A 6 meter ladder...

Question

A $6$ meter long ladder leans with a vertical wall and top of the ladder is 3 meters above the ground.If it slips at a rate of $2$ m/s then how fast the level is decreasing from the wall?

My attempt:First i draw the picture which is right triangle with hypotenuse $6$ and opposite $3$ then by pythagorean theorem i found base is $3\sqrt3$.If i suppose base is x and opposite is y,then what i have to calculate $\frac{dx}{dt}$ or $\frac{dy}{dt}$ and further how i can do this?

Answer 1

The picture shows your problem (note: not necessarily to scale). $x$ is the distance from the base of the ladder to the wall, and $y$ is the height of the ladder on the wall. Also, you know by the Pythagorean theorem that $x^2+y^2=36$. You may differentiate both sides with respect to $t$ (I presume you know how to do this, keeping in mind that $x$ and $y$ are functions of $t$). You will get an equation in terms of $x$, $y$, $\frac {dx}{dt}$, and $\frac {dy}{dt}$. You know three of these, so solving for the one you seek is trivial at this point.

Let me know if you have further questions!

Answer 2

Let $x=x(t)$ be the distance of the foot of the ladder from the foot of the building, and let $y$ be the height of the top of the ladder. Then by the Pythagorean Theorem we have $x^2+y^2=36$.

Differentiation of $x^2+y^2=6^2$ gives $2x\frac{dx}{dt}+2y \frac{dy}{dt}=0$, and now we can answer any of the typical qeustions.

Answer 3

Thanks so what i understand is

$x^2+y^2=36$

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$

$x=3\sqrt{3}\hspace{0.03cm},y=3\hspace{0.03cm},\frac{dx}{dt}=2$

Therefore $\hspace{0.03cm}$$\frac{dy}{dt}=-2\sqrt{3}$

Answer 4

I'd like to give a full solution $\dot{y}(t)$ not only for the initial situation at time $t=0$.

Let's also start using the Pythagorean theorem:

$$ x^2 + y^2 = 36 \Rightarrow x\cdot\dot{x} + y\cdot\dot{y} = 0 $$

But now let's recall the fact that $x=x(t)$ and $y=y(t)$ are not constant but are actually functions of the time. Therefore deriving another time leads to:

$$ \dot{x}^2 + \dot{y}^2 + y\cdot\ddot{y} = 0 $$

Using the first equation we can re-express y in terms of $x, \dot{x}, \dot{y}$ :

$$ y = -\frac{x\cdot\dot{x}}{\dot{y}} \Rightarrow \dot{x}^2 + \dot{y}^2 - \frac{x\cdot\dot{x}}{\dot{y}}\cdot\ddot{y} = 0 $$

Since we know that $\dot{x}=2$ is constant we can write $x = x_0 + \dot{x}\cdot t$ and we obtain:

$$ \frac{\dot{x}^2\cdot t + x_0\cdot\dot{x}}{\dot{y}}\cdot\ddot{y} - \dot{y}^2 - \dot{x}^2 = 0 \hspace{1cm}, \hspace{0.3cm} \textrm{where} \hspace{0.5cm} \dot{x} = 2 \hspace{0.3cm} \textrm{and} \hspace{0.3cm} x_0 = 3\sqrt{3} $$

The constant solution $\dot{y} = c$ leads to $c = 2i$ which is imaginary and therefore not what we are interested in. Two other solutions are (using Mathematica or WolframAlpha):

$$ \dot{y}_{1,2}(t) = \pm \frac{i\cdot\dot{x}^2\cdot \exp(c\cdot\dot{x}^2)\cdot(x_0 + \dot{x}\cdot t)}{\sqrt{x_0^2\cdot\dot{x}^2\cdot\exp(c\cdot 2\dot{x}^2) - 1 + 2x_0\cdot\dot{x}^3\cdot\exp(c\cdot 2\dot{x}^2)\cdot t + \dot{x}^4\cdot\exp(c\cdot 2\dot{x}^2)\cdot t^2}} $$

We are only interested in the $+$ solution since the ladder will move downwards the wall not upwards.

There are three remarkable things about the solution:

• the solution includes a constant $c$ which has to be determined using the boundary conditions, i.e. $\dot{y}(0) = -2\sqrt{3}$ as the others already calculated
• the numerator of the solution is imaginary but actually the solution isn't. Depending on the constant $c$ the argument of the square root will become negative and therefore the denominator also imaginary $\rightarrow$ the solution is real. Of course there are values of $t$ for which the denominator becomes real and the solution imaginary but those values of $t$ corresponds to times after the ladder already hit the ground. After the ladder hit the ground (and if it's still advanced in x-direction) our initial assumptions about the triangle become invalid and so does the solution.

• as the solution is an increasing function the speed with which the right end of the ladder moves down the wall also increases. This seems perfectly fine if you recall where we started from: $$ x^2 + y^2 = 36 \Rightarrow y = \sqrt{36 - x^2} $$

  

  if you regard the plot of this function you will see that $y$ does not change linear with $x$ but that the change increases. What we've calculated above is the derivative of this function. But wait - did we actually calculate the derivative of $y$ in a direct way? No, we went a bit around using the differential equation obtained from the Pythagorean theorem.

So we calculated the derivative of $y$ using a differential equation. But what if we do it straight away using the above expression for $y$? Well, we'll end up with the same result with a little less effort (since the solving of differential equations is not really straight forward in some cases):

$$ \dot{y} \equiv \frac{d}{dt}y = \frac{d}{dt}\sqrt{36 - x^2} = -\frac{x\cdot\dot{x}}{\sqrt{36-x^2}} $$

After recalling again that $\dot{x}$ is constant and therefore $x = x_0 + \dot{x}\cdot t$ we obtain:

$$ \dot{y}(t) = -\frac{(x_0 + \dot{x}\cdot t)\cdot\dot{x}}{\sqrt{36 - (x_0 + \dot{x}\cdot t)^2}} $$

Note that we get the same dependence on $t$ as already with the above approach! Only the parameters are different but after evaluating the constant $c$ of the above approach you will see that both solutions are actually the same!