Question

I need help figuring out a related rates problem.
A ladder 13feet long is leaning against the side of a building. If the foot of the ladder is pulled away from the building at a constant rate of 2/3 feet per second, how fast is the area of the triangle formed by the ladder, the building and the ground changing (in feet squared per second) at the instant when the top of the ladder is 12 feet above the ground?

Answer 1

got a formula for the area of the triangle?

Answer 2

I have in mind first using Pythagorean formula to find the height of the triangle formed by the 13 feet long leaning against the side of the building.

Answer 3

A = 1/2 b*h

Answer 4

Since I don't have the height, I have to use pythagorean to find h, right?

Answer 5

ok lets use \(x\) and \(y\) instead

Answer 6

making \[A=\frac{1}{2}xy\] you are looking for
\[A'\]

Answer 7

you got a formula for \(A'\)?

Answer 8

uhm not yet

Answer 9

don't think too hard, just write it (product rule)

Answer 10

I have to find the derivative of A=1/2xy or A=1/2x(sqrt144-x^2) ?

Answer 11

oh no just \(A(t)=\frac{1}{2}x(t)y(t)\)

Answer 12

or just
\[A=\frac{1}{2}xy\] i guess you can do it the other way too, but it is going to be annoying
i can show you why they are the same when we are done the easy way

Answer 13

ok

Answer 14

so what is \(A'\) ?

Answer 15

\[\frac{ dA }{ dt } =\frac{ 1 }{ 2 } y\frac{ dx }{ dt } + \frac{ 1 }{ 2 } x\frac{ dy }{ dt } \]

Answer 16

wow

Answer 17

ups

Answer 18

yes, i would have written
\[A'=\frac{1}{2}(xy'+yx')\]same thing only quicker

Answer 19

now you are almost done
you know 3 out of those four numbers

Answer 20

what is y though?

Answer 21

well we didn't say
you want to call y the base or the height?

Answer 22

oh, well seeing it as b * h I suppose the height

Answer 23

probably should have specified at the beginning but up until now all has been symmetric, so it didn't matter

Answer 24

ok so we call \(y\) the height

Answer 25

then you are told the following
the foot of the ladder is pulled away from the building at a constant rate of 2/3 feet per second
makes
\[x'=\frac{2}{3}\]

Answer 26

yes

Answer 27

you are told
\[y=12\]

Answer 28

pythagoras tells you \(x=5\)

Answer 29

oo I see

Answer 30

the only thing missing is \(y'\)

Answer 31

so so in the case of "and the
ladder changing at the instant the bottom of the ladder is 12 feet from the wall?" I would use y= sqrt(144-x^2) right

Answer 32

i wouldn't but you can

Answer 33

then that 12 would just be x instead of y

Answer 34

i would use
\[x^2+y^2=13\\
2xx'+2yy'=0\\
y'=-\frac{x}{y}x'\]

Answer 35

no no it would be correct, what you wrote

Answer 36

\[y= \sqrt{169-x^2}\] actually

Answer 37

and therefore
\[y'=-\frac{xx'}{\sqrt{169-x^2}}\]

Answer 38

oh yes lol

Answer 39

in any case that is exactly what i wrote

Answer 40

\[y'=-\frac{x}{y}x'\]

Answer 41

because the denominator is \(y\)

Answer 42

now you are good to find \(y'\)

Answer 43

What is y' here?

Answer 44

I know x' is the constant rate

Answer 45

and you have all the numbers you need for
\[A'=\frac{1}{2}(xy'+yx')\]

Answer 46

to find \(y'\) you can use this
\[y'=-\frac{xx'}{\sqrt{169-x^2}}\] what you wrote with \(x=5\) and \(x'=\frac{2}{3}\) or you can use
\[y'=-\frac{x}{y}x'\]with \(x=5,y=12,x'=\frac{2}{3}\)

Answer 47

clear?

Answer 48

Yes, I'm solving it now and see what I get as result.

Answer 49

good luck
it is pretty much arithmetic from here on in
so be careful with it, that is where all the mistake are usually made
at least on my end

Answer 50

@satellite73 is the answer negative?

Answer 51

@satellite73 also how would this be different if "the
ladder changing at the instant the bottom of the ladder is 12 feet from the wall?" from "the instant when the top of the ladder is 12 feet above the ground?"

Answer 52

nvm I just figured out the difference in between both.

Answer 53

I still have the doubt though about the answer. If I solve it when the ladder at the bottom is 12 then I get a negative answer but when I solve it when the ladder at the top is 12 then I get a positive answer. Even though I am searching for the area of the triangle, why is it affected by both?

Answer 54

I just figured out my answer is correct. Now I am just wondering why the area is negative when the bottom is 12 and positive when top is 12.

Answer 55

or why it decreases when bottom is 12 and increases when top is 12.