

Triangle geometry helps us understand spatial relationships within triangles, particularly right triangles in this case. The sliding ladder forms a right triangle where:

• The base \(x\) is the distance from the wall,
• The height \(y\) is how high the ladder reaches up the wall,
• The hypotenuse \(L\) is the ladder itself with a fixed length.

Understanding these components allows one to visualize changes as the ladder moves. The area of this triangle can also be calculated using:\[ A = \frac{1}{2} \cdot x \cdot y \]Differentiating with respect to time gives us the rate at which the area changes:\[ \frac{dA}{dt} = \frac{1}{2} \left(x \frac{dy}{dt} + y \frac{dx}{dt}\right) \]Plugging in values illustrates how the triangle's area dynamically changes at \( -34 \) ft²/sec as the ladder slides, reflecting the interplay of its geometric components.

trigonometric relationships

Trigonometry comes into play when analyzing angles in geometry. In our ladder scenario, the angle \(\theta\) between the ladder and the ground shifts as the ladder base moves out.Expressed as:\[ \cos(\theta) = \frac{x}{L} \]where \(x\) is the base of the ladder and \(L\) is the hypotenuse. Differentiating this relationship provides insights into how the angle changes:\[ -\sin(\theta) \frac{d\theta}{dt} = \frac{1}{L} \frac{dx}{dt} \]Knowing \( \sin(\theta) = \frac{5}{13} \) enables us to solve for the rate of change of \(\theta\):\[ \frac{d\theta}{dt} = -\frac{5}{12} \text{ radians/sec} \]This negative value indicates the angle is decreasing. Understanding these trigonometric relationships helps comprehend how angles and their rates of change are influenced by the movement of the ladder, crucial for applying derivatives in geometry.