Question

Moving Shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light as shown below. How fast is the ball's shadow moving along the ground 1\(/ 2\) sec later? (Assume the ball falls a distance \(s=16 t^{2}\) in \(t\) sec. $)

Short answer

The ball's shadow is moving along the ground at a speed of 8 ft/sec towards the pole.

Step-by-step solution

Set Up the Problem

In establishing a relationship between the variables, let's denote the distance of the ball from the ground as \(y\) (which changes as the ball falls), the distance of the shadow from the base of the pole as \(x\) and the height of the pole as \(H = 50\) ft. We know that a right triangle is formed by the pole, the ball and the point directly under the ball on the ground. Another right triangle is formed by the pole, the point directly under the ball on the ground and the end of the shadow of the ball. These two right triangles are similar since they share an angle and their right angles.

Derive a Relationship between Variables

By similarity of triangles, we have \(\frac{H-y}{x-30} = \frac{H}{x}\). Cross multiplication gives us \(Hx - yx = H(H-y)\), which simplifies to \((H-y)x = H(H-y)\). We have \(y = H - s\) (since \(s\) is the distance fallen by the ball), and \(x = \frac{H^2}{H-y}\).

Differentiate the Equation

Differentiating the equation \(x = \frac{H^2}{H-y}\) with respect to \(t\), we obtain \(\frac{dx}{dt} = -\frac{H^2 \cdot ds}{dt}\times\frac{1}{(H-y)^2}\). We know that \(\frac{ds}{dt} = 32t\) (differentiating \(s = 16t^2\)) and we substitute \(y = H - s\). So, \(\frac{dx}{dt} = -\frac{H^2 \cdot 32t}{(H - H + 16t^2)^2}\).

Solve for x

Substituting the given values \(H = 50\) ft and \(t = 0.5\) sec into the equation, we get the speed at which the shadow along the ground is moving: \(\frac{dx}{dt} = -\frac{50^2 \cdot 32 \cdot 0.5}{(50 - 50 + 16 \cdot 0.5^2)^2} = 8\) ft/sec. The negative sign indicates the direction of motion, which in this case means that the shadow is moving towards the pole.

Key concepts

Similarity of Triangles

Understanding the concept of similar triangles is essential in solving many calculus problems, especially in related rates scenarios involving geometric shapes. Two triangles are considered similar if they have the same shape but not necessarily the same size. This occurs when their corresponding angles are equal and the sides have the same ratio.

In the context of the given problem, the similarity of two right triangles comes into play. The first triangle is formed by the pole, the ball's current position, and the ball's projection onto the ground. The second triangle includes the pole, the ball's projection on the ground, and the tip of the shadow on the ground. These triangles are similar as they both have a right angle and share the angle at the pole, meaning the ratios of their corresponding sides are proportional.

To use this to our advantage, we set up a proportion based on the 'similar triangles', leading to the key equation \(\frac{H-y}{x-30} = \frac{H}{x}\). This proportion is the backbone of finding the relationship between various changing quantities in the problem at hand, and it's through this relationship that we can eventually find the rate at which the ball's shadow moves along the ground. The understanding of similarity is critical because it provides a bridge to connect different quantities in a changing scene, which are often the focus in related rates problems.

Differentiating Equations

When we differentiate equations in related rates problems, we're applying the concept of finding the instantaneous rate of change of one variable with respect to another. This process is at the heart of calculus and is pivotal for solving related rates problems.

In the initial solution to the exercise, differentiating the equation \(x = \frac{H^2}{H-y}\) requires us to use the chain rule—a fundamental instrument in differentiation. The chain rule helps us deal with the derivative of a composite function. In layman's terms, if a variable we're interested in is buried inside a function which is within another function (like a Russian doll), the chain rule helps us unpack and differentiate it systematically.

The differentiation step in our problem leads to \(\frac{dx}{dt}\) which is the rate at which the shadow's length changes over time. We use known derivatives such as \(\frac{ds}{dt}\) from given equations, like \(s = 16t^2\), to find the required rate of change \(\frac{dx}{dt}\). Precise differentiation is crucial, as any mistakes in this step can lead to incorrect outcomes, thwarting the purpose of the exercise.

Shadow Movements in Calculus

Shadow problems in calculus demonstrate the real-world applications of related rates, as they showcase how various rates of change influence each other in a dynamic setting. The movement of a shadow can be a complex interplay of different factors such as the light source, the object casting the shadow, and the relative positions of these elements.

In the provided exercise, the movement of the ball's shadow is driven by the falling motion of the ball. As the ball drops, the triangle it forms with its projection on the ground and the tip of the shadow changes shape, causing the shadow to move. The rate at which the shadow moves along the ground is what we need to find, which is a classic example of a related rates problem in calculus. To solve this, we use the principles of similar triangles to establish the relationship between the distances, then employ calculus to differentiate and find the rate at which the shadow is shrinking - moving closer to the pole.

These problems require us to visualize the scenario, understand the geometric relationships, implement the correct differentiation techniques, and finally interpret the sign and magnitude of our result to describe the motion of the shadow accurately.