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The question is as follows: A 10-foot ladder is sliding down a wall. The top of the ladder is moving at a constant rate of 3 feet per second. What is the rate of change in the angle that the ladder makes with the ground when the base of the ladder is 6 feet from the wall?

I get -5/18 rad/s as my answer but that does not seem to be correct.

I have x/10 = cos(θ), the first derivative being d(θ)/dt = dx/dt*(1/-10sin(θ)). Sin(θ) = 8/10 making the full thing dθ/dt= 3(-1/(10(8/10))) which gives me -5/18 rad/s.

Where am I going wrong?

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Having in mind that from

$$ x^2+y^2=l^2\\ x = l\cos\theta\\ y = l\sin\theta $$

we have

$$ \dot x x + \dot y y = 0\\ \dot x =-l\sin\theta\dot \theta $$

so

$$ \dot x x_0 + \dot y_0 y = 0\Rightarrow \dot x = -\frac{\dot y_0}{x_0}y $$

here $x_0 = 6,\dot y_0 = -3, l = 10$ so $$ l\sin\theta\dot\theta = \frac{\dot y_0}{x_0}y = \frac{\dot y_0}{x_0}l\sin\theta $$

and finally

$$ \dot\theta = \frac{\dot y_0}{x_0} = -\frac{3}{6}=-\frac 12 $$

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Let $y$ be the distance from the top of the ladder to the ground. Then $\sin \theta=y/10$ or $\theta= \arcsin(y/10)$. Differentiating by $t$ we get $\large{\frac{d \theta}{dt}=\frac{1}{\sqrt{100-y^2}}\frac{dy}{dt}=-\frac{3}{\sqrt{100-y^2}}=-\frac{1}{2}}$ (when $y=8$)

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  • $\begingroup$ When the base of the ladder is $6~\text{ft}$ from the wall, $y = 8~\text{ft}$. $\endgroup$ – N. F. Taussig Oct 17 '18 at 0:44
  • $\begingroup$ @N.F.Taussig: Thank you for spotting my mistake! $\endgroup$ – Vasya Oct 17 '18 at 2:41
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The mistakes seems to be you are given the top of the ladder is moving at a constant of $3$ feet per second.

We are not given that $\frac{dx}{dt}=3.$

Note that we have:

$$x^2+y^2=10^2$$

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

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  • $\begingroup$ So dy/dt = 3 then? Given the equation you have how would I use that to get an angle? $\endgroup$ – ovil101 Oct 16 '18 at 16:53
  • $\begingroup$ You just have to find the value of $\frac{dx}{dt}$ wrong right. Find $\frac{dx}{dt}$ from $\frac{dy}{dt}, x, y$ and then substitute it back to your original working. $\endgroup$ – Siong Thye Goh Oct 16 '18 at 16:58

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