# relation rate word problem

• Jul 22nd 2007, 11:41 AM
driver327
relation rate word problem
2. A ladder 13 ft. long rest against a vertical wall and is sliding down the wall at the rate of 3ft/s at the instant the foot of the ladder is 5 ft. from the base of the wall. At this instant how fast is the foot( or base) of the ladder moving away from the wall? Side note, this problem looks like a pythag. theorem: x^+y^2=r^2. the answer comes out to 7.2 ft./s. though I know what formula to use for the certain situation, I really do not know the steps in solving this problem.

And also, I tried isolating dr/dt and it gave 2x+2y-2r=dr/dt and getting the value for x and y(which i know does not make any sense).
• Jul 22nd 2007, 11:54 AM
Jhevon
Quote:

Originally Posted by driver327
2. A ladder 13 ft. long rest against a vertical wall and is sliding down the wall at the rate of 3ft/s at the instant the foot of the ladder is 5 ft. from the base of the wall. At this instant how fast is the foot( or base) of the ladder moving away from the wall? Side note, this problem looks like a pythag. theorem: x^+y^2=r^2. the answer comes out to 7.2 ft./s. though I know what formula to use for the certain situation, I really do not know the steps in solving this problem.

Yes! it is using Pythagoras. i hope you drew a diagram to see what's going on (i won't, i usually do, but not this time :D).

Introducing r gets you in trouble, we only need x and y here. r is a constant, we don't need to replace it with a variable, since it does not change! the ladder will always be 13 ft long

Let $x$ be the distance the foot of the ladder is away from the wall.
Let $y$ be the vertical distance from the foot of the wall to the top of the ladder
Let $x'$ be the rate at which the horizontal distance is increasing
Let $y'$ be the rate at which the vertical distance is decreasing

We want $x'$ when $y' = -3$ and $x = 5$

By Pythagoras, we have

$x^2 + y^2 = 13^2$ ..............(1)

from the above, when $x = 5$ , we have $y = 12$

going back to equation (1), we proceed using implicit differentiation

$\Rightarrow 2x~x' + 2y~y' = 0$

$\Rightarrow x' = - \frac {y~y'}{x}$

We care about the instant where $x = 5$, $y' = -3$, $y = 12$

Thus we have:

$x' = - \frac {(12)(-3)}{5} \implies \boxed { x' = 7.2 }$

Note, $x' = \frac {dx}{dt}$ and $y' = \frac {dy}{dt}$