Question

Could someone help me with this problem, it is regarding related rates: A 5 meter long ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate if 0.4 m/sec. Determine how fast the top of the ladder is descending when the foot of the ladder is 3 meters from the house

Answer 1

i am stuck when it comes to figuring out how to put the values into the derived equation: 2x(dx)/(dt)+2y(dy)/(dt)=0

Answer 2

Okay imagine a ladder leaning against a wall making a triangle.. the ladder is hypothenuse, and is z, the wall is one leg, x and the floor is the other leg, y.
z^2 = x^2 + y^2

Answer 3

ok

Answer 4

2z z' = 2x x' + 2y y'
or
z z' = x x' + y y'
Now z' is the change in length of z, but z does not change, it's a ladder so z' = 0

Answer 5

z', x', y' are all derivatives with respect to t by the way, it's just that z' is easier to write than dz/dt

Answer 6

where the the two's go

Answer 7

divide everything by 2

Answer 8

gotcha

Answer 9

so now if z' = 0
then
z z' = x x' + y y'
0 = x x' + y y'

Answer 10

actually draw this.. instead of calling the wall x, call the wall y..
So draw a ladder against a wall, or simply a triangle with the ladder being z, wall is y and floor is x

Answer 11

ok

Answer 12

The foot of the ladder is pulled away at the rate of 0.4m/s
so x is increasing, so x' = +0.4m/s

Answer 13

Now you need to find y' when x = 3m

Answer 14

so:
x x' + y y' = 0
3 * 0.4 + y * y' = 0
or
y * y' = -1.2

Answer 15

are you okay so far?

Answer 16

yes i follow

Answer 17

okay so now all you need to know is what y is when x = 3
to find y, do:
z^2 = x^2 + y^2
z = 5 (length of the ladder) and x = 3

Answer 18

25 = 9 + y^2
25 - 9 = y^2
y^2 = 16, y = 4

Answer 19

ok

Answer 20

plug that into the previous equation:
y * y' = -1.2
4y' = -1.2
y' = -1.2/4 = -0.3 m/s

Answer 21

So the top of the ladder is DESCENDING at 0.3 m/s
as you can see we got a negative value so we are right since the top is descending so the height, or y, is decreasing basically..

Answer 22

this is great thanks, i appreaciate it

Answer 23

do you have any tips for solving these problems
?

Answer 24

Well always draw the graph and look at what's changing and what's not...

Answer 25

Correct graph in these problems is like.. 30% problem solved..
Correct graph and knowing what's changing(increasing/decreasing) is about 90% problem is solved.. all you need to do after that is take what you have and plug it all in..

Answer 26

thanks