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From "Calculus made easy" by Thompson. A ladder of fixed length is against a horizontal wall.Height of ladder is y,distance of base of ladder from wall is x.For positive increment to x, there will be a negative increment to y.Some numerical values were given and $\frac{dy}{dx}$ was derived.

" $\frac{dy}{dx}= -0.11$."

"It is easy to see that except at one particular point, dy will be of a different size from dx".What point will that be?

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  • $\begingroup$ Going forward, with your future MathSE postings, please make more of an effort to proofread your postings. The problem that you posted is impossible to solve unless the Height of the ladder is specified. You omitted this critical information in your posting. The only reason that MathSE reviewers could attack the problem is that they were able to track down the source of the problem and find the information that you omitted. $\endgroup$ Commented Dec 19, 2022 at 22:54

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With the ladder seated inclined against a vertical wall: $$x^2+y^2=l^2$$, where x and y are the horizontal and vertical projections of the ladder and l is the length of it.

Therefore $$\left\{\begin{matrix} x&=&l\cos(\theta)\\y&=&l\sin(\theta)\end{matrix}, 0\le\theta\le \frac{\pi}{2}\right.$$

From here $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=-\cot(\theta)$$

The cotangent function is positive on $(0, \frac{\pi}{2})$ however the negative derivative accounts for the decrease of y while x increases.

Therefore $|dy|=|dx|\Rightarrow \cot(\theta)=1$: the only position possible is with the ladder is seated at a $45^\circ$ angle.

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The point will roughly be x = 128 (and more exactly around 127.986327395), which is when the slope of the graph is -1.

Or in simple terms, when the horizontal distance from the wall is the same as the distance to the upper point of the ladder resting on the wall.

I got it visually with Desmos, but you could achieve the same algebraically by taking the function that determines the high point of the ladder as a function of the horizontal distance from the wall:

$$ f\left(x\right)=\sqrt{181^{2}-x^{2}} $$

Taking the derivative:

$$ f'\left(x\right)=\frac{-x}{\sqrt{181^{2}-x^{2}}} $$

And then setting $ f'(x) $ equal to -1. Also, in case you wonder, the slope 1 is physically meaningless because the ladder can't be at -128 inches from the wall.

enter image description here

Link to Desmos: https://www.desmos.com/calculator/crydn9wrlo

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  • $\begingroup$ You seem to be inferring that the height of the ladder is $(181)$. What is the basis for this conclusion? $\endgroup$ Commented Dec 18, 2022 at 22:28
  • $\begingroup$ The book they mention: "Calculus made Easy" $\endgroup$
    – Jon
    Commented Dec 19, 2022 at 22:25
  • $\begingroup$ I don't own that book. Please explain. $\endgroup$ Commented Dec 19, 2022 at 22:26
  • $\begingroup$ The OP is referring to an example of book where the author explicitly mentions that the ladder is 181 inches.(calculusmadeeasy.org/3.html). $\endgroup$
    – Jon
    Commented Dec 19, 2022 at 22:29
  • $\begingroup$ As I mentioned in another comment. It's a fair critic that the OP didn't specify some of the details in their question. I just had extra context due to having the book. The answer by @WindSoul is better as it's more generic $\endgroup$
    – Jon
    Commented Dec 19, 2022 at 22:31
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$\frac{dy}{dx} = -0.11$ this means $y'(x) = -0.11$ integrating we get that $y(x) = -0.11x + L$ where L is the heigth of the ladder, all you have to do now is to check where $y = -0.11x + L$ intersects $y=x$

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  • $\begingroup$ This is incorrect because the derivative is -0.11 at a specific point, not at every point. $\endgroup$
    – Jon
    Commented Dec 18, 2022 at 20:33
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    $\begingroup$ @Jon To be fair, the original post is unclear in this regard and does not provide the necessary context (e.g. the actual height of the wall and length of the ladder) to derive $y’$ for other values of $x$. I thought they meant $dy/dx$ is constant at first glance too. In your own post, you seem to pull ‘$181$’ out of thin air $\endgroup$
    – FShrike
    Commented Dec 18, 2022 at 20:49
  • $\begingroup$ That a fair criticism, @FShrike. I have the book and that's why I have the full context. $\endgroup$
    – Jon
    Commented Dec 19, 2022 at 22:24

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