# Question 35f3f

Mar 24, 2015

The rate is $\frac{65}{6} \text{m/min"=10 5/6 "m/min"~~10.83 "m/min}$

This is a nice example of a Related Rates problem. (Perhaps better called by the full name "Related Rates of Change")

Solution:
$h$ is the distance from the floor to the end of the ladder.

$l$ is the length of the ladder.
$\frac{\mathrm{dl}}{\mathrm{dt}} = 10 \text{m/min}$

We need to find $\frac{\mathrm{dh}}{\mathrm{dt}}$

The relationship between the variables, $h$ and $l$ is given by the Pythagorean Theorem:

${h}^{2} + {5}^{2} = {l}^{2}$

We need to find $\frac{\mathrm{dh}}{\mathrm{dt}}$

To find the relationship between the rates of change, differentiate (implicitly) with respect to $t$.

$2 h \frac{\mathrm{dh}}{\mathrm{dt}} + 0 = 2 l \frac{\mathrm{dl}}{\mathrm{dt}}$

So:

$h \frac{\mathrm{dh}}{\mathrm{dt}} = l \frac{\mathrm{dl}}{\mathrm{dt}}$

Substitute what we know and solve for the desired value.

We are told that $\frac{\mathrm{dl}}{\mathrm{dt}} = 10 \text{m/min}$ and we are interested in
the instant when $l = 13 \text{m}$, but we also need $h$ at that instant.

Using Pythagoras to find $h$ when $l = 13 \text{m}$
${h}^{2} + {5}^{2} = {13}^{2}$
${h}^{2} = 144$
$h = 12 \text{m}$

Now we can finish:

$h \frac{\mathrm{dh}}{\mathrm{dt}} = l \frac{\mathrm{dl}}{\mathrm{dt}}$ so at the instant we were asked about:

12"m"(dh)/(dt)=13"m"(10"m/min")#

Thus
$\frac{\mathrm{dh}}{\mathrm{dt}} = \frac{130}{12} \text{m/min"=65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min}$