Urgent class test question (1 Viewer)

[RG11]

A ladder of length 3 metres is leaning against a verticle wall. The foot of the ladder is pulled away from the wall at the rate of 0.5 metres per second. How fast is the top of the ladder moving down the wall when the foot is 1 metre away from the wall?

all i have is the equation B^2 + H^2=9 will help where b is the distance between the foot of the ladder to the wall and h is the distance between the top of the ladder and the ground.

Im unsure what to do next though.

Thanks in advance!

 

[enigma_1]

Using rates of change, db/dt=0.5 which is given to you

We want to find dh/dt when b=1 metre

So constructing an equation:
dh/dt = dh/db x dh/dt

The equation that ypu figured out above, make h the subject and then differentiate it. You'll get -b(9-b^2)^-1/2

Then sub it in:

dh/dt = -b(9-b^2)^-1/2 x db/dt

sub in db/dt=0.5

dh/dt = -b(9-b^2)^-1/2 x 0.5

And then sub in b=1

dh/dt = -1(9-1)^-1/2 x 0.5

= -0.18 m/s