Question #35f3f

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Question #35f3f

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Answer 1 · 2015-03-24T13:35:04

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

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

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

$l$ $l$ is the length of the ladder.
$\frac{d l}{d t} = 10 m/min$ $(dl)/dt=10"m/min"$

We need to find $\frac{d h}{d t}$ $(dh)/(dt)$

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

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

We need to find $\frac{d h}{d t}$ $(dh)/(dt)$

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

$2 h \frac{d h}{d t} + 0 = 2 l \frac{d l}{d t}$ $2h(dh)/(dt)+0=2l(dl)/(dt)$

So:

$h \frac{d h}{d t} = l \frac{d l}{d t}$ $h(dh)/(dt)=l(dl)/(dt)$

Substitute what we know and solve for the desired value.

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

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

Now we can finish:

$h \frac{d h}{d t} = l \frac{d l}{d t}$ $h(dh)/(dt)=l(dl)/(dt)$ so at the instant we were asked about:

$12 m \frac{d h}{d t} = 13 m (10 m/min)$ $12"m"(dh)/(dt)=13"m"(10"m/min")$

Thus
$\frac{d h}{d t} = \frac{130}{12} m/min = \frac{65}{6} m/min = 10 \frac{5}{6} m/min \approx 10.83 m/min$ $(dh)/(dt)=130/12 "m/min"=65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"$

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Question #35f3f

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