# relation rate word problem

• Jul 22nd 2007, 10:41 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, 10:54 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 $\displaystyle x$ be the distance the foot of the ladder is away from the wall.
Let $\displaystyle y$ be the vertical distance from the foot of the wall to the top of the ladder
Let $\displaystyle x'$ be the rate at which the horizontal distance is increasing
Let $\displaystyle y'$ be the rate at which the vertical distance is decreasing

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

By Pythagoras, we have

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

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

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

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

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

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

Thus we have:

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

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