Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)

Answer 1

check the picture below.

so, when r = 10 and x = 8, the vertical distance is 6, namely in pythagorean theorem lingo, b = 6.

let's keep in mind that, the ladder is not growing any longer or shrinking, and therefore is constantly always just 10, that matters, since is just a scalar value and also because the derivative of a constant is 0.

$\bf cos(\theta )=\cfrac{x}{r}\implies cos(\theta )=\cfrac{1}{10}x\implies \stackrel{chain~rule}{-sin(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{10}\cdot\stackrel{chain~rule}{\cfrac{dx}{dt}\cdot 1}$

$\bf -sin(\theta )\cfrac{d\theta }{dt}=-\cfrac{1}{10}\cdot \cfrac{dx}{dt}\implies \cfrac{d\theta }{dt}=-\cfrac{1}{-10sin(\theta )}\cdot \cfrac{dx}{dt} \\\\\\ \begin{cases} \frac{dx}{dt}=1.3\\ sin(\theta )=\frac{6}{10} \end{cases}\implies \cfrac{d\theta }{dt}=-\cfrac{1}{10\cdot \frac{6}{10}}\cdot 1.3\implies \cfrac{d\theta }{dt}=-\cfrac{1.3}{6}~radians$

Answer 2

The angle's rate of change when the bottom of the ladder is 8ft from the wall is

$\dfrac{-1.3}{6} rad/s$

It is given the Length of Ladder $(h)$ is $10ft.$. and the distance between the bottom of the ladder to the wall $(r)$ is $8ft.$ as shown in the below figure.

By using the Pythagoras Theorem

$b=\sqrt{10^{2}-8^{2} }\\=\sqrt{36} \\=6ft.$

and

$cos(\theta)= r/h\\cos(\theta)=r/10$

Differentiating both sides with respect to $'t'$ by using the chain rule

$-sin(\theta)\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\dfrac{1}{10} \dfrac{\mathrm{d}r }{\mathrm{d} t}\\\\\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\dfrac{1}{10} \dfrac{\mathrm{d}r }{\mathrm{d} t} \dfrac{1}{-sin(\theta)} } ......(eq. 1)$

given

$\dfrac{\mathrm{d}r }{\mathrm{d} t}=1.3ft./s\\sin(\theta)=\frac{6}{10}$

putting this in eq.1, we get

$\dfrac{\mathrm{d} \theta}{\mathrm{d} t} = \dfrac{1.3}{10} (\dfrac{1}{-\frac{6}{10} }) \\\dfrac{\mathrm{d} \theta}{\mathrm{d} t} =\dfrac{-1.3}{6} rad/s$

So the angle's rate of change when the bottom of the ladder is 8ft from the wall is$\dfrac{-1.3}{6} rad/s$.

Know more about Pythagoras Theorem here:

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