Question

An angler hooks a trout and reels in his line at 4 in / s. Assume the tip of the fishing rod is 12 ft above the water and directly above the angler, and the fish is pulled horizontally directly toward the angler (see figure). Find the horizontal speed of the fish when it is \(20 \mathrm{ft}\) from the angler.

Short answer

Answer: The horizontal speed of the fish when it is 20 feet away from the angler is -11/30 feet per second.

Step-by-step solution

Convert all measurements to the same unit

To make calculations easier, we will convert all measurements to feet. We are given that the angler is reeling in his line at a rate of 4 in/s, which is equal to \(\frac{1}{3} \mathrm{ft/s}\) (\(\frac{4}{12} = \frac{1}{3} \) since there are 12 inches in a foot).

Set up the Pythagorean theorem

Let the distance between the angler and the fish be x, and the distance between the tip of the fishing rod and the fish be y. Then, we have the following equation using the Pythagorean theorem: $$y^2 = x^2 + 144$$

Differentiate the equation with respect to time

Now, let's differentiate both sides of the equation with respect to time (t): $$\frac{d}{dt}(y^2) = \frac{d}{dt}(x^2 + 144)$$ We get: $$2y\frac{dy}{dt} = 2x\frac{dx}{dt}$$

Plug in the known values and solve for the horizontal speed

We are given that the angler is reeling in his line at a rate of \(\frac{1}{3}\,ft/s\), and we want to find out the horizontal speed of the fish, which is \(\frac{dx}{dt}\). We can set up the following equation: $$2y\left(-\frac{1}{3}\right) = 2x\frac{dx}{dt}$$ Since we want to find the horizontal speed of the fish when it is 20 ft from the angler, we plug in \(x = 20\,ft\) and solve for \(y\) using the Pythagorean theorem: $$y^2 = 20^2 + 144$$ So, $$y = \sqrt{20^2 + 144} = 22\,ft$$ Now, we can plug in \(x=20\) and \(y=22\) into our equation and solve for \(\frac{dx}{dt}\): $$2(22)\left(-\frac{1}{3}\right) = 2(20)\frac{dx}{dt}$$ Now, solve for \(\frac{dx}{dt}\): $$\frac{dx}{dt} = \frac{22}{20}\left(-\frac{1}{3}\right) = -\frac{11}{30}\,ft/s$$

Interpret the result

The horizontal speed of the fish when it is 20 ft away from the angler is \(-\frac{11}{30}\,ft/s\). The negative sign indicates the fish is moving towards the angler, which is expected since the angler is reeling in his line.

Key concepts

Related Rates

Related rates are a powerful tool in calculus used to find the rate at which one quantity changes with respect to another related quantity. When two or more quantities are linked through formulas or relationships, altering one quantity will affect the others.

In the given problem, the distance from the fishing rod's tip to the fish and the distance of the fish from the angler are connected via the Pythagorean theorem. As the angler reels in the line, the lengths change simultaneously over time. Therefore, we can say that these quantities' rates of change are linked. This is the core idea behind related rates.

• Calculate the rate of change in one variable using another related variable.
• Set up an equation involving all related variables.
• Differentiation with respect to time reveals how these related quantities change over time.

Related rates problems often involve using differentiation rules and algebra to find unknown changing rates that impact the scenario being examined.

Differentiation with Respect to Time

To discover how quantities vary over time, such as distances or speeds, we use differentiation with respect to time. This method involves taking the derivative of each component of an equation individually concerning time.

In our exercise, the equation from the Pythagorean theorem, \( y^2 = x^2 + 144 \) represents the relationship between the diagonal (line being reeled in) and the horizontal distance of the fish from the angler. By differentiating both sides concerning time \( t \), we obtain:

\( \frac{d}{dt}(y^2) = \frac{d}{dt}(x^2 + 144) \)

This process results in the following:

• \( 2y\frac{dy}{dt} = 2x\frac{dx}{dt} \)

This equation shows us how the vertical and horizontal rate of change of the line relates. By substituting known values, we can solve for the rate of change of the x-distance, thereby finding the fish's horizontal speed. Differentiation allows us to calculate related rates effectively, turning static equations into dynamic ones.

Horizontal Speed Calculation

Calculating horizontal speed in the context of this problem involves solving for the rate at which the distance between the fish and the angler changes as the fish is reeled in. Once the related rates are established through differentiation, we can plug in the given values to find the specific horizontal speed.

After differentiating the equation, substituting the known values \( y = 22 \) ft, \( \frac{dy}{dt} = -\frac{1}{3} \) ft/s, and \( x = 20 \) ft into:

• \( 2(22)\left(-\frac{1}{3}\right) = 2(20)\frac{dx}{dt} \)
We simplify to find the horizontal speed or \( \frac{dx}{dt} \):

\( \frac{dx}{dt} = -\frac{11}{30} \) ft/s

The negative sign indicates the fish is moving toward the angler, which makes sense since the angler is reeling the line in. Calculating horizontal speed in related rates problems involves evaluating the outcome of changes over time using given or derived simultaneous rates in the setup.