A semi-cylindrical groove of radius r = 20 cm is made on a horizontal floor. An ant wants to cross t

Q: Why does the vertex of the wires lie exactly on the semi-cylinder?

By Thales's theorem, the angle subtended by a diameter at any point on the circumference of a semicircle is exactly $90^\circ$. Since the wires form a right angle and rest on the edges (which are $2r$ apart), the vertex must lie on the semicircle.

Q: Why did we choose the downward path to be $x$ and not $y$?

The ant travels at $2v$ downwards and $v$ upwards. To minimize the total time, the ant should cover the larger distance at the faster speed. Thus, we assign the larger segment to the downward journey.

Q: Why did the derivative give a maximum instead of a minimum?

The function for time is a sum of a linear term and a concave-down square root term. Its second derivative is strictly negative, meaning the curve is shaped like an inverted U. The critical point where the slope is zero naturally corresponds to the peak (maximum time).

Q: How did we find the boundary limits for $x$?

The boy only has wires of length $l = 38 \text{ cm}$. Therefore, neither segment inside the groove can exceed $l$. Since $x^2 + y^2 = 4r^2$, setting $y \le l$ forces $x \ge \sqrt{4r^2 - l^2}$.

Q: What if the wires were infinitely long?

If the wires were infinitely long, the minimum time would occur as $x$ approaches $2r$ and $y$ approaches $0$. The ant would essentially drop straight down the diameter and climb up a negligible distance, taking time $\frac{2r}{2v} = \frac{r}{v}$.

Question

A semi-cylindrical groove of radius $r = 20 	ext{ cm}$ is made on a horizontal floor. An ant wants to cross the groove. A boy decides to help the ant making a bridge consisting of straight wire segments. But all the wires available are of length $l = 38 	ext{ cm}$, so the boy rigidly connects two wires at right angle and places the bridge in the groove as shown in the figure. If the ant can crawl up a wire segment at speed $v = 0.5 	ext{ cm/s}$ and down a wire segment at speed $2v$, in what minimum time can the ant cross the groove with the help of this bridge?

Sigma Percentile · Accepted Answer

### The Geometric Illusion

Imagine a semi-cylindrical groove carved into a horizontal floor. The distance between its edges is exactly its diameter, $2r = 40 	ext{ cm}$. A boy places two wires, each of length $l = 38 	ext{ cm}$, joined at a right angle, to help an ant cross.

Here is a beautiful piece of geometry: because the wires form a $90^\circ$ angle and rest on the edges of the groove, Thales's theorem dictates that the vertex of this right angle must lie exactly on the surface of the semi-cylindrical groove! The locus of points subtending a right angle at the ends of a diameter is the semicircle itself.

### Setting Up the Kinematics

Let the lengths of the wire segments inside the groove be $x$ and $y$. Since they form a right-angled triangle with the diameter as the hypotenuse, we can apply the Pythagorean theorem:

$$x^2 + y^2 = (2r)^2 = 4r^2$$

The ant wants to cross as fast as possible. It crawls down at a speed of $2v$, and up at a speed of $v$. To minimize time, it should obviously travel the longer distance downwards. So, the total time $t$ is the sum of the time taken for each segment:

$$t = \frac{x}{2v} + \frac{y}{v}$$

Let's express $y$ in terms of $x$ using our constraint. Substituting $y = \sqrt{4r^2 - x^2}$, we get our time function entirely in terms of $x$:

$$t(x) = \frac{x}{2v} + \frac{\sqrt{4r^2 - x^2}}{v}$$

### The Calculus Trap

To find the minimum, we differentiate $t$ with respect to $x$ and set it to zero:

$$\frac{dt}{dx} = \frac{1}{2v} - \frac{x}{v\sqrt{4r^2 - x^2}} = 0$$

Rearranging the terms and squaring both sides, we find that $x = \frac{2r}{\sqrt{5}}$. But wait... is this really a minimum?

There is a massive catch here. If you check the second derivative, you will realize that this value of $x$ actually gives the **maximum** time, not the minimum! The calculus has found the absolute worst possible path for the ant. The function is shaped like an inverted U, and the critical point is its peak.

### The Boundary Solution

Since the maximum is in the middle, the minimum must lie at the extreme boundaries of our domain. What are the limits for $x$? The wires are only $l$ centimeters long. So $y$ cannot exceed $l$, which means $x$ must be at least $\sqrt{4r^2 - l^2}$.

So, what should we do to minimize the time? Since the ant is faster going downwards, we want it to cover as much distance as possible on the downward stretch. Therefore, we will set $x$ to its maximum possible value, which is the full length of the wire, $l$. This gives us our expression for the minimum time:

$$t_{min} = \frac{l}{2v} + \frac{\sqrt{4r^2 - l^2}}{v}$$

Let's substitute the values and get the answer. We plug in $l = 38$, $r = 20$, and $v = 0.5$:

$$t_{min} = \frac{38}{1} + \frac{\sqrt{1600 - 1444}}{0.5} = 38 + 2\sqrt{156} \approx 63 	ext{ s}$$

And with that, the ant makes it across in the fastest possible way!

Visualized Solution

\text{Geometric Setup}

\text{Right Angle in a Semicircle}

\text{Constraint Equation}

\text{Time of Travel}

\text{Single Variable Function}

\text{Finding Critical Points}

\text{Solving for } x

\text{The Calculus Trap}

\text{Checking Boundaries}

\text{Minimum Time}

\text{Final Calculation}

The Sigma Insight: Motion in a Straight Line

The Geometric Illusion

Setting Up the Kinematics

The Calculus Trap

The Boundary Solution

Similar Questions

A ball is thrown up with a certain velocity, so that it reaches a height $h$ . Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

A material point moving along a straight line enters an $87.5 m$ segment with speed $5.0 m/s$ and leaves with speed $10 m/s$ . The particle crosses the segment with unidirectional acceleration that never exceeds $1.0 m/s^{2}$ . Find range of average acceleration of the point on this segment?

A small block slides without friction down an inclined plane starting from rest. Let $s_{n}$ be the distance travelled from $t = n - 1$ to $t = n$ . Then, $\frac{s _{n}}{s _{n + 1}}$ is

In a car race on a straight path, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the cars start from rest and travel with constant acceleration $a_{1}$ and $a_{2}$ respectively. Then 'v' is equal to

A helicopter rises from rest on the ground vertically upwards with a constant acceleration $g$ . A food packet is dropped from the helicopter when it is at a height $h$ . The time taken by the packet to reach the ground is close to (Here, $g$ is the acceleration due to gravity).