A stick AB of length l stands vertically on a horizontal floor leaning on a wall. A beetle P starts

Q: Why do we maximize the square of the height instead of the height itself?

Maximizing a positive function $f(x)$ and its square $(f(x))^2$ yields the exact same critical points. Squaring removes the messy square root, making the differentiation much simpler and significantly less prone to algebraic errors.

Q: How did we get the equation $y_P = \frac{s}{l} y_A$?

Imagine a large right triangle formed by the stick, the wall, and the floor. The beetle's position forms a smaller, similar right triangle inside it. The ratio of their hypotenuses (the beetle's distance $s$ to the total stick length $l$) is equal to the ratio of their vertical sides ($y_P$ to $y_A$).

Q: What happens if the beetle reaches the top of the stick?

The problem states the upper end A does not leave the wall. If the beetle reaches A, it can't climb any higher relative to the stick. From that point on, its height is just the height of point A, which is continuously decreasing as the stick slides down.

Q: Why is there a condition $u \le v\sqrt{2}$?

Calculus gives us the mathematical peak of the parabolic trajectory of the beetle's height. However, the stick has a finite length $l$. If the beetle climbs too fast, it runs out of stick before this mathematical peak is reached in time.

Q: Can the beetle's height be zero?

Yes! At $t=0$, the beetle is at the floor, so its height is zero. Also, if the stick completely falls flat (which takes time $l/v$), the height would be zero again, assuming the beetle hasn't been squashed!

Question

A stick AB of length $l$ stands vertically on a horizontal floor leaning on a wall. A beetle P starts climbing the stick from the floor. When the beetle starts climbing, the lower end B of the stick is made to move away from the wall with a constant velocity $v$. The beetle climbs the stick with a constant speed $u$ relative to the stick. If the upper end A does not leave the wall, what maximum height can the beetle rise?

Sigma Percentile · Accepted Answer

### Setting the Stage: The Sliding Stick

Imagine a stick of length $l$ leaning against a wall. Suddenly, the bottom end starts slipping away at a constant velocity $v$. This is a classic constrained motion problem. To tackle it, we first need to establish a solid coordinate system. Let's place the corner where the wall meets the floor right at the origin $(0,0)$.

As the lower end B moves away, its position on the x-axis at any time $t$ is simply $x_B = vt$. Because the stick is rigid and its length $l$ is constant, the upper end A must slide down the wall. Using the Pythagorean theorem, we can express the y-coordinate of point A as:

$$y_A = \sqrt{l^2 - x_B^2} = \sqrt{l^2 - v^2t^2}$$

This equation perfectly captures the downward trajectory of the top end as time ticks on.

### The Beetle's Journey: Kinematics in Action

Now, let's introduce our protagonist: a little beetle P. It starts at the bottom end B and climbs up the stick with a constant speed $u$ relative to the stick itself. The distance it covers along the stick in time $t$ is $s = ut$.

We want to find the maximum height the beetle reaches above the floor. Let's call this height $y_P$. If we look closely, the stick, the wall, and the floor form a large right-angled triangle. The beetle's position forms a smaller, similar right-angled triangle. By the properties of similar triangles, the ratio of the beetle's height to the top end's height is equal to the ratio of its distance along the stick to the total length of the stick:

$$y_P = \frac{s}{l} y_A = \frac{ut}{l} \sqrt{l^2 - v^2t^2}$$

This is our master equation. It describes the beetle's height at any given moment.

### The Calculus of Maximum Height

To find the peak of the beetle's trajectory, we need to maximize $y_P$. Differentiating a function with a square root can be algebraically messy. A brilliant mathematical trick is to maximize the square of the function instead. Let's define $f(t) = y_P^2$:

$$f(t) = \frac{u^2}{l^2} (l^2 t^2 - v^2 t^4)$$

Now, we take the derivative of $f(t)$ with respect to time and set it to zero to find the critical points:

$$f'(t) = \frac{u^2}{l^2} (2l^2 t - 4v^2 t^3) = 0$$

Factoring out $2t$, we get $2t (l^2 - 2v^2 t^2) = 0$. Since we are interested in a time $t > 0$, we solve the term inside the parentheses:

$$t = \frac{l}{v\sqrt{2}}$$

Substituting this critical time back into our height equation yields the unconstrained maximum height:

$$h_{max} = \frac{u}{l} \left(\frac{l}{v\sqrt{2}}\right) \sqrt{l^2 - v^2 \left(\frac{l^2}{2v^2}\right)} = \frac{lu}{2v}$$

### The Hidden Catch: Physical Constraints

Calculus is powerful, but it doesn't know the physical limits of our stick! The mathematical peak we just found is only valid if the beetle is actually still on the stick at that exact moment. The distance $s$ must be less than or equal to the total length $l$.

Let's check this constraint by plugging in our critical time:

$$ut \le l \implies u \left(\frac{l}{v\sqrt{2}}\right) \le l \implies u \le v\sqrt{2}$$

This tells us that the beetle will only experience that smooth, parabolic peak if its climbing speed $u$ is relatively slow compared to the stick's sliding speed $v$.

### The Fast Beetle Scenario

What if the beetle is a speed demon and $u > v\sqrt{2}$? In this scenario, the beetle reaches the very top of the stick (point A) before the stick has had a chance to slide down to the mathematical peak.

The maximum height is therefore achieved at the exact moment the beetle reaches the top, which occurs at time $t = \frac{l}{u}$. The height at this moment is simply the height of point A:

$$h_{max} = y_A\left(\frac{l}{u}\right) = \sqrt{l^2 - v^2\left(\frac{l}{u}\right)^2} = l\sqrt{1 - \frac{v^2}{u^2}}$$

Combining both scenarios gives us a beautifully complete, piecewise solution that perfectly models the physics of the situation.

Visualized Solution

$C oor d ina t e S e t u p$

$P os i t i o n o f U pp er E n d A$

$P os i t i o n o f B ee tl e P$

$H e i g h t o f t h e B ee tl e$

$M a x imi z in g t h eH e i g h t$

$D i f f er e n t ia t in g t h e F u n c t i o n$

$T im eo f M a x im u m H e i g h t$

$U n co n s t r ain e d M a x im u m H e i g h t$

$T h e P h y s i c a l C o n s t r ain t$

$B o u n d a r y C o n d i t i o n$

$F ina l P i ece w i se S o l u t i o n$

The Sigma Insight: Relative Velocity

Setting the Stage

The Beetle's Journey

The Calculus of Maximum Height

The Hidden Catch

The Fast Beetle Scenario

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Comprehension Passage

When and for how long can the particle C fulfil the given condition?

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What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_{1}$ . If he remains stationary on a moving escalator, then the escalator takes him up in time $t_{2}$ . The time taken by him to walk up on the moving escalator will be

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CoordinateSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PositionofUpperEndA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PositionofBeetleP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } HeightoftheBeetle

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UnconstrainedMaximumHeight

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