1
$\begingroup$

A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.

Is my solution wrong? How come I can't get the correct answer?

x=horizontal axis , y=vertical axis
Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s 
Y=2A/X 
dy/dx=-2A/x^2.....A=6,x=3
     =-12/9

dy/dt=dy/dx * dx/dt
     =-12/9 * 0.4
     =-0.533333
$\endgroup$
1
$\begingroup$

You can't use an area relation because we're not told what $\dfrac{dA}{dt}$ is. If you draw out the situation, we see that we can actually use a different relation to solve this problem.

The equation that you want to use comes directly from the Pythagorean theorem; in particular, if $x$ is the distance the base of the ladder is from the wall and $y$ is the the height of the ladder along the wall, then for the 5 m long ladder, we have that $x^2+y^2 = 25$.

Differentiating this will then give you the appropriate equations involving $\dfrac{dx}{dt}$ (i.e. the rate the ladder is sliding away from the wall) and $\dfrac{dy}{dt}$ (i.e. the rate the ladder is sliding down the wall). You can then plug in all the known values at the appropriate instant in time (when $x=3$, $y=\ldots$ [which can be found using the Pythagorean theorem], and $dx/dt = 0.4\text{ m/s}$) and then solve for $\dfrac{dy}{dt}$.

Hopefully this is enough information to help you get the correct solution. :-)

$\endgroup$
  • $\begingroup$ thank you for your information! but , since the triangle applies the pythagorean theorem, why we cannot apply the area relation on this question? $\endgroup$ – nid your hlp Dec 25 '13 at 14:02
  • $\begingroup$ If $A=\frac{1}{2}xy$, then $$\frac{dA}{dt} = \frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}{dt}$$ We can't proceed with solving for $\dfrac{dy}{dt}$ without knowing what $\dfrac{dA}{dt}$ equals. Unfortunately, there's no way we can determine the value of $\dfrac{dA}{dt}$ from the given information in the problem; this is why we instead use the Pythagorean relation to set up the appropriate relation (since it relates the sides of the triangle to each other). In fact, all other sliding ladder problems are set up in this manner. I hope this helps! $\endgroup$ – Christopher Toni Dec 25 '13 at 14:13
  • $\begingroup$ You can apply the area relation. But you have to understand that the area $A$ is a function of $x$ and $y$. So when you write $$ Y = \frac{2A}{x}, $$ you should be writing $$ Y = \frac{2A(x, y)}{x} $$ so that to compute $dy/dx$, you also need to differentiate $A$ rather than treating it as a constant the way you did in your work. $\endgroup$ – John Hughes Dec 25 '13 at 14:13
  • $\begingroup$ John, Why cant we treat the A as a constant? if it applies the Pythagorean relation, the Area should be a constant right? it doesnt matter how u move the bottom of the ladder , the Area will still remain the same. $\endgroup$ – nid your hlp Dec 25 '13 at 14:21
  • $\begingroup$ @nidyourhlp The area won't remain the same as the ladder slides down the wall; consider the right triangle with sides $1,2\sqrt{6},5$ and $3,4,5$ (where 5 is the hypotenuse in both cases). The area of the first triangle is $\sqrt{6}$ whereas the second triangle has area $6$, which clearly isn't the same area as the first triangle. Hence, the area of the triangle enclosed by the ladder, wall, and ground is constantly changing; this is why you can't treat area as a constant. $\endgroup$ – Christopher Toni Dec 26 '13 at 1:48
1
$\begingroup$

First we use Pythagorean Theorem

$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$

Next we use that we can write $x=0.4t=\frac{2}{5}t$: $$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$

Now we calculate the derivative (using the chain rule) $$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\right)}{2\sqrt{25-\frac{4}{25}t^2}} = \frac{-\frac{8}{25}t}{2\sqrt{25-\frac{4}{25}t^2}}.$$

This can be simplified further, but it isn't necessary. Now for the last step: evaluating the derivative at $t=7.5$ (which comes from solving $0.4t=3$).

$$\frac{dy}{dt}|_{t=7.5} =\frac{-\frac{8}{25}\cdot 7.5}{2\sqrt{25-\frac{4}{25}\cdot7.5^2}}=-0.3. $$ So the conclusion is that the speed, when the bottom is $3\text{m}$ from the wall, is $-0.3 \text{ m}/\text{s}$.

$\endgroup$
-1
$\begingroup$

This problem can be solved 'numerically' without resorting to other methods. First we have:

$$ w^2 + g^2 = L $$ $$ (w + f)^2 + (g + p)^2 = L $$

where $w$ is the wall, $g$ is the ground, $L$ is the ladder squared, $f$ is the fall rate and $p$ is the pull rate. Note that the units of f and p are metres - time is implicit. Also note that we do not presume that the fall rate is negative - it should come out that way. Equating, canceling common terms and solving the quadratic in $f$ yields:

$$ f = \sqrt{w^2 - 2gp - p^2} - w $$

Now we can calculate a series of increasingly accurate approximations of the fall rate if the pull rate is $0.4 m/s$. Here is a Python function that does this:

def fallrate(wall, ground, pullrate):  
  for i in [2**n for n in range(15)]:  
    print(i * ((wall**2 - 2 * ground * (pullrate/i) - (pullrate/i)**2)**.5 - wall))  
  input("\nEnter to exit. ")  

fallrate(4, 3, 0.4)  

Note that if we halve the distance we use to calculate the pull rate we will get a rate for half the distance. We must therefore double the result for the purpose of comparison - hence the i * etc at the start of the approximation formula. The fall rate seems to converge on $-0.3 m/s$. This approach can be seen as a type of calculus - if we divide the rates on both sides of the approximation formula by $i$ we are in effect turning them into infinitesimals; and it agrees with the result given by Eff worked out using conventional methods.

$\endgroup$
  • $\begingroup$ I've done the job of fixing the formatting of the post. I think I got the indentation correctly but I might have missed something. $\endgroup$ – Doktoro Reichard Dec 29 '13 at 5:48
  • $\begingroup$ Thanks, that's better. I just indented all the contents of the function - can't use more spaces because of line length. $\endgroup$ – user117644 Dec 29 '13 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.