Calculus, using the chain rule in a real life problem.

Let the side opposite the angle, θ, we are interested in = x

The hypotenuse = 25

So we have

sin θ = x / 25            differentiate both sides with respect to time, t

(d/dt ) sin θ   =     ( d / dt )  ( x /  25)

Remember that  on the left side

d / dt  = d / dθ * dθ / dt        where  d/dθ sinθ = cos θ         

And  on the right side

 d / dt   =  d / dx *  dx / dt          where   d/dx (x/25)  =  1/25

So we have....

 cos θ  * ( dθ / dt)   =  (1/25) * dx / dt            [ note that  dx/dt   = 2ft/s ]

dθ / dt    =     (1 / cos θ)  * (1/25) (2) 

dθ / dt  =   (2 / 25)  sec θ

When the ladder is 7 ft from the wall..... sec θ   =  

25 / √[ 25^2 - 7^2]  =    25 / √[625 - 49 ]  =  25 /  √576    =  25 / 24

So

dθ / dt   =      (2 / 25 )  * ( 25 / 24) rads / sec  =  1 / 12 rads/ sec ≈  4.77° / sec

Thanks Chris,

I get the same answer.  I hope you do not mind Chris, I cannot resist presenting it.

But it is the same as yours :)

\(\frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot \frac{dx}{dt}\\ \frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot 2\\ \)

NOW

\(sin\theta = \frac{x}{25}\\ x=25sin\theta\\ \frac{dx}{d\theta}=25cos\theta\\ When\;\;x=7 \qquad cos\theta = \frac{24}{25}\\ so\\ When\;\;x=7 \qquad \frac{dx}{d\theta}=25\cdot \frac{24}{25}=24\\ so\\ When\;\;x=7 \qquad \frac{d\theta}{dx}=\frac{1}{24}\\ so\\ When\;\;x=7 \qquad \frac{d\theta}{dt}=\frac{1}{24}\cdot 2=\frac{1}{12}\;\;radians/sec\\\)

There are pi radians in 180 degrees

so 1 radian = 180/pi degrees

and   1/12 radians = 180/(12pi) = 4.77 degrees. 

I used this diagram to get the situation when x=7

A ladder 25 feet long is leaning against a wall.
The base of the ladder is sliding away from the wall at 2 feet per second.
Find the rate at which the angle between the ladder and the wall is changing
when the base of the ladder is 7 feet from the wall.

\(\begin{array}{|rcll|} \hline \sin(\phi) &=& \dfrac{x}{25\ \text{feet}} \\\\ \phi &=& \arcsin\left(\dfrac{x}{25\ \text{feet}} \right) \\\\ \dfrac{d\phi}{dx} &=& \dfrac{1} { \sqrt{ 25^2\ \text{feet}^2-x^2} } \quad | \quad \cdot dx \\\\ d\phi &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} }~dx \quad | \quad :dt \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} } \dfrac{dx}{dt} \quad | \quad \dfrac{dx}{dt}=2\dfrac{feet}{s},\quad x=7\ \text{feet} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-7^2\ \text{feet}^2} }\cdot 2\dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{2}{24\ \text{feet}}\cdot \dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot \dfrac{rad}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot\dfrac{180^{\circ}}{\pi\ \text{rad}} \dfrac{\text{rad}}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{ 15^{\circ}}{\pi s} \\\\ \mathbf{\dfrac{d\phi}{dt}} & \mathbf{=} & \mathbf{ \dfrac{ 4.77464829276^{\circ}}{s} } \\ \hline \end{array} \)

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