+0

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

0
271
4

What a terribly rude guest.

Other people use questions and answers to learn from!

Anyway, thanks to Heureka we still have the question:

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.

(reinstated by Melody)

Nov 21, 2018
edited by Guest  Nov 21, 2018
edited by Melody  Nov 22, 2018

#1
+3

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   Nov 21, 2018
#2
+1

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 Nov 21, 2018
#3
+10

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.

$$\begin{array}{|rcll|} \hline \sin(\phi) &=& \dfrac{x}{25\ \text{feet}} \\\\ \phi &=& \arcsin\left(\dfrac{x}{25\ \text{feet}} \right) \\\\ \dfrac{d\phi}{dx} &=& \dfrac{1} { \sqrt{ 25^2\ \text{feet}^2-x^2} } \quad | \quad \cdot dx \\\\ d\phi &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} }~dx \quad | \quad :dt \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} } \dfrac{dx}{dt} \quad | \quad \dfrac{dx}{dt}=2\dfrac{feet}{s},\quad x=7\ \text{feet} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-7^2\ \text{feet}^2} }\cdot 2\dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{2}{24\ \text{feet}}\cdot \dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot \dfrac{rad}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot\dfrac{180^{\circ}}{\pi\ \text{rad}} \dfrac{\text{rad}}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{ 15^{\circ}}{\pi s} \\\\ \mathbf{\dfrac{d\phi}{dt}} & \mathbf{=} & \mathbf{ \dfrac{ 4.77464829276^{\circ}}{s} } \\ \hline \end{array}$$ Nov 21, 2018
edited by heureka  Nov 21, 2018
#4
0

A note to ALL forum users.

NEVER delete your question.  It is the height of rudeness!

If you have good reason to edit it that might be ok but ALWAYS indicate very clearly what you have changed.

Thank you.

Nov 22, 2018