Question

Do dogs know calculus? A mathematician stands on a beach with his dog at point \(A\). He throws a tennis ball so that it hits the water at point \(B\). The dog, wanting to get to the tennis ball as quickly as possible, runs along the straight beach line to point \(D\) and then swims from point \(D\) to point \(B\) to retrieve his ball. Assume \(C\) is the point on the edge of the beach closest to the tennis ball (see figure). a. Assume the dog runs at speed \(r\) and swims at speed \(s,\) where \(r>s\) and both are measured in meters per second. Also assume the lengths of \(B C, C D,\) and \(A C\) are \(x, y,\) and \(z,\) respectively. Find a function \(T(y)\) representing the total time it takes for the dog to get to the ball. b. Verify that the value of \(y\) that minimizes the time it takes to $$\text { retrieve the ball is } y=\frac{x}{\sqrt{r / s+1} \sqrt{r / s-1}}$$ c. If the dog runs at \(8 \mathrm{m} / \mathrm{s}\) and swims at \(1 \mathrm{m} / \mathrm{s}\), what ratio \(y / x\) produces the fastest retrieving time? d. A dog named Elvis who runs at \(6.4 \mathrm{m} / \mathrm{s}\) and swims at \(0.910 \mathrm{m} / \mathrm{s}\) was found to use an average ratio of \(y / x\) of 0.144 to retrieve his ball. Does Elvis appear to know calculus? (Source: T.Pennings, Do Dogs Know Calculus? The College Mathematics Journal, \(34,3,\) May 2003 )

Short answer

Answer: Elvis appears to intuitively minimize his time retrieving the ball as his observed ratio of 0.144 is very close to the calculated ideal ratio of 0.138.

Step-by-step solution

Find distance AD

Note that the triangles \(\triangle ACD\) and \(\triangle CBD\) are similar, as \(\angle ACD = \angle CBD\) (both are right angles) and \(\angle ACB = \angle DCB\). Thus, we have: $$ \frac{AD}{CD} = \frac{CD}{BC} \Longrightarrow AD = CD^2/BC = y^2/x $$

Write the total time function

The total time it takes for the dog to get to the ball can be expressed as the sum of the time it takes to run from A to D and swim from D to B. Thus, we have: $$ T(y) = \frac{AD}{r} + \frac{DB}{s} = \frac{y^2/x}{r} + \frac{\sqrt{x^2 - y^2}}{s} $$ b. Verify that the value of \(y\) that minimizes the time it takes to retrieve the ball is \(y=\frac{x}{\sqrt{r / s+1} \sqrt{r / s-1}}\)

Differentiate T(y) with respect to y

First, let's find the derivative of \(T(y)\) with respect to \(y\): $$ \frac{dT}{dy} = \frac{2y}{xr} - \frac{y}{s\sqrt{x^2 - y^2}} $$

Set the derivative equal to zero and solve for y

To find the minimum time, we set the derivative equal to zero, and solve for \(y\): $$\frac{2y}{xr} - \frac{y}{s\sqrt{x^2 - y^2}} = 0$$ Multiply through by \(xrs\sqrt{x^2 - y^2}\): $$ 2y^2s(x^2 - y^2) - yr^2x\sqrt{x^2 - y^2} = 0 $$ Divide by \(y(x^2 - y^2)\), and rearrange: $$ 2s = r^2\frac{\sqrt{x^2 - y^2}}{x} $$ Square both sides and solve for \(y\): $$ y^2 = \frac{x^2}{\left(\frac{r^2}{2s}\right)^2 + 1} $$ And finally, take the square root: $$ y = \frac{x}{\sqrt{r / s+1} \sqrt{r / s-1}} $$ Thus, we have verified the given formula for \(y\). c. If the dog runs at 8\(/text{m}//text{s}\) and swims at 1\(/text{m}//text{s}\), what ratio \(y / x\) produces the fastest retrieving time?

Calculate y/x using the formula

Plug in \(r=8\) and \(s=1\) into the formula from part b: $$ \frac{y}{x} = \frac{1}{\sqrt{8 + 1} \sqrt{8 - 1}} = \frac{1}{\sqrt{9} \sqrt{7}} $$ d. A dog named Elvis who runs at 6.4\(/text{m}//text{s}\) and swims at 0.910\(/text{m}//text{s}\) was found to use an average ratio of \(y / x\) of 0.144 to retrieve his ball. Does Elvis appear to know calculus?

Calculate the ideal y/x ratio for Elvis

Plug in \(r=6.4\) and \(s=0.910\) into the formula from part b: $$ \frac{y}{x} = \frac{1}{\sqrt{7.037 + 1} \sqrt{7.037 - 1}} \approx 0.138 $$

Compare Elvis's ratio to the ideal ratio

Elvis's observed ratio of 0.144 is very close to the calculated ideal ratio of 0.138. This suggests that Elvis appears to "know" calculus, or at least intuitively, be able to minimize his time retrieving the ball.

Key concepts

Understanding Related Rates

Related rates are a fascinating application of derivatives in calculus, particularly when we observe variables that change over time. An excellent illustration of this concept is the scenario of a mathematician's dog fetching a ball. We have the dog's path divided into two parts: a run on the beach and a swim in the water. As the dog aims for the shortest time to the ball, the rates at which various distances change are interconnected. This leads to a related rates problem where the distance the dog runs and swims are functions of time.

When facing such a problem, it is crucial to establish a relationship between the variables involved. In our example, this includes the distances the dog travels on sand and water and the speeds of running and swimming. To solve the problem, one must write an equation that relates these variables and take its derivative with respect to time. By recognizing how changing one distance impacts the time it takes to fetch the ball, we gain insight into the optimal path the dog should take.

Solving Minimization Problems

Minimization problems are at the core of optimization. They involve finding the lowest possible value of a function, such as time, cost, or distance. Taking the dog's fetch scenario, the goal is to determine the path that minimizes the time to retrieve the ball. To solve such a problem, we first express the quantity to be minimized, in this case, time, as a function of variables we can control, like the distance the dog runs before it swims.

In our example, we achieve minimization using calculus by finding the function's derivative and locating the point where this derivative is zero, indicating a potential minimum. After checking the second derivative for confirmation, we can pinpoint the exact pathway, defined by certain distance ratios, which provides the quickest fetch. While calculus helps find the optimal solution mathematically, the overarching objective is to deduce a practical, applicable course of action that yields the best possible outcome under the given constraints.

Applying Derivatives

Derivatives are more than just a mathematical tool—they're a way to express how something changes in response to various factors. In the context of our example with the dog fetching a ball, the application of derivatives provides a method to calculate the rate of change in time with respect to the distance run on the beach.

By deriving the function that represents the total time taken, we can analyze how slight adjustments in the dog's path alter the total fetch time. This application of derivatives lets us predict and optimize outcomes, exemplifying the powerful role that calculus plays in both theoretical and practical problem-solving. Furthermore, by comparing the dog's actual approach to the fetch with the calculated optimal path, we psychologically probe into whether the dog's instincts align with mathematical efficiency, cheekily questioning if the dog 'knows' calculus.